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#### Page No 268:

#### Question 1:

Draw the graph of each of the following equations:

(i) *x* = 5

(ii) *y* = −2

(iii) *x* + 6 = 0

(iv) *x* + 7 = 0

(v) *y* = 0

(vi) *x* = 0

#### Answer:

(i) x = 5

(ii) y = -2

(iii) x + 6 = 0 ⇒ x = -6

(iv) x + 7 = 0 ⇒ x = -7

(v) y = 0

(vi) x = 0

#### Page No 269:

#### Question 2:

Draw the graph of the equation *y* = 3*x*. From your graph, find the value of *y* when *x* = −2.

#### Answer:

Given equation: y = 3x

When x = -2, y = -6.

When x = -1, y = -3.

Thus, we have the following table:

x | -2 | -1 |

y | -6 | -3 |

Now plot the points (-2,-6), (-1, -3) on a graph paper.

Join the points and extend the line in both the directions.

The line segment is the required graph of the equation.

Also, when x = -2, y = -6.

#### Page No 269:

#### Question 3:

Draw the graph of the equation *x* + 2*y* − 3 = 0. From your graph, find the value of *y* when *x* = 5.

#### Answer:

Given equation: x + 2y - 3 = 0

Or, x + 2y = 3

When y = 0, x + 0 = 3 ⇒ x = 3

When y = 1, x + 2 = 3 ⇒ x = 3-2 = 1

When y = 2, x + 4 = 3 ⇒ x = 3 - 4 = -1

Thus, we have the following table:

x | 3 | 1 | -1 |

y | 0 | 1 | 2 |

Now plot the points (3,0) ,(1,1) and (-1,2) on the graph paper.

Join the points and extend the line in both the directions.

The line segment is the required graph of the equation.

When x = 5,

$y=\frac{3-x}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{3-5}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow y=-1$

#### Page No 269:

#### Question 4:

Draw the graph of each of the following equations:

(i) *y* = *x*

(ii) *y* = −*x*

(iii) *y* + 3*x* = 0

(iv) 2*x* + 3*y* = 0

(v) 3*x* − 2*y* = 0

(vi) 2*x* + *y* = 0

#### Answer:

(i) Given equation: y = x

When x = 0, y = 0.

When x = 1, y = 1.

Thus, we have the following table:

x | 0 | 1 |

y | 0 | 1 |

Now, plot the points (0,0) and (1,1).

Join the points and extend the line in both the directions.

The line segment is the required graph of y = x.

(ii) Given equation: y = -x

When x = 0, y = 0.

When x = 1, y = -1.

Thus, we have the following table:

x | 0 | 1 |

y | 0 | -1 |

Now, plot the points (0,0) and (1,-1).

Join the points and extend the line in both the directions.

The line segment is the required graph of y = -x.

(iii) Given equation: y + 3x = 0

When x = 0, y = 0.

When x = 1, y = -3.

Thus, we have the following table:

x | 0 | 1 |

y | 0 | -3 |

Now, plot the points (0,0) and (1,-3)

Join the points and extend the line in both the directions.

Thus, the line segment is the required graph of y + 3x = 0.

(iv) Given equation: 2x+ 3y = 0

When x = 0, y = 0.

When x = 3, y = -2.

Thus, we have the following table:

x | 0 | 3 |

y | 0 | -2 |

Now, plot the points (0,0) and (3,-2).

Join the points and extend the line in both the directions.

The line segment is the required graph of 2x+ 3y = 0

(v) Given equation: 3x - 2y = 0

When x = 0, y = 0.

When x = 2, y = 3.

Thus, we have the following table:

x | 0 | 2 |

y | 0 | 3 |

Now, plot the points (0,0) and (2,3).

Join the points and extend the line in both the directions.

The line segment is the required graph of 3x - 2y = 0

(vi) Given equation: 2x + y = 0

When x = 0, y = 0.

When x = 1, y = -2.

Thus, we have the following table:

x | 0 | 1 |

y | 0 | -2 |

Now, plot the points (0,0) and (1,-2)

Join the points and extend the line in both the directions.

The line segment is the required graph of 2x + y = 0.

#### Page No 269:

#### Question 5:

Draw the graph of the equation 2*x* − 3*y* = 5. From the graph, find (i) the value of *y* when *x* = 4 and (ii) the value of *x* when *y* = 3.

#### Answer:

Given equation :

$2x-3y=5\phantom{\rule{0ex}{0ex}}\Rightarrow 2x=3y+5\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{3y+5}{2}$

When, $y=-1,x=\frac{-3+5}{2}=\frac{2}{2}=1$

When, $y=-3,x=\frac{-9+5}{2}=\frac{-4}{2}=-2$

Thus, we have the following table:

$\text{x}$ | 1 | -2 |

$y$ | -1 | -3 |

(i) When x = 4:

$4=\frac{3y+5}{2}\Rightarrow 8=3y+5\phantom{\rule{0ex}{0ex}}\Rightarrow 3y=8-5=3\phantom{\rule{0ex}{0ex}}\Rightarrow 3y=3\phantom{\rule{0ex}{0ex}}\Rightarrow y=1$

(ii) When y = 3:

$x=\frac{3y+5}{2}=\frac{14}{2}=7$

#### Page No 269:

#### Question 6:

Draw the graph of the equation 2*x* + *y* = 6. Find the coordinates of the point, where the graph cuts the *x*-axis.

#### Answer:

Given equation:

$2x+y=6\phantom{\rule{0ex}{0ex}}\Rightarrow y=6-2x\phantom{\rule{0ex}{0ex}}$

When, $x=0$, $y=6-0=6$.

When, $x=1$, $y=6-2=4$.

When, $x=2$, $y=6-4=2$.

Thus, we have the following table:

$\text{x}$ | 0 | 1 | 2 |

$y$ | 6 | 4 | 2 |

Plot the points $(0,6),(1,4)$ and $(2,2)$ on the graph paper. Join these points and extend the line.

Clearly, the graph cuts the $x-axis$ at $P(3,0)$.

#### Page No 269:

#### Question 7:

Draw the graph of the equation 3*x* + 2*y* = 6. Find the coordinates of the point, where the graph cuts the *y*-axis.

#### Answer:

Given equation: $3x+2y=6$. Then,

$2y=6-3x\Rightarrow y=\frac{6-3x}{2}$

When $x=2$, $y=\frac{6-6}{2}=0$

When $x=4$, $y=\frac{6-12}{2}=-3$

Thus, we get the following table:

$\text{x}$ | 2 | 4 |

$y$ | 0 | -3 |

Plot the points $(2,0),(4,-3)$ on the graph paper. Join the points and extend the graph in both the directions.

Clearly, the graph cuts the $y-axis$ at P(0,3).

#### Page No 269:

#### Question 1:

*x* = 0 is the equation of

(a) the* x*-axis

(b) the *y*-axis

(c) a line parallel to the *x*-axis

(d) a line parallel to the *y*-axis

#### Answer:

(b) the *y*-axis

x = 0 is the equation of the y-axis.

So, the graph of x = 0 is YOY^{'}.

#### Page No 270:

#### Question 2:

*y* = 0 is the equation of

(a) the *x*-axis

(b) the *y*-axis

(c) a line parallel to the *x*-axis

(d) a line parallel to the *y*-axis

#### Answer:

(a) the *x*-axis

y = 0 is the equation of the x- axis.

So, the graph of y = 0 is X^{'}^{}OX.

#### Page No 270:

#### Question 3:

*x* + 3 = 0 is the equation of a line

(a) parallel to the *x*-axis and passing through (−3, 0)

(b) parallel to the *y*-axis and passing through (−3, 0)

(c) parallel to the *y*-axis and passing through (0, −3)

(d) None of these

#### Answer:

(b) parallel to the *y*-axis and passing through (−3, 0)

Given equation: $x+3=0$

Or, $x=-3$

The equation of the line is parallel to the $y$-axis and passes through $(-3,0)$.

#### Page No 270:

#### Question 4:

*y* − 4 = 0 is the equation of a line

(a) parallel to the *x*-axis and passing through (4, 0)

(b) parallel to the *x*-axis and passing through (0, 4)

(c) parallel to the *y*-axis and passing through (0, 4)

(d) None of these

#### Answer:

(b) parallel to the *x*-axis and passing through (0, 4)

Given equation: $y-4=0$

Or , $y=4$

The equation of the line is parallel to the $\text{x}$-axis and passes through $(0,4)$.

#### Page No 270:

#### Question 5:

The point of the form (*a*, *a*), where *a* ≠ 0, lies on

(a) the *x*-axis

(b) the *y*-axis

(c) the line *y* = *x*

(d) the line *x* + *y* = 0

#### Answer:

(c) the line *y* = *x*

Given, a point of the form $(a,a)$, where $a$≠ 0 .

When $a=1$, the point is (1,1)

When $a=2$, the point is (2,2).....and so on.

Plot the points (1,1) and (2,2).......and so on. Join the points and extend them in both the direction. You will get the equation of the line $y=x$.

#### Page No 270:

#### Question 6:

The point of the form (*a*, −*a*), where *a* ≠ 0, lies on

(a) the* x*-axis

(b) the *y*-axis

(c) the line *y* = *x*

(d) the line *x* + *y* = 0

#### Answer:

(d) the line *x* + *y* = 0

Given, a point of the form $(a,-a)$, where $a$ ≠ 0.

When $a=1$, the point is (1,-1).

When $a=2$, the point is (2,-2).

When $a=3$, the point is (3,-3).......and so on.

Plot these points on a graph paper. Join these points and extend them in both the directions.

You will get the equation of the line $x+y=0$.

#### Page No 270:

#### Question 7:

The linear equation 3*x* − 5*y* = has

(a) a unique solution

(b) two solutions

(c) infinitely many solutions

(d) no solution

#### Answer:

(c) infinitely many solutions

Given linear equation: $3x-5y=15$

Or, $x=\frac{5y+15}{3}$

When $y=0$, $x=\frac{15}{3}=5$.

When $y=3$, $x=\frac{30}{3}=10$.

When $y=-3$, $x=\frac{0}{3}=0$.

Thus, we have the following table:

$\text{x}$ | 5 | 10 | 0 |

$y$ | 0 | 3 | -3 |

Plot the points $A(5,0),B(10,3)$ and $C(0,-3)$. Join the points and extend them in both the directions.

We get infinite points that satisfy the given equation.

Hence, the linear equation has infinitely many solutions.

#### Page No 270:

#### Question 8:

The graph of the linear equation 3*x* + 2*y* = 6 cuts the *y*-axis at the point

(a) (2, 0)

(b) (0, 2)

(c) (0, 3)

(d) (3, 0)

#### Answer:

(c) (0, 3)

Given linear equation: $3x+2y=6$

Or, $y=\frac{6-3x}{2}$

Putting $x=0,$ we get: $y=\frac{6}{2}=3$

Thus, the graph cuts the y-axis at (0,3).

#### Page No 270:

#### Question 9:

The graph of the linear equation 4*x* + 3*y* = 12 cuts the *y*-axis at the point

(a) (4, 0)

(b) (0, 4)

(c) (0, 3)

(d) (3, 0)

#### Answer:

(d) (3, 0)

Given equation: $4x+3y=12$

$\text{Or},x=\frac{12-3y}{4}$

When, y = 0, x = $\frac{12-0}{4}=3$

Thus, the graph cuts the x-axis at (3,0).

#### Page No 270:

#### Question 10:

The graph of the line *x* = 3 passes through the point

(a) (0, 3)

(b) (2, 3)

(c) (3, 2)

(d) None of these

#### Answer:

(c) (3, 2)

The graph of line x = 3 is a line parallel to the y-axis.

Hence, its passes through (3,2), satisfying x =3.

#### Page No 270:

#### Question 11:

The graph of the line *y* = 2 passes through the point

(a) (2, 0)

(b) (2, 3)

(c) (5, 2)

(d) None of these

#### Answer:

(c) (5, 2)

The graph of the line y = 2 is a line parallel to the x-axis.

Thus, it passes through (5,2), satisfying y =2.

#### Page No 270:

#### Question 12:

The graph of the line *y* = −3 does not pass through the point

(a) (2, −3)

(b) (3, −3)

(c) (0, −3)

(d) (−3, 2)

#### Answer:

(d) (−3, 2)

The graph of the line y = -3 does not pass through (-3,2) since (-3,2) does not satisfy y = -3.

#### Page No 271:

#### Question 13:

A linear equation in two variables *x* and *y* is of the form *ax* + *by* + *c* = 0, where

(a) *a* ≠ 0, b ≠ 0

(b) *a* ≠ 0, *b* = 0

(c) *a* = 0, *b* ≠ 0

(d) *a* = 0, *c* = 0

#### Answer:

(a) *a* ≠ 0, b ≠ 0

A linear equation in two variables $\text{x}$ and $y$ is of the form $ax+by+c=0$, where $a$ ≠ 0 and $b$ ≠ 0.

#### Page No 271:

#### Question 14:

Any point on the *x*-axis is of the form

(a) (*x*, *y*), where *x* ≠ 0 and *y* ≠ 0

(b) (0, *y*), where *y* ≠ 0

(c) (*x*, 0), where *x* ≠ 0

(d) (*y*, *y*), where *y* ≠ 0

#### Answer:

(c) (*x*, 0), where *x* ≠ 0

Any point on the $\text{x}$-axis is of the form $(\text{x,0)}$, where $\text{x}$ ≠ 0.

#### Page No 271:

#### Question 15:

Any point on the *y*-axis is of the form:

(a) (*x*, 0), where *x* ≠ 0

(b) (0, *y*), where *y* ≠ 0

(c) (*x*, *x*), where *x* ≠ 0

(d) None of these

#### Answer:

(b) (0, *y*), where *y* ≠ 0

Any point on the $y$-axis is of the form $(0,y)$, where $y$ ≠ 0.

#### Page No 271:

#### Question 16:

How many linear equations in *x* and *y* can be satisfied by *x* = 2, *y* = 3?

(a) Only one

(b) Only two

(c) Infinitely many

(d) None of these

#### Answer:

(c) Infinitely many

Infinite linear equations are satisfied by $x=2,y=3$.

#### Page No 271:

#### Question 17:

The graph of the linear equation 3*x* + 2*y* = 6 is the line which meets the *x*-axis at the point

(a) (0, 3)

(b) (2, 0)

(c) (2, 3)

(d) (3, 2)

#### Answer:

(b) (2, 0)

Given equation: $3x+2y=6$

Or, $x=\frac{6-2y}{3}$

When y = 0, $x=\frac{6-0}{3}=2$.

Thus, the graph meets the *x*-axis at (2,0).

#### Page No 271:

#### Question 18:

The graph of the linear equation 2*x* + 5*y* = 10 is the line which meets the *y*-axis at the point

(a) (0, 2)

(b) (5, 0)

(c) $\left(\frac{1}{2},2\right)$

(d) (2, 1.2)

#### Answer:

(a) (0, 2)

Given equation: $2x+5y=10$

$y=\frac{10-2x}{5}$

When $x=0$, $y=\frac{10}{5}=2$

Thus, the graph meets the *y*-axis at (0,2).

#### Page No 271:

#### Question 19:

If each of (−2, 2), (0, 0) and (2, −2) is a solution of a linear equation in *x* and *y*, the equation is

(a) *x* − *y* = 0

(b) *x* + *y* = 0

(c) −2*x* + *y* = 0

(d) −*x* + 2*y* = 0

#### Answer:

(b) *x* + *y* = 0

Given points: $(-2,2),(0,0)$ and $(2,-2)$

We have to check which equation satisfies the given points.

Let us check for (a) $x-y=0$

Substituting $(-2,2)$ in the equation, we get: $x-y=-2-2=-4$

Substituting $(0,0)$ in the equation, we get: $x-y=0-0=0$

Substituting $(2,-2)$ in the equation, we get: $x-y=2+2=4$

So, the given points do not satisfy the equation.

Now, let us check (b) $x+y=0$

Substituting $(-2,2)$ in the equation, we get: $x+y=-2+2=0$

Substituting $(0,0)$ in the equation, we get: $x+y=0+0=0$

Substituting $(2,-2)$ in the equation, we get: $x+y=2-2=0$

So, the given points satisfy the equation.

Similarly, we can check other equations and find that the given points will not satisfy the equations.

Hence, each of $(-2,2),(0,0)$ and $(2,-2)$ is a solution of the linear equation $x+y=0$.

#### Page No 271:

#### Question 20:

The graph of the linear equation *x* − *y* = 0 passes through the point

(a) $\left(\frac{-1}{2},\frac{1}{2}\right)$

(b) $\left(\frac{3}{2},\frac{-3}{2}\right)$

(c) (0, −1)

(d) (1, 1)

#### Answer:

(d) (1, 1)

Given equation: $x-y=0$ or, $x=y$

When $x=1$, $y=1$

When $x=2$, $y=2$... and so on

Thus, we get the following table:

$\text{x}$ | 1 | 2 |

$y$ | 1 | 2 |

Plot the points on the graph paper. Join the points and extend them in both the directions.

We can see the linear equation $x-y=0$ passes through the point (1,1).

#### Page No 272:

#### Question 21:

**Assertion:** *x* = 3 is a line parallel to the *y*-axis.

**Reason:** The equation of a line parallel to the *y*-axis is *x* = *a*.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

(c) Assertion is true and Reason is false.

(d) Assertion is false and Reason is true.

#### Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

We know that equation of a line parallel to the $y$-axis is $x=a$. So, Reason (R) is true.

Also, $x=3$ is a line parallel to the $y$-axis. So, Assertion (A) is true.

Thus, Reason (R) and Assertion (A) are true and Reason (R) is a correct explanation of Assertion (A).

#### Page No 272:

#### Question 22:

**Assertion:** *y* = *mx* represents a line passing through the origin.

**Reason:** Any line parallel to *x*-axis is *y* = *b*.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

(c) Assertion is true and Reason is false.

(d) Assertion is false and Reason is true.

#### Answer:

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

We know that $y=mx$ is the equation of a line passing through the origin.

To check we will put the value of origin(0,0) in the equation:

LHS: y = 0

RHS : m(0) = 0

Since, LHS = RHS = 0

Therefore, origin satisfies the given equation.

Moreover, we know that $y=b$ is the equation of the line parallel to the $\text{x}$-axis.

So, Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

#### Page No 272:

#### Question 23:

**Assertion:** *x* + *y* = 5 is the equation of a line passing through the origin.

**Reason:** *y* = *mx* is the equation of a line passing through the origin.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

(c) Assertion is true and Reason is false.

(d) Assertion is false and Reason is true.

#### Answer:

(d) Assertion is false and Reason is true.

Given equation: $x+y=5$

Or, $y=5-x$

When

$x=1,y=5-1=4$

$x=2,y=5-2=3\phantom{\rule{0ex}{0ex}}x=3,y=5-3=2\phantom{\rule{0ex}{0ex}}x=4,y=5-4=1\phantom{\rule{0ex}{0ex}}x=5,y=5-5=0$

Thus, we get the following table:

$\text{x}$ | 1 | 2 | 3 | 4 | 5 |

$y$ | 4 | 3 | 2 | 1 | 0 |

Plot the points $(1,4),(2,3),(3,2),(4,1)$ and $(5,0)$. Join these points and extend them in both the directions.

We get a line that does not pass through the origin.

Moreover, we know that $y=mx$ is the equation of the line passing through the origin.

So, Assertion (A) is false and Reason (R) is true.

#### Page No 272:

#### Question 24:

Match the following columns:

Column I |
Column II |

(a) The equation of a line parallel to the x-axis is |
(p) y = mx |

(b) The equation of a line parallel to the y-axis is |
(q) $\frac{5}{2}$ |

(c) The equation of a line through the origin is | (r) x = k |

(d) If the point (2, 3) lies on the graph of the equation 3y = ax + 4, then a = |
(s) y = k |

(b) ......,

(c) ......,

(d) ......,

#### Answer:

(a) The equation of a line parallel to the $x-$axis is $y=k$.

(b)The equation of a line parallel to the $y$-axis is $x=k$.

(c) The equation of a line through the origin is $y=mx$.

(d) The point (2, 3) lies on the graph of $3y=ax+4$. Then,

$3\times 3=a\times 2+4\phantom{\rule{0ex}{0ex}}\Rightarrow 9=2a+4\phantom{\rule{0ex}{0ex}}\Rightarrow 2a=9-4=5\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{5}{2}$

Column I | Column II |

(a) The equation of a line parallel to the x- axis is | (s) y = k |

(b) The equation of a line parallel to the y- axis is | (r) x = k |

(c) The equation of a line through the origin is | (p) y = mx |

(d) If the point (2, 3) lies on the graph of the equation 3y = ax + 4, then a = |
(q) $\frac{5}{2}$ |

#### Page No 272:

#### Question 25:

Write each of the following in the form *ax* + *by* + *c* = 0.

(i) *x* = −2

(ii) *y* = 6

#### Answer:

(i) $x=-2$

$\Rightarrow x+2=0\phantom{\rule{0ex}{0ex}}\mathrm{Since}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{y}\mathrm{term},\mathrm{we}\text{have}:\phantom{\rule{0ex}{0ex}}a=1,b=0,c=2\phantom{\rule{0ex}{0ex}}\text{U}\mathrm{sin}\text{gthesevalueswe}\mathrm{get}\mathrm{the}\mathrm{following}\mathrm{equation}:\phantom{\rule{0ex}{0ex}}x+0.y+2=0\phantom{\rule{0ex}{0ex}}$

(ii) $y=6$

$\Rightarrow y-6=0\phantom{\rule{0ex}{0ex}}\mathrm{Since}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{x}\mathrm{term},\mathrm{we}\text{have}:\phantom{\rule{0ex}{0ex}}a=0,b=1,c=-6\phantom{\rule{0ex}{0ex}}\text{U}\mathrm{sin}\text{g}\text{these}\text{values,we}\mathrm{get}\mathrm{the}\mathrm{following}\mathrm{equation}:\phantom{\rule{0ex}{0ex}}0.x+y-6=0\phantom{\rule{0ex}{0ex}}$

#### Page No 272:

#### Question 26:

Write each of the following in the form *ax* + *by* + *c* = 0.

(i) 3*x* = 5

(ii) 5*y* = 4

#### Answer:

(i) $3x=5$ or $x=\frac{5}{3}$

$\Rightarrow x-\frac{5}{3}=0\phantom{\rule{0ex}{0ex}}\mathrm{Since}\mathrm{there}\mathrm{is}\mathrm{no}\text{'}\mathