Rd Sharma Xi (2018) Solutions for Class 12 Science Math Chapter 15 Linear Inequations are provided here with simple step-by-step explanations. These solutions for Linear Inequations are extremely popular among class 12 Science students for Math Linear Inequations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi (2018) Book of class 12 Science Math Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi (2018) Solutions. All Rd Sharma Xi (2018) Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

Page No 15.10:

Question 1:

Solve: 12x < 50, when

(i) x ∈ R
(ii) x ∈ Z
(iii) x ∈ N

Answer:

We have, 12x<50x<5012                  Dividing both the sides by 12x<256(i) xRThen, the solution of the given inequation is -,256.(ii) xZThen, the solution of the given inequation is ..........-3, -2, -1, 0, 1, 2, 3, 4.(iii)  xNThen, the solution of the given inequation is  1, 2, 3, 4.

Page No 15.10:

Question 2:

Solve: −4x > 30, when

(i) x ∈ R
(ii) x ∈ Z
(iii) x ∈ N

Answer:

-4x>30x<-304         (Dividing both the sides by -4)x<-152(i) xRThen, the solution of the given inequation is -, -152.(ii) xZThen, the solution of the given inequation is ..........-9, -8.(iii) xNThen, the solution of the given inequation is ϕ.

Page No 15.10:

Question 3:

Solve: 4x − 2 < 8, when

(i) x ∈ R
(ii) x ∈ Z
(iii) x ∈ N

Answer:

We have, 4x-2<84x<8+2                (Transposing -2 to the RHS)4x<10x<104           (Dividing both the sides by 4)x<52(i) xRThen, the solution of the given inequation is -, 52.(ii) xZThen, the solution of the given inequation is .......-3, -2, -1, 0, 1, 2.(iii) xNThen, the solution of the given inequation is 1, 2.

Page No 15.10:

Question 4:

3x − 7 > x + 1

Answer:

3x-7>x+13x-x>1+7      Transposing x to LHS and -7 to RHS2x>8x>4                Dividing both sides by 2Hence, the  solution set of the given inequation is 4, 

Page No 15.10:

Question 5:

x + 5 > 4x − 10

Answer:

We have, x+5>4x-105+10 > 4x-x                 (Transposing x to the RHS and -10 to the LHS)15>3x3x<15x<5           Dividing both sides by 3x-, 5Hence, the solution set of the given inequation is -, 5.

Page No 15.10:

Question 6:

3x + 9 ≥ −x + 19

Answer:

3x+9-x+193x+x19-9             (Transposing -x to the LHSand 9 to the RHS)4x10x52              (Dividing both the sides by 4)Hence, the solution set of the given inequation is [52, ).

Page No 15.10:

Question 7:

23-xx5+4

Answer:

23- xx5+46-2xx5+46-4x5+2x        Transposing -2x to the RHS and 4 to the LHS2   11x511x52x1011            Mltiplying both the sides by 511Thus, the solution set of the given inequation is (-, 1011].

Page No 15.10:

Question 8:

3x-254x-32

Answer:

3x-254x-3223x-254x-36x-420x-15-4+1520x-6x          (Transposing 6x to the RHS and -15 to the LHS)11  14x 14x11x1114          (Dividing both the sides by 14)Hence, the solution set of the given inequation is [1114, ).

Page No 15.10:

Question 9:

−(x − 3) + 4 < 5 − 2x

Answer:

-x-3+4<5-2x-x+3+4<5-2x-x+7<5-2x-x+2x<5-7        Transposing -2x to LHS and 7 to RHSx<-2Hence, the solution set of the given inequation is -, -2.

Page No 15.10:

Question 10:

x5<3x-24-5x-35

Answer:

x5<3x-24-5x-3520×x5<20×3x-24-5x-35   Multiplying both the sides by 204x<53x-2-45x-34x<15x-10-20x+124x<-5x+24x+5x<2                     (Transposing -5x to the LHS) 9x<2x<29       (Dividing both the sides by 9)Hence, the solution set of the given inequation is  -, 29.

Page No 15.10:

Question 11:

2x-1532+x7

Answer:

2x-1532+x714x-1152+x 14x-1430+15x30+15x14x-1415x-14x-14-30             (Transposing 14x to the LHS and 30 to the RHS)x-44Hence, the solution to the given inequation is  [-44, ).

Page No 15.10:

Question 12:

5x2+3x4394

Answer:

5x2+3x439410x+3x439410x+3x3913x39x3913            (Dividing both the sides by 13)x3Hence, the solution set of the given inequation is [3, ).

Page No 15.10:

Question 13:

x-13+4<x-55-2

Answer:

x-13+4<x-55-2x-13-x-55<-2-4    Transposing 4 to the RHS and x-55 to the LHS5x-1-3x-515<-65x-5-3x+1515<-62x+1015<-62x+10<-902x<-90-10              Transposing 10 to the RHS2x<-100  x<-1002            Dividing both the sides by 2x<-50Hence, the solution of the given inequation is -, -50.

Page No 15.10:

Question 14:

2x+34-3<x-43-2

Answer:

2x+34-3<x-43-22x+34-x-43<-2+3      (Transposing  x-43 to the LHSand -3 to the RHS)32x+3-4x-412<132x+3-4x-4<12        (Multiplying both the sides by 12)6x+9-4x+16<122x+25<122x<12-252x<-13x<-132      (Dividing both the sides by 2)Hence, the solution of the given inequation is -, -132.

Page No 15.10:

Question 15:

5-2x3<x6-5

Answer:

5-2x3<x6-55-2x3-x6<-5             Transposing x6to the LHS25-2x-x6<-5 10-4x-x6<-5 10-5x6<-5 10-5x<-3010+30<5x40<5x5x>40x>405x>8Hence, the solution set of the given inequality is 8, .

Page No 15.10:

Question 16:

4+2x3x2-3

Answer:

4+2x3x2-34+2x3-x2-38+4x-3x6-38+x-18                  Multiplying both the sides by 6x-26              Transposing 8 to the RHSThus, the solution set of the given inequation is [-26, ).

Page No 15.10:

Question 17:

2x+35-2<3x-25

Answer:

2x+35-2<3x-252x+35-3x-65<2    Transposing  3x-25to the LHS and -2 to the RHS2x+3-3x+65<22x+3-3x+6<10            Multiplying both the sides by 5-x+9<10-x<1x>-1       Multiplying both the sides by -1Hence, the solution set of the given inequation is -1, .

Page No 15.10:

Question 18:

x-2 5x+83

Answer:

x-25x+833x-25x+8      Multiplying both the sides by 33x-65x+85x+83x-65x-3x-6-8      Transposing 3x to the LHS and 8 to the RHS2x-14x-7     Dividing both the sides by 2Hence, the solution of the given inequation is [-7,).

Page No 15.10:

Question 19:

6x-54x+1<0

Answer:

6x-54x+1<0Equating 6x-5 and 4x+1 to zero, we obtain x=56 and -14 as the critical points.


 x-14,56

Page No 15.10:

Question 20:

2x-33x-7>0

Answer:

2x-33x-7>0Equating 2x-3 and 3x-7 to zero, we obtain x=32, 73 as the critical points.


 x-,3273,

Page No 15.10:

Question 21:

3x-2<1

Answer:

3x-2<13x-2-1<03-x+2x-2<0-x+5x-2<0x-5x-2>0



x-,25,

Page No 15.10:

Question 22:

1x-12

Answer:

1x-121x-1-201-2x+2x-10-2x+3x-102x-3x-10


x-,1[32,)

Page No 15.10:

Question 23:

4x+32x-5<6

Answer:


4x+32x-5<64x+32x-5-6<04x+3-62x-52x-5<04x+3-12x+302x-5<0-8x+332x-5<08x-332x-5>0


x-,52338,

Page No 15.10:

Question 24:

5x-6x+6<1

Answer:


5x-6x+6<15x-6x+6-1<05x-6-x-6x+6<04x-12x+6<0x-3x+6<0



 x-6, 3

Page No 15.10:

Question 25:

5x+84-x<2

Answer:

We have, 5x+84-x<25x+84-x-2<05x+8-24-x4-x<05x+8-8+2x4-x<07x4-x<07xx-4>0



 x-,04,

Page No 15.10:

Question 26:

x-1x+3>2

Answer:

We have, x-1x+3>2x-1x+3-2>0x-1-2x+3x+3>0x-1-2x-6x+3>0-x-7x+3>0                     x+7x+3<0



∴ x-7,-3

Page No 15.10:

Question 27:

7x-58x+3>4

Answer:

We have,

7x-58x+3>47x-58x+3-4>07x-5-48x+38x+3>07x-5-32x-128x+3>0-25x-178x+3>025x+178x+3<0        (Multiplying by -1 to make the coefficient of x in the LHSpositive)




x-1725,-38

Page No 15.10:

Question 28:

xx-5>12

Answer:

xx-5>12xx-5-12>02x-x+52(x-5)>0x+52(x-5)>0x+5x-5>0



 x-,-55,



Page No 15.15:

Question 1:

Solve each of the following system of equations in R.

1. x + 3 > 0, 2x < 14

Answer:

x+3>0x>-3x-3, ...iAlso, 2x<14x<7       Dividing both the sides by 2x-,7 ...iiThus, the solution of the given set of inequalities is the intersection of i and ii.-3,-,7=-3,7 x-3,7Thus, the solution of the given set of inequalities is -3,7.

Page No 15.15:

Question 2:

Solve each of the following system of equations in R.

2. 2x − 7 > 5 − x, 11 − 5x ≤ 1

Answer:

We have, 2x-7>5-x2x+x>5+73x>12x>4 x4, ...iAlso, 11-5x15x11-1x2x[2,) ...iiSolution set of the given set of inequations is intersection of i and ii4,[2,)=4,Thus, the solution set of the given set of inequations is 4,.

Page No 15.15:

Question 3:

Solve each of the following system of equations in R.

3. x − 2 > 0, 3x < 18

Answer:

We have, x-2>0x>2x2, ...iAlso, 3x<18x<6x-,6 ...iiSolution of the given set of the inequations is intersection of i and ii2,-,6=(2,6)Thus, (2,6) is the solution of the given set of inequalities.

Page No 15.15:

Question 4:

2x + 6 ≥ 0, 4x − 7 < 0

Answer:

We have, 2x+602x-6x-3x[-3,) ...iAlso, 4x-7<04x<7x<74x-,74 ...iiThus, the solution of the given inequations is the intersection of i and ii.[-3,)-74=[-3,74)Thus, the solution of the given inequations is [-3,74).

Page No 15.15:

Question 5:

Solve each of the following system of equations in R.

5. 3x − 6 > 0, 2x − 5 > 0

Answer:

3x-6>03x>6x>2x2, ...iAlso, 2x-5>02x>5x>52x52, ...iiSolution of the given set of inequalities is the intersection of i and ii.2,52,=52,Thus, the solution of the given set of inequalities is 52,.

Page No 15.15:

Question 6:

Solve each of the following system of equations in R.

6. 2x − 3 < 7, 2x > −4

Answer:

2x-3<72x<7+3x<5x-, 5      ...(i)Also, 2x>-4x>-2 x-2,     ...(ii)Thus, the solution of the given set of inequalities is the intersection of (i) and (ii).-, 5 -2,=-2,5Thus, the solution of the given set of inequalities is -2,5.

Page No 15.15:

Question 7:

Solve each of the following system of equations in R.

7. 2x + 5 ≤ 0, x − 3 ≤ 0

Answer:

We have, 2x+502x-5x-52x(-, -52]               ...  (i)Also,  x-30x3x(-, 3]                  ...   (ii)Thus, the solution of the given set of inequalities is the intersection of (i) and  (ii).(-,-52] (-,3]  = (-,-52]Thus, the solution of the given set of inequalities is (-,-52].

Page No 15.15:

Question 8:

Solve each of the following system of equations in R.

8. 5x − 1 < 24, 5x + 1 > −24

Answer:

5x-1<245x<24+1x<5 x-,5         ...  (i)Also,  5x+1>-245x>-24-1x>-5x(-5,)           ...  (ii)Hence, the solution of the given set of inequalities is the intersection of (i) and (ii).-,5 -5,=-5, 5Thus, the solution of the given set of inequalities is -5, 5.

Page No 15.15:

Question 9:

Solve each of the following system of equations in R.

9. 3x − 1 ≥ 5, x + 2 > −1

Answer:

3x-153x5+1x2x[2,)                 ... (i)Also, x+2>-1x>-1-2x>-3x(-3,)            ... (ii)Hence, the solution of the given set of inequalities is the intersection of (I) and (ii).[2,)(-3,)= [2,)Thus, the solution of the given set of inequalities is [2,).

Page No 15.15:

Question 10:

Solve each of the following system of equations in R.

10. 11 − 5x > −4, 4x + 13 ≤ −11

Answer:

We have,

11-5x>-4-5x>-4-11-5x>-155x<15         Multiplying both sides by -1x<155 x<3x(-,3)               ... (i)Also,  4x+13-114x-11-134x-24x-6x(-,-6]          ... (ii)Hence, the solution of the given set of inequalities is the intersection of (i) and (ii).(-3)(-,-6]=(-,-6]Hence, the solution of the given set of inequalities is (-,-6].

Page No 15.15:

Question 11:

Solve each of the following system of equations in R.

11. 4x − 1 ≤ 0, 3 − 4x < 0

Answer:

We have, 4x-104x1x14    (Dividing both the sides by 4)x(-,14]          ...(i)Also,   3-4x<00> 3-4x4x>3x>34       Dividing both sides by 4x34,            ...(ii)Hence, the solution of the given set of inequalities is the intersection of (i) and (ii).But, -1434, =ϕThus, the given set of inequations has no solution.

Page No 15.15:

Question 12:

Solve each of the following system of equations in R.

12. x + 5 > 2(x + 1), 2 − x < 3 (x + 2)

Answer:

x+5>2x+1x+5>2x+22x+2 < x+52x-x<5-2x<3x(-, 3)            ... (i)Also,  2-x<3x+22-x<3x+63x+6>2-x3x+x>2-64x>-4x>-1x(-1, )           ... (ii)Hence, the solution of the given set of inequations is the intersection of (i) and (ii).(-, 3)  (-1, )=(-1, 3)Thus, the solution of the given set of inequations is (-1, 3).

Page No 15.15:

Question 13:

Solve each of the following system of equations in R.

13. 2 (x − 6) < 3x − 7, 11 − 2x < 6 − x

Answer:

2x-6<3x-72x-12<3x-73x-7 > 2x-123x-2x>-12+7x>-5x-5,            ... (i)Also,  11-2x<6-x6-x>11-2x2x-x>11-6x>5x5,              ... (ii)Hence, the solution of the given inequation is the intersection of (i) and (ii).-5,5,=5,Hence, the solution of the given inequation is 5,.

Page No 15.15:

Question 14:

Solve each of the following system of equations in R.

14. 5x-7<3x+3, 1-3x2x-4

Answer:

5x-7<3x+35x-7<3x+95x-3x<9+72x<16x<8x-,8                ... (i)Also,  1-3x2x-4x-41-3x2x+3x21+42x+3x255x10 x2 x(-,2]              ... (ii)Hence, the solution of the given set of inequalities is the intersection of (i) and (ii).-,8(-,2]=(-,2]Hence, the solution of the given set of inequalities is (-,2].

Page No 15.15:

Question 15:

Solve each of the following system of equations in R.

15. 2x-34-24x3-6, 22x+3<6x-2+10

Answer:

2x-34-24x3-62x-34-4x3-6+232x-3-16x12-46x-9-16x-48-10x-3910x39      Multiplying both sides by -1x3910x(-,3910]               ... (i)Also,    22x+3< 6x-2+104x+6<6x-12+104x+6<6x-26x-2>4x+66x-4x>6+22x>8x>4  x4,                    ... (ii)Hence, the solution of the given set of inequalities is the intersection of (i) and (ii),(-,3910]4,= which is an empty set.Thus, there is no solution of the given set of inequations.

Page No 15.15:

Question 16:

Solve each of the following system of equations in R.

16. 7x-12<-3, 3x+85+11<0

Answer:

7x-12<-37x-1<-67x<-6+1x<-57x-,-57            ...  (i)Also,   3x+85+11<03x+8+555<03x+63<03x<-63x<-21 x-,-21            ...  (ii)Hence, the solution to the given set of inequations is the intersection of (i) and (ii). -,-57--21 = -,-21Hence, the solution to the given set of inequations is -,-21.

Page No 15.15:

Question 17:

Solve each of the following system of equations in R.

17. 2x+17x-1>5, x+7x-8>2

Answer:

2x+17x-1>52x+17x-1-5>02x+1-35x+57x-1>06-33x7x-1>0x17,211             ...(i)Also, x+7x-8>2x+7x-8-2>0x+7-2x+16x-8>023-xx-8>0x(-,8)23,         .....(ii)Hence, the solution to the given set of inequalities is the intersection of (i) and (ii), which is empty. x

Page No 15.15:

Question 18:

Solve each of the following system of equations in R.

18. 0<-x2<3

Answer:

0<-x2<30<-x<6     (Multiplying throughout by 2)0>x>-6      (Multiplying throughout by -1)x(-6,0)  Hence, the interval (-6,0) is the solution of the given set of inequations.

Page No 15.15:

Question 19:

Solve each of the following system of equations in R.

19. 10 ≤ −5 (x − 2) < 20

Answer:

10-5x-2<2010-5x+10<2010-10-5x<20-10       (Substracting 10 from all three terms)0-5x<100x>-2       (Dividing all three terms by -5 )x(-2,0]

Hence, the interval (-2,0] is the solution of the given set of inequations.

Page No 15.15:

Question 20:

Solve each of the following system of equations in R.

20. −5 < 2x − 3 < 5

Answer:

-5<2x-3<5-5+3<2x<5+3      Adding 3 throughout-2<2x<8-1<x<4    Dividing throughout by 2x-1,4

Hence, the interval (-1,4) is the solution of the given set of inequaltions.



Page No 15.16:

Question 21:

Solve each of the following system of equations in R.
4x+136x+1, x>0

Answer:

4x+136x+1, x>04x+13 and 36x+1Now,4x+134x+1-30 4-3x-3x+10 1-3xx+103x-1x+10x-, -1[13, )

 

Thus, the solution set of the inequation is -, -1[13, ).

And 6x+136x+1-306-3x-3x+103-3xx+103x-3x+10x(-1, 1]



Thus, the solution set of the inequation is (-1, 1].

The common values of x in both the inequation is 13, 1.

Hence, the solution set of both the inequation is 13, 1.



Page No 15.22:

Question 1:

Solve x+13>83

Answer:

As, x+13>83x+13<-83 or x+13>83           As, x>ax<-a or x>ax<-13-83 or x>83-13x<-93 or x>73x<-3 or x>73 x-, -373, 

Page No 15.22:

Question 2:

Solve 4-x+1<3

Answer:

As, 4-x+1<34-x<3-14-x<2-2<4-x<2         As, x<a-a<x<a-2-4<-x<2-4-6<-x<-22<x<6 x2, 6

Page No 15.22:

Question 3:

Solve 3x-42512

Answer:

As, 3x-42512-5123x-42512          As, xa-axa-563x-456-56+43x56+4-5+2463x5+2461963x2961918x2918 x1918, 2918

Page No 15.22:

Question 4:

Solve x-2x-2>0

Answer:

We have,x-2x-2>0As, x-2=x-2, x22-x, x<2And x-2x-2>0 for x>2So, x>2 x2, 

Page No 15.22:

Question 5:

Solve 1x-3<12

Answer:

As, 1x-3<121x-3-12<02-x-32x-3<02-x+3x-3<05-xx-3<0Case I: When x0, x=x5-xx-3<05-x<0 and x-3>0 or 5-x>0 and x-3<0x>5 and x>3 or x<5 and x<3x>5 and x<3x[0, 3)5, Case II: When x0, x=-x,5+x-x-3<0x+5x+3>0x+5>0 and x+3>0 or x+5<0 and x+3<0x>-5 and x>-3 or x<-5 and x<-3x>-3 and x<-5x-, -5(-3, 0]So, from both the cases, we getx-, -5(-3, 0][0, 3)5,  x-, -5-3, 35, 

Page No 15.22:

Question 6:

Solve x+2-xx<2

Answer:

As, x+2-xx<2x+2-xx-2<0x+2-x-2xx<0x+2-3xx<0Case I: When x-2, x+2=x+2,x+2-3xx<02-2xx<0-2x-1x<0x-1x>0x-1>0 and x>0 or x-1<0 and x<0x>1 and x>0 or x<1 and x<0x>1 or x<0x[-2, 0)1, Case II: When x-2, x+2=-x+2,-x+2-3xx<0-x-2-3xx<0-4x-2x<0-22x+1x<02x+1x>02x+1>0 and x>0 or 2x+1<0 and x<0x>-12 and x>0 or x<-12 and x<0x>0 or x<-12x(-, -2]0, So, from both the cases, we getx[-2, 0)1, (-, -2]0,  x-,01, 

Page No 15.22:

Question 7:

Solve 2x-1x-1>2

Answer:

As, 2x-1x-1>22x-1x-1<-2 or 2x-1x-1>2      As, x>2x<-2 or x>22x-1x-1+2<0 or 2x-1x-1-2>02x-1+2x-2x-1<0 or 2x-1-2x+2x-1>04x-3x-1<0 or 1x-1>04x-3x-1<0 or x-1>04x-3>0 and x-1<0 or 4x-2<0 and x-1>0 or x-1>0x>34 and x<1 or x<34 and x>1 or x>134<x<1 or ϕ or x>134<x<1 or x<134<x<1 or x>1 x34, 11, 

Page No 15.22:

Question 8:

Solve x-1+x-2+x-36

Answer:

We have, x-1+x-2+x-36       .....iAs, x-1=x-1, x11-x, x<1;x-2=x-2, x22-x, x<2andx-3=x-3, x33-x, x<3Now,Case I: When x<1,1-x+2-x+3-x66-3x63x0x0So, x(-, 0]Case II: When 1x<2,x-1+2-x+3-x64-x6x4-6x-2So, xϕCase III: When 2x<3,x-1+x-2+3-x6x6So, xϕCase IV: When x3,x-1+x-2+x-363x-663x12x123x4So, x[4, )So, from all the four cases, we getx(-, 0][4, )

Page No 15.22:

Question 9:

Solve x-2-1x-2-20                                                                                                                                                       [NCERT EXEMPLAR]

Answer:

As, x-2-1x-2-20Case I: When x2, x-2=x-2,x-2-1x-2-20x-3x-40x-30 and x-4>0 or x-30 and x-4<0x3 and x>4 or x3 and x<4ϕ or 3x<43x<4So, x[3, 4)Case II: When x2, x-2=2-x,2-x-12-x-201-x-x0x-1x0x-10 and x>0 or x-10 and x<0x1 and x>0 or x1 and x<00<x1 or ϕ0<x1So, x(0, 1] From both the cases, we getx(0, 1][3, 4)

Page No 15.22:

Question 10:

Solve 1x-312                                                                                                                                                        [NCERT EXEMPLAR]

Answer:

As, 1x-3121x-3-1202-x-32x-302-x+32x-305-xx-30Case I: When x0, x=x,5-xx-305-x0 and x-3>0 or 5-x0 and x-3<0x5 and x>3 or x5 and x<3x5 or x<3x0, 3[5, )Case II: When x<0, x=-x,5+x-x-30x+5x+30x+5>0 and x+3>0 or x+5<0 and x+3<0x>-5 and x>-3 or x<-5 and x<-3x>-3 or x<-5x-, -5-3, So, from both the cases, we getx-, -5-3, 0, 3[5, ) x(-, -5]-3, 3[5, )

Page No 15.22:

Question 11:

Solve x+1+x>3                                                                                                                                                   [NCERT EXEMPLAR]

Answer:

We have, x+1+x>3As, x+1=x+1, x-1-x+1, x<-1and x=x, x0-x, x<0Case I: When x<-1,x+1+x>3-x+1-x>3-2x-1>3-2x>4x<4-2x<-2So, x-, -2Case II: When -1x<0,x+1+x>3x+1-x>31>3, which is not possibleSo, xϕCase III: When x0,x+1+x>3x+1+x>32x+1>32x>2x>22x>1So, x1, From all the cases, we getx-, -21, 

Page No 15.22:

Question 12:

Solve 1x-23                                                                                                                                               [NCERT EXEMPLAR]

Answer:

As, 1x-23x-21 and x-23x-2-1 or x-21 and -3x-23         As, xax-a or xa; and xa-axax1 or x3 and -3+2x3+2x1 or x3 and -1x5x(-, 1][3, ) and x-1, 5 x-1, 13, 5

Page No 15.22:

Question 13:

Solve 3-4x9                                                                                                                                  [NCERT EXEMPLAR]

Answer:

As, 3-4x93-4x-9 or 3-4x9      As, xax-a or xa-4x-9-3 or -4x9-3-4x-12 or -4x6x-12-4 or x6-4x3 or x-32 x(-, -32][3, )



Page No 15.24:

Question 1:

Find all pairs of consecutive odd positive integers, both of which are smaller than 10, such that their sum is more than 11.

Answer:

Let the smaller odd positive integer be x. Then, the other odd positive integer shall be x + 2.
Therefore, as per the given conditions:

x+2<10 and x+x+2>11x<8 and 2x>9x<8 and x>92Since x is an odd integer,Therefore, x=5,7Hence, pairs are (5,7),(7,9).

Page No 15.24:

Question 2:

Find all pairs of consecutive odd natural number, both of which are larger than 10, such that their sum is less than 40.

Answer:

Let x be the smaller of the two odd natural numbers. Then, the other odd natural number will be x + 2.
Therefore, as per the given conditions:

x>10 and x+x+2<40x>10 and  2x+2<40x>10 and x<1910<x<19 x11, 13, 15, 17Hence, the pairs are (11,13), (13,15), (15,17), (17,19).

Page No 15.24:

Question 3:

Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23.

Answer:

Let x be the smaller even integer. Then, the other even integer shall be x + 2.
Therefore, as per the given condition:

x>5 and x+x+2<23x>5 and 2x+2<23x>5 and 2x<21x>5 and x<212 x6, 8, 10Hence, the pairs are (6,8), (8,10),(10,12).

Page No 15.24:

Question 4:

The marks scored by Rohit in two tests were 65 and 70. Find the minimum marks he should score in the third test to have an average of at least 65 marks.

Answer:

Let x be the minimum  marks he scores in the third test.

 Then, 65+70+x365135+x365135+x195     (Multiplying both the sides by 3)x195-135x60Hence, the minimum marks Rohit should score in the third test should be 60.

Page No 15.24:

Question 5:

A solution is to be kept between 86° and 95°F. What is the range of temperature in degree Celsius, if the Celsius (C)/ Fahrenheit (F) conversion formula is given by F=95C+32.

Answer:

Suppose the temperature of the solution is x degree Celsius.
x in Fahrenheit = 95x+32
Then, as per the given condition:

86<95x+32<9586-32<95x<95-32      (Subtratcting 32 throughout)54<95x<6359×54<59×95x<59×63      (Multiplying by 59 throughout)30<x<35Hence, the range of the temperature in degree Celsius is between 30°C and 35°C.

Page No 15.24:

Question 6:

A solution is to be kept between 30°C and 35°C. What is the range of temperature in degree Fahrenheit?

Answer:

Let x degree Fahrenheit be the temperature of the solution.

Then, 30°C<x°F<35°CNow, F = 95C+3295×30+32<x°F<95×35+3254+32<x<63+3286°<x°<95°Hence, the range of the temperature in Fahrenheit is between 86° and 95°.

Page No 15.24:

Question 7:

To receive grade 'A' in a course, one must obtain an average of 90 marks or more in five papers each of 100 marks. If Shikha scored 87, 95, 92 and 94 marks in first four paper, find the minimum marks that she must score in the last paper to get grade 'A' in the course.

Answer:

Let x be the minimum marks scored in the last paper.


  Then, 9087+95+92+94+x510090368+x5100450368+x500450-368368+x-368500-36882x132But x can not be more than 100 82x100Hence, the minimum marks that Shikha must score in the fifth paper is 82.

Page No 15.24:

Question 8:

A company manufactures cassettes and its cost and revenue functions for a week are C=300+32x and R = 2x respectively, where x is the number of cassettes produced and sold in a week. How many cassettes must be sold for the company to realize a profit?

Answer:

To realise profit, revenue must be greater than the cost.

 2x>300+32x2x-32x>30012x>300x>600Thus, the company must sell more than 600 cassettes in a week to realise profit. 



Page No 15.25:

Question 9:

The longest side of a triangle is three times the shortest side and third side is 2 cm shorter than the longest side if the perimeter of the triangles at least 61 cm, find the minimum length of the shortest-side.

Answer:

Let the shortest side of the triangle be x cm.
Then, the longest side will be 3x and the third side will be 3x − 2.

 Perimeter of the triangle61x+3x+3x-2617x61+2x9    (Dividing throughout by 7)Hence, the minumum length of the shortest side is 9 cm.

Page No 15.25:

Question 10:

How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?

Answer:

Let x litres of water be added to the 1125 litres of 45% solution of the acid.
Total quantity of mixture is (1125+x) litres.
Total acid content in 1125 litres of mixture = 45% of 1125
  It is given that the acid content in the resulting mixture must be more than 25% and less than 30%. 25% of1125+x<45%×1125<30% of (1125+x)25100×1125+x<45100×1125<30100×1125+xMultiplying throughout by 100:28125+25x<50625<33750+30xx<50625-2812525 and  x>50625-3375030x<900 and x>562.5Thus, the water to be added should be more than 562.5 litres but less than 900 litres.

Page No 15.25:

Question 11:

A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If there are 640 litres of the 8% solution, how many litres of 2% solution will have to be added?

Answer:

Suppose x litres of 2% solution is added in the existing solution of 8% of boric acid.
Resulting mixture = (640 + x) L
Therefore, as per given conditions:
4% of (640+x)<8% of  640+2% of  x<6% of (640+x)4100(640+x)<8100×640+2100×x<6100(640+x)Multiplying throughout by 100:2560+4x<5120+2x<3840+6x2560+4x<5120+2x and 5120+2x<3840+6x2x<2560  and 4x>1280x<1280 and x>320320<x<1280Thus, the amount of solution must be less than 1280 litres but more than 320 litres.

Page No 15.25:

Question 12:

The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between 7.2 and 7.8. If the first two pH reading are 7.48 and 7.85, find the range of pH value for the third reading that will result in the acidity level being normal.

Answer:

Let x be the third  pH value.


Then, 7.2<7.48+7.85+x3<7.87.2<15.33+x3<7.821.6<15.33+x<23.4      Multiplying throughout by 321.6-15.33<15.33+x-x<23.4-15.336.27<x<8.07Hence, the range for the pH value for the third reading must be between 6.27 and 8.07



Page No 15.28:

Question 1:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

1. x + 2yy ≤ 0

Answer:

We have,x+2y-y0

Converting the given inequation to equation, we obtain x + 2y - y = 0, i.e x + y = 0
Putting y = 0 and x = 0 in this equation, we obtain x = 0 and y = 0.
So, this line intersects the x-axis and the y-axis at (0,0).
We draw the line of the equation x + y = 0
Now we take a point (1, 1) ( any point which does not lie on the line x + y = 0 )
(1, 1) does not satisfy the inequality. So, the region not containing (1, 1)
is represented by the following figure.

Hence, the shaded region represents the in equation.

Page No 15.28:

Question 2:

Represent to solution set of each of the following in equations graphically in two dimensional plane:

2. x + 2y ≥ 6

Answer:

Converting the in equation to equation, we obtain x + 2y = 6, i.e x + 2y - 6 = 0.
Putting y = 0 and x = 0 in this equation, we obtain x = 6 and y = 3.
So, this line meets x-axis at (6,0) and y-axis at (0,3).
We plot these points and join them by a thick line. This divides the xy plane into two parts.
To determine the region represented by the given inequality, consider point O(0,0). Clearly,
(0,0) does not satisfy inequality x+2y6.
So, the region that does not contain the origin is represented by the given inequality.
Hence, the shaded region is the solution to the in equation.

Page No 15.28:

Question 3:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

3. x + 2 ≥ 0

Answer:

Converting the inequation to equation, we obtain x + 2 = 0, i.e x = -2.
Clearly, it is a parallel line to y-axis at a distance of -2 units from it. This line divides the xy plane into two parts, viz LHS  of x = -2 and RHS of x = -2. To determine the region represented by the given inequality, consider point O(0,0). Clearly, (0,0)  does not satisfy the inequality. So, the region that does not contain the origin is represented by the given inequality.
Hence, the shaded region is the solution to the inequation.

Page No 15.28:

Question 4:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

4. x − 2y < 0

Answer:

Converting the inequation to equation, we obtain x - 2y = 0
Putting y = 0 and x = 0 in this equation, we obtain x = 0 and y = 0 respectively. So, this line meets the x-axis at (0,0) and the y- axis at (0,0).
If x = 1, then y = 1/2.
So, we have another point (1,1/2).
We plot these points and join them by a thin line. This divides the xy plane into two parts.
We take a point (0, 2) and it does not satisfy the inequation.

Therefore, we shade the region which is opposite to the point (0, 2)
The shaded region is the solution to the inequation.

Page No 15.28:

Question 5:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

5. −3x + 2y ≤ 6

Answer:

Converting the inequation to equation, we obtain -3x + 2y = 6, i.e -3x + 2y - 6 = 0
Putting y = 0 and x = 0 in this equation, we obtain x = -2 and y = 3 respectively. So, this line meets the x-axis at (-2,0) and the y-axis at (0,3). We plot these points and join them by a thick line.This divides the xy plane in two parts. To determine the region represented by the given inequality, consider point O(0,0). Clearly, (0,0) satisfy the inequality -3x+2y6. So, the region containing the origin is represented by the given inequality.
Hence, the shaded region is the solution to the inequation.

Page No 15.28:

Question 6:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

6. x ≤ 8 − 4y

Answer:

Converting the inequation to equation, we obtain x + 4y - 8 = 0
Putting y = 0 and x = 0 in this equation, we obtain x = 8 and y = 2 respectively. So, this line meets the x-axis at (8,0) and y-axis at (0,2). We plot these points and join them by a thick line. This divides the xy plane into two parts. To determine the region represented by the given inequality, consider point O(0,0). Clearly, (0,0) satisfies the inequality. So, the region containing the origin is represented by the given inequality.
Hence, the shaded region is the solution to the inequation.

Page No 15.28:

Question 7:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

7. 0 ≤ 2x − 5y + 10

Answer:

Converting the inequation to equation, we obtain 2x - 5y+10 = 0
Putting y = 0 and x = 0 in this equation, we obtain x = - 5 and y = 2 respectively.
So, this line meets the x-axis at (-5,0) and the y-axis at (0,2).
We plot these points and join them by a thick line.
This divides the xy plane into two parts. To determine the region represented by the given inequality, consider point O(0,0).
Clearly, (0,0) satisfies the inequality.
So, the region containing the origin is represented by the given inequality.
Hence, the shaded region is the solution to the inequation.

Page No 15.28:

Question 8:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

8. 3y ≥ 6 − 2x

Answer:

Converting the inequation to equation, we obtain 3y+2x - 6 = 0
Putting y = 0 and x = 0 in this equation, we obtain x = 3 and y = 2 respectively.
So. this line meets the x-axis at (3,0) and y-axis at (0,2).
We plot these points and join them by a thick line.
This divides the xy plane into two parts. To determine the region represented by the given inequality, consider point O(0,0).
Clearly, (0,0) does not satisfy the inequality.
So, the region not containing the origin is represented by the given inequality.
Hence, the shaded region is the solution to the inequation.

Page No 15.28:

Question 9:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

9. y ≥ 2x − 8

Answer:

Converting the inequation to equation, we obtain 2x - y - 8 =0
Putting y = 0 and x = 0 in this equation, we obtain x = 4 and y = - 8 respectively.
So, this line meets the x-axis at (4,0) and y-axis at (0, -8).
We plot these points and join them by a thick line.This divides the xy plane into two parts.
To determine the region represented by the given inequality, consider point O(0,0).
Clearly, (0,0) does satisfy the inequality.
So, the region containing the origin is represented by the given inequality.
Hence, the shaded region is the solution to the inequation.

Page No 15.28:

Question 10:

Represent to solution set of each of the following inequations graphically in two dimensional plane:

10. 3x − 2yx + y − 8

Answer:

Converting the inequation to equation, we obtain 3x - 2y - x - y + 8 = 0, i.e 2x - 3y + 8= 0
Putting  y =0 and x = 0 in this equation, we obtain x- 4 and y = 8/3 respectively.
So, this line meets the x-axis at (-4, 0) and y-axis at (0,8/3).
We plot these points and join them by a thick line.
This divides the xy plane into two parts.
To determine the region represented by the given inequality, consider point O(0,0).
Clearly, (0,0) does not satisfy the inequality.
So, the region that does not contain the origin is represented by the given inequality.
Hence, the shaded region is the solution to the inequation.



Page No 15.30:

Question 1:

Solve the following systems of linear inequations graphically:

(i) 2x + 3y ≤ 6, 3x + 2y ≤ 6, x ≥ 0, y ≥ 0
(ii) 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0
(iii) xy ≤ 1, x + 2y ≤ 8, 2x + y ≥ 2, x ≥ 0, y ≥ 0
(iv) x + y ≥ 1, 7x + 9y ≤ 63, x ≤ 6, y ≤ 5, x ≥ 0, y ≥ 0
(v) 2x + 3y ≤ 35, y ≥ 3, x ≥ 2, x ≥ 0, y ≥ 0

Answer:

(i) Converting the inequations to equations, we obtain:
2x + 3y = 6, 3x + 2y = 6, x = 0, y = 0

2x + 3y =6:  This line meets the x-axis at (3,0) and the y-axis at (0, 2). Draw a thick line joining these points.
We see that the origin (0, 0) satisfies the inequation 2x + 3y ≤ 6
So, the portion containing  the origin represents the solution set of the inequation 2x + 3y ≤ 6

3x+2y =6:  This line meets the x-axis at (2, 0) and the y-axis at (0, 3). Draw a thick line joining these points.
We see that the origin (0,0) satisfies the inequation 3x + 2y ≤ 6.
So, the portion containing  the origin represents the solution set of the inequation 3x + 2y ≤ 6

Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
Hence. the shaded region in the figure represents the solution set of the given set of inequations.

(ii)  Converting the inequations to equations, we obtain:
2x + 3y = 6, x + 4y = 4, x = 0, y = 0

2x + 3y =6:  This line meets the x-axis at (3, 0) and the y-axis at (0, 2). Draw a thick line joining these points.
We see that the origin (0,0) satisfies the inequation 2x + 3y ≤ 6.
So, the portion containing  the origin represents the solution set of the inequation 2x + 3y ≤ 6

x + 4y = 4:  This line meets the x-axis at (4, 0) and the y-axis at (0, 1). Draw a thick line joining these points.
We see that the origin (0,0) satisfies the inequation x + 4y ≤ 4.
So, the portion containing  the origin represents the solution set of the inequation x + 4y ≤ 4

Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
Hence, the shaded region in the figure represents the solution set of the given set of inequations.


(iii)  Converting the inequations to equations, we obtain:
x - y=1, x +2y =8, 2x + y = 2, x = 0, y = 0

x - y = 1:  This line meets the x-axis at (1, 0) and the y-axis at (0, - 1). Draw a thick line joining these points.
We see that the origin (0, 0) satisfies the inequation x - y ≤ 1 So, the portion containing  the origin represents the solution set of the inequation x - y ≤ 1

x + 2y =8:  This line meets the x-axis at (8, 0) and the y-axis at (0, 4). Draw a thick line joining these points.
We see that the origin (0, 0) satisfies the inequation x + 2y ≤ 8 So, the portion containing  the origin represents the solution set of the inequation x + 2y ≤ 8

2x + y =2:  This line meets the x-axis at (1, 0) and the y-axis at (0, 2). Draw a thick line joining these points.
We see that the origin (0, 0) does not satisfy the inequation 2xy 2 So, the portion that does not contain the origin represents the solution set of the inequation 2xy 2

Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
Hence, the shaded region in the figure represents the solution set of the given set of inequations.


(iv) Converting the inequations to equations, we obtain:
x + y =1, 7x + 9y = 63, x = 6, y = 5

x + y=1:  This line meets the x-axis at (1, 0) and the y-axis at (0, 1). Draw a thick line joining these points.
We see that the origin (0, 0)  does not satisfy the inequation xy 1 So, the portion not containing  the origin represents the solution set of the inequation xy 1


7x + 9y =63:  This line meets the x-axis at (9, 0) and the y-axis at (0, 7). Draw a thick line joining these points.
We see that the origin (0, 0) satisfies the inequation 7x + 9y ≤ 63 So, the portion containing  the origin represents the solution set of the inequation 7x + 9y ≤ 63

x = 6:  This line is parallel to the x-axis at a distance 6 units from it.
We see that the origin (0, 0) satisfies the inequation x ≤ 6 So, the portion containing  the origin represents the solution set of the inequation x ≤ 6

y = 5:  This line is parallel to the y-axis at a distance 5 units from it.
We see that the origin (0,0) satisfies the inequation y ≤ 5 So, the portion containing  the origin represents the solution set of the inequation
y ≤ 5

Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
Hence, the shaded region in the figure represents the solution set of the given set of inequations.


(v) Converting the inequations to equations, we obtain:
2x + 3y = 35, x = 0, y = 0

2x + 3y = 35:  This line meets the x-axis at (17.5, 0) and the y-axis at (0, 35/3). Draw a thick line joining these points.
We see that the origin (0, 0) satisfies the inequation 2x + 3y ≤ 35 So, the portion containing  the origin represents the solution set of the inequation 2x + 3y ≤ 35

x = 2:  This line is parallel to the x-axis at a distance 2 units from it.
We see that the origin (0, 0) does not satisfy the inequation x 2 So, the portion that does not contain  the origin represents the solution set of the inequation x 2

y = 3:  This line is parallel to the y-axis at a distance 3 units from it.
We see that the origin (0, 0) does not satisfies the inequation y ≥ 3 So, the portion opposite to the origin represents the solution set of the inequation y ≥ 3

Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
Hence, the shaded region in the figure represents the solution set of the given set of inequations.

Page No 15.30:

Question 2:

Show that the solution set of the following linear inequations is empty set:

(i) x − 2y ≥ 0, 2xy ≤ −2, x ≥ 0, y ≥ 0
(ii) x + 2y ≤ 3, 3x + 4y ≥ 12, y ≥ 1, x ≥ 0, y ≥ 0

Answer:

(i) Converting the inequations to equations, we obtain:
x - 2y = 0, 2x - y = -2, x = 0, y = 0
x - 2y = 0:  This line meets the x-axis at (0, 0) and y-axis at (0, 0). If x = 1, then y = 1/2,
so we have another point (1, 1/2). Draw a thick line through (0, 0) and (1, 1/2).
We see that the origin (1, 0) satisfies the inequation x + 2y ≤ 3 So, the portion containing the (1, 0) represents the solution set of the inequation x - 2y ≤ 0

2x - y = -2:  This line meets the x-axis at (-1, 0) and y-axis at (0, 2). Draw a thick line joining these points.
We see that the origin (0, 0) does not satisfy the inequation 2x - y -2 So, the portion not containing the origin represents the solution set of the inequation 2x - y -2

Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
We see in the figure that there is no common region in all the lines. Hence, the solution set to the given set of inequations is empty.




(ii) Converting the inequations to equations, we obtain:
x + 2y = 3, 3x + 4y =12,  y = 1

x + 2y = 3:  This line meets the x-axis at (3, 0) and y-axis at (0, 3/2). Draw a thick line joining these points.
We see that the origin (0, 0) satisfies the inequation x + 2y ≤ 3. So, the portion containing the origin represents the solution set of the inequation x + 2y ≤ 3

3x + 4y =12:  This line meets the x-axis at (4, 0) and y-axis at (0, 3). Draw a thick line joining these points.
We see that the origin (0, 0) does not satisfy the inequation 3x + 4y 12 So, the portion not containing the origin represents the solution set of the inequation 3x + 4y 12

y = 1:  This line is parallel to x-axis at a distance of 1 unit from it.
We see that the origin (0, 0) does not satisfy the inequation y  1 So, the portion not containing the origin represents the solution set of the inequation y  1

Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
We see in the figure that there is no common region in all the lines. Hence, the solution set to the given set of inequations is empty.

Page No 15.30:

Question 3:

Find the linear inequations for which the shaded area in Fig. 15.41 is the solution set. Draw the diagram of the solution set of the linear inequations:

Figure

Answer:

Considering the line 2x + 3y = 6, we find that the shaded region and the origin (0, 0) are on the opposite side of this line and (0, 0) does not satisfy the inequation 2x + 3y 6 So, the first inequation is 2x + 3y 6

Considering the line 4x + 6y = 24, we find that the shaded region and the origin (0, 0) are on the same side of this line and (0, 0) satisfies the inequation 4x + 6y 24 So, the corresponding inequation is 4x + 6y 24

Considering the line x - 2y = 2, we find that the shaded region and the origin (0, 0) are on the same side of this line and (0, 0) satisfies the inequation x - 2y 2 So, the corresponding inequation is x - 2y 2

Considering the line -3x + 2y = 3, we find that the shaded region and the origin (0, 0) are on the same side of this line and (0, 0) satisfies the inequation -3x + 2y 3 So, the corresponding inequation is -3x + 2y 3

Also, the shaded region is in the first quadrant. Therefore, we must have x 0 and y0

Thus, the linear inequations comprising the given solution set are given below:
2x + 3y 6, 4x + 6y 24, x - 2y 2, -3x + 2y 3, x0 and y0



Page No 15.31:

Question 4:

Find the linear inequations for which the solution set is the shaded region given in Fig. 15.42

Figure

Answer:

Considering the line x + y = 4, we find that the shaded region and the origin (0, 0) are on the same side of this line and (0, 0) does not satisfy the inequation x + y 4 So, the first inequation is x + y 4

Considering the line y = 3, we find that the shaded region and the origin (0, 0) are on the same side of this line and (0, 0) satisfies the inequation  y 3 So, the corresponding inequation is y 3

Considering the line x = 3, we find that the shaded region and the origin (0, 0) are on the same side of this line and (0, 0) satisfies the inequation x 3 So, the corresponding inequation is x 3

Considering the line x + 5y = 4, we find that the shaded region and the origin (0, 0) are on the opposite side of this line and (0, 0) does not satisfy the inequation x + 5y 4 So, the corresponding inequation is x + 5y 4

Considering the line 6x + 2y = 8, we find that the shaded region and the origin (0, 0) are on the opposite side of this line and (0, 0) does not satisfy the inequation 6x + 2y 8 So, the corresponding inequation is 6x + 2y 8

Also the shaded region is in the first quadrant. Therefore, we must have x  0 and y  0

Thus, the linear inequations comprising the given solution set are given below:
  x + y 4, y 3, x 3, x + 5y 4, 6x + 2y 8, x  0 and y  0

Page No 15.31:

Question 5:

Show that the solution set of the following linear in equations is an unbounded set:
x + y ≥ 9
3x + y ≥ 12
x ≥ 0, y ≥ 0

Answer:

Converting the inequations to equations, we obtain:
x + y = 9, 3x + y = 12 , x = 0, y = 0

x + y = 9:  This line meets the x-axis at (9, 0) and y-axis at (0, 9). Draw a thick line through these points.
Now, we see that the origin (0, 0) does not satisfy the inequation x + y 9 Therefore, the potion that does not contain the origin is the solution set to the inequaltion x + y 9

3x + y = 12:  This line meets the x-axis at (4, 0) and y-axis at (0, 12). Draw a thick line through these points.
Now, we see that the origin (0, 0) does not satisfy the inequation 3x + y 12. Therefore, the potion that does not contain the origin is the solution set to the inequaltion 3x + y 12.

Also, x ≥ 0, y ≥ 0  represents the first quadrant. Hence, the solution set lies in the first quadrant.

We see that in this solution set, the shaded region is unbounded (infinite). Hence, the solution set to the given set of inequalities is an unbounded set.

Page No 15.31:

Question 6:

Solve the following systems of inequations graphically:

(i) 2x + y ≥ 8, x + 2y ≥ 8, x + y ≤ 6
(ii) 12x + 12y ≤ 840, 3x + 6y ≤ 300, 8x + 4y ≤ 480, x ≥ 0, y ≥ 0
(iii) x + 2y ≤ 40, 3x + y ≥ 30, 4x + 3y ≥ 60, x ≥ 0, y ≥ 0
(iv) 5x + y ≥ 10, 2x + 2y ≥ 12, x + 4y ≥ 12, x ≥ 0, y ≥ 0

Answer:

(i) Converting the inequations to equations, we obtain:
2x + y = 8, x + 2y = 8, x + y = 6

2x + y = 8:  This line meets the x-axis at (4, 0) and y-axis at (0, 8). Draw a thick line through these points.
Now, we see that the origin (0, 0) does not satisfy the inequation 2x + y  8.
Therefore, the region that does not contain the origin is the solution of the inequality 2x + y 8

x + 2y = 8:  This line meets the x-axis at (8, 0) and y-axis at (0, 4). Draw a thick line through these points.
Now, we see that the origin (0, 0) does not satisfy the inequation x + 2y 8
Therefore, the region that does not contain the origin is the solution of the inequality x + 2y 8

x + y = 6:  This line meets the x-axis at (6, 0) and y-axis at (0, 6). Draw a thick line through these points.
Now, we see that the origin (0, 0) satisfies the inequation x + y 6 Therefore, the region containing the origin is the solution of the inequality  x + y 6

Hence, the solution to the inequalities is the intersection of the above three solutions. Thus, the shaded region represents the solution set of the given set of inequalities.


(ii) Converting the inequations to equations, we obtain:
12x + 12y = 840,  3x + 6y = 300, 8x + 4y = 480, x = 0, y = 0

12x + 12y = 840:  This line meets the x-axis at (70, 0) and y-axis at (0, 70). Draw a thick line through these points.
Now, we see that the origin (0, 0) satisfies the inequation 12x + 12y 840
Therefore the region containing the origin is the solution of the inequality 12x + 12y 840

3x + 6y =300:  This line meets the x-axis at (100, 0) and y-axis at (0, 50). Draw a thick line through these points.
Now, we see that the origin (0, 0) satisfies the inequation 3x + 6y 300
Therefore, the region containing the origin is the solution of the inequality 3x + 6y 300

8x + 4y = 480:  This line meets the x-axis at (60, 0) and y-axis at (0, 120). Draw a thick line through these points.
Now, we see that the origin (0, 0) satisfies the inequation 8x + 4y 480 Therefore, the region containing the origin is the solution of the inequality 8x + 4y 480

Also, x  0, y  0 represens the first quadrant. So, the solution set must lie in the first quadrant.

Hence, the solution to the inequalities is the intersection of the above three solutions. Thus, the shaded region represents the solution set of the given set of inequalities.


(iii)  Converting the inequations to equations, we obtain:
x + 2y = 40, 3x + y = 30, 4x + 3y = 60

x + 2y = 40:  This line meets the x-axis at (40, 0) and y-axis at (0, 20). Draw a thick line through these points.
We see that the origin (0,0) satisfies the inequation x + 2y 40
Therefore, the region containing the origin is the solution of the inequality x + 2y 40

3x + y = 30:  This line meets the x-axis at (10, 0) and y-axis at (0, 30). Draw a thick line through these points.
We see that the origin (0, 0) does not satisfy the inequation 3x + y 30
Therefore, the region that does not contain the origin is the solution of the inequality 3x + y 30

4x + 3y = 60:  This line meets the x-axis at (15, 0) and y-axis at (0, 20). Draw a thick line through these points.
We see that the origin (0, 0) does not satisfy the inequation 4x + 3y 60 Therefore, the region that does not contain the origin is the solution of the inequality 4x + 3y 60

Also, x ≥ 0, y ≥ 0 represents the first quadrant. So, the solution set must be in the first quadrant.

Hence, the solution to the inequalities is the intersection of the above three solutions. Thus, the shaded region represents the solution set of the given set of inequalities.


(iv)  Converting the inequations to equations, we obtain:
5x + y = 10, 2x + 2y = 12, x + 4y = 12


5x + y =10:  This line meets the x-axis at (2, 0) and y-axis at (0, 10). Draw a thick line through these points.
We see that the origin (0, 0) does not satisfy the inequation 5x + y 10
Therefore, the region that does not contain the origin is the solution of the inequality 5x + y 10

2x + 2y = 12:  This line meets the x-axis at (6, 0) and y-axis at (0, 6). Draw a thick line through these points.
We see that the origin (0, 0) does not satisfy the inequation 2x + 2y 12
Therefore, the region that does not contain the origin is the solution of the inequality 2x + 2y 12

x + 4y = 12:  This line meets the x-axis at (12, 0) and y-axis at (0, 3). Draw a thick line through these points.
We see that the origin (0, 0) does not satisfy the inequation x + 4y 12
Therefore, the region that does not contain the origin is the solution of the inequality
x + 4y 12

Also, x ≥ 0, y ≥ 0 represents the first quadrant. So, the solution set must be in the first quadrant.

Hence, the solution to the inequalities is the intersection of the above three solutions. Thus, the shaded region represents the solution set of the given set of inequalities.
Here, the solution set is unbounded region.

Page No 15.31:

Question 7:

Show that the following system of linear equations has no solution: x+2y3, 3x+4y12, x0, y1

Answer:

We have,x+2y3                 .....i3x+4y12            .....iix0                         .....iiiy1                         .....iv

As, the points satisfying x + 2y = 3 are:
 

x 3 0 5
y 0 1.5 -1

Also, the points satisfying 3x + 4y = 12 are:
 
x 0 4 8
y 3 0 -3

Now, the region representing the given inequalities is as follows:


Since, there is no common region.

So, the given system of inequalities has no solution.
x 3 0 5
y 0 1.5 -1
 

Page No 15.31:

Question 8:

Show that the solution set of the following system of linear inequalities is an unbounded region: 2x+y8, x+2y10, x0, y0.

Answer:

We have,2x+y8          .....ix+2y10        .....iix0                   .....iiiy0                   .....iv

As, the solutions of the equation 2x + y = 8 are:
 

x 0 4 2
y 8 0 4

As, the solutions of the equation x + 2y = 10 are:
 
x 0 10 2
y 5 0 4

Now, the graph represented by the inequalities (i), (ii), (iii) and (iv) is as follows:


Since, the common shaded region is the solution set of the given set of inequalities.

So, the solution set of the given linear inequalities is an unbounded region.

Page No 15.31:

Question 1:

Mark the correct alternative in each of the following:
If x<7, then
(a) -x<-7
(b) -x-7
(c) -x>-7
(d) -x-7

Answer:

x<7subtracting x on both sides, we getx-x<7-x0<7-xsubtracting 7 on both sides, we get0-7<7-x-7-7<-x-x>-7

Hence, the correct option is (c).

Page No 15.31:

Question 1:

Write the solution of the inequation x2x-2>0.

Answer:

We have,x2x-2>0Equating both the numerator and the denominator with zero, we obtain x=0 and x=2 as critical points.Plotting these points on the real line, we see that the real line is divided into three regions.


Therefore, the solution set of the given inequality is x(2,).

Page No 15.31:

Question 2:

Write the solution set of the inequation x+1x2.

Answer:

We have,x+1x2x2+1x2x2+1x-20x2-2x+1x0(x-1)2x0Either (x-1)20 and x > 0 or (x-1)2<0 and x<0.But, (x-1)2 is always greater than zero. (x-1)20 and x > 0x > 0x0, 

Page No 15.31:

Question 3:

Write the set of values of x satisfying the inequation (x2 − 2x + 1) (x − 4) < 0.

Answer:

We have:Therefore, the solution of the given inequality is x(2,)

x2-2x+1x-4<0(x-1)2(x-4)<0Equating each one to zero, we obtain x=1 and x= 4.Therefore, 1 and 4 are critical points.Drawing the number lines, we get:


Therefore, the solution set of the given inequality is x-, 11, 4herefore, the solution of the given inequality is x(2,)

Page No 15.31:

Question 4:

Write the solution set of the equation |2 − x| = x − 2.

Answer:

We have,2-x=x-2Now 2 cases arise.CASE 1: When 2-x0, then 2-x=2-x2-x=x-22-x=x-22x=4x=2So, this condition is satisfied when x =2.CASE 2: When 2-x<0 i.e. when x>2, then 2-x=-(2-x)2-x=x-2-(2-x)=x-2-2+x=x-2 -2=-2So, this condition is satisfied when x>2Hence, from the given two cases, the solution set of the given equation is [2,)

Page No 15.31:

Question 5:

Write the set of values of x satisfying |x − 1| ≤ 3 and |x − 1| ≥ 1.

Answer:

We have,x-13 and x-11We know:x-ara-rxa+rAnd, x-arxa-r or xa+r 1-3x1+3  and x1-1 or x1+1-2x4  and x0 or x2x[-2,4] and x(-,0] [2,)x-2, 0 U 2, 4

Page No 15.31:

Question 6:

Write the solution set of the inequation 1x-2>4

Answer:

We have:1x-2>4Here, two cases arise.CASE 1: When 1x-2>0, then 1x-2=1x-2 1x-2>4 1x-2-4>01x>6x0, 16                 ...(i)CASE 2: When 1x-2<0, then 1x-2=-1x-2 -1x+2>4-1x>21x<-2x-,-12            ...(ii)Hence, the solution set of the given inequation is the union of (i) and (ii). x-,-120, 16

Page No 15.31:

Question 7:

Write the number of integral solutions of x+2x2+1>12.

Answer:

We have,  x+2x2+1>12x+2x2+1-12>02x+2-x2+12x2+1>02x+4-x2-12x2+1>0-x2+2x+32x2+1>0To make the fraction of the left side positive, either the numerator or thedenominator should be positive or both should be negative.Since, it is clear that the denominator is positive, the numerator must be positive.-x2+2x+3>0x2-2x-3<0x-3x+1<0Now, to make the left side negative, one of these i.e. (x-3) or (x+1) should be positive and the other should be negative.Also,  x+1>x-3 x+1>0 and x-3<0x>-1 and x<3x-1, 3The integral solution of x is 0, 1, 2.Hence, there are 3 integral solutions of the given inequation.

Page No 15.31:

Question 8:

Write the set of values of x satisfying the inequations 5x + 2 < 3x + 8 and x+2x-1<4.

Answer:

We have,5x+2<3x+8 and x+2x-1<42x<6  and  x+2x-1-4<0x<3   and  x+2-4x+4x-1<0x(-,3) and -3x+6x-1<0x(-,3) and -x+2x-1<0For -x+2x-1<0, critical points are 1 and 2.x(2,) (-,1) x(-,1)(2,3)

Page No 15.31:

Question 9:

Write the solution of set of x+1x>2.

Answer:

We have:x+1x>2x+1x-2>0CASE 1: When x+1x>0, then x+1x=x+1xNow,  x+1x-2>0x+1x-2>0x2+1-2xx>0(x-1)2x>0x>0 and x 1x(0,1)U(1,)    ...iCASE 2: When x+1x<0, then x+1x=-(x+1x)Now,  x+1x-2>0-x-1x-2>0-x2-1-2xx>0x2+1+2xx<0       Multiplying both sides by -1(x+1)2x<0x<0 and x-1x(-,-1)U(-1, 0)    ...iiThus, the solution set of the given inequation is the union of i and ii (0,1)U(1, )(-,-1)U(-1, 0)=R--1, 0, 1 xR--1, 0, 

Page No 15.31:

Question 10:

Write the solution set of the inequation |x − 1| ≥ |x − 3|.

Answer:

We have:x-1x-3x-1-x-30The LHS of the inequation has two seperate modulus. Equating these to zero, we obtain x=1,3 as critical points.These points divide the real line in three regions, i.e (-,1], [1,3], [3,).CASE 1: When -<x1, then x-1=-(x-1) and x-3=-(x-3) x-1-x-30-(x-1)--(x-3)0-x+1+x-30-20But this is not possible.CASE 2: When 1x3, then x-1=x-1and x-3 =-(x-3) x-1-x-30x-1+x-302x4x2x[2,)CASE 3: When 3x<, then x-1=x-1 and x-3 =x-3 x-1-x-30x-1-x+30 20This is true.Hence, the solution to the given inequality comes from cases 2 and 3.[2,)U [3,)= [2,) x[2,)



Page No 15.32:

Question 2:

Mark the correct alternative in each of the following: 
If − 3x+17<-13, then
(a) x(10, )
(b) x[10, )
(c) x(-, 10]
(d) x[-10, 10)

Answer:

-3x+17<-13Subtracting 17 on both sides, we get-3x+17-17<-13-17-3x<-30Dividing -3 on both sides, we get-3x-3>-30-3x>10x10, 

Hence, the correct option is (a).

Page No 15.32:

Question 3:

Mark the correct alternative in each of the following:
Given that xy and are real numbers and x<yb>0, then
(a) xb<yb
(b) xbyb
(c) xb>yb
(d) xbyb

Answer:

Given that xy and are real numbers and x<yb>0.

Both sides of an inequality can be multiplied or divided by the same positive number.

 xb<yb

Hence, the correct option is (a).

Page No 15.32:

Question 4:

Mark the correct alternative in each of the following:
If is a real number and x<5, then
(a) x5
(b) -5<x<5
(c) x-5
(d) -5x5

Answer:

If is a real number.
x<5
-5<x<5

Hence, the correct option is (b).
 

Page No 15.32:

Question 5:

Mark the correct alternative in each of the following:
If and are real numbers such that a>0 and x>a, then
(a) x(-a)
(b) x[-a]
(c) x(-aa)
(d) x(--a (a)

Answer:

If and are real numbers such that a>0.

x>a
x>a or x<-ax-, -a  a, 

Hence, the correct option is (d).

Page No 15.32:

Question 6:

Mark the correct alternative in each of the following:
If x-1>5, then
(a) x(-4, 6)
(b) x[-4, 6]
(c) x(--4)  (6, )
(d) x(--4)  [6. )

Answer:

x-1>5x-1>5 or x-1<-5x>5+1 or x<-5+1x>6 or x<-4x-, -4  6, 

Hence, the correct option is (c).
 

Page No 15.32:

Question 7:

Mark the correct alternative in each of the following:
If x+29, then
(a) x(-7, 11)
(b) x[-11, 7]
(c) x(--7)  (11, )
(d) x(--7)  [11, )

Answer:

x+29-9x+29-9-2x+2-29-2-11x7x-11, 7

Hence, the correct option is (b).

Page No 15.32:

Question 8:

Mark the correct alternative in each of the following:
The inequality representing the following graph is 
(a) x<3
(b) x3
(c) x>3
(d) x3

Answer:

As according to the graph,

x lies between -3 and 3-3x3x3

Hence, the correct option is (b).

Page No 15.32:

Question 9:

Mark the correct alternative in each of the following:
The linear inequality representing the solution set given in Fig. 15.44 is 
(a) x<5
(b) x>5
(c) x5
(d) x5

Answer:

As according to the graph,

x lies between (-, -5] and [5, )x5 or x-5x5

Hence, the correct option is (c).

Page No 15.32:

Question 10:

Mark the correct alternative in each of the following:
The solution set of the inequation x+25 is
(a) (-7, 5)
(b) [-7, 3]
(c) [-5, 5]
(d) (-7, 3)

Answer:

x+25-5x+25-5-2x+2-25-2-7x3x-7, 3

Hence, the correct option is (b).

Page No 15.32:

Question 11:

Mark the correct alternative in each of the following:
If x-2x-20, then
(a) x[2, )
(b) x(2, )
(c) x(-, 2)
(d) x(-, 2]

Answer:

x-2x-20x-2>0x>2x2, 

Hence, the correct option is (b).

Page No 15.32:

Question 12:

Mark the correct alternative in each of the following:
If x+310, then
(a) x(-13, 7]
(b) x(-13, 7)
(c) x(--13)  (7, )
(d) x(--13]  [7, )

Answer:

x+310x+310 or x+3-10x10-3 or x-10-3x7 or x-13x(-, -13]  [7, )

Hence, the correct option is (d).



View NCERT Solutions for all chapters of Class 12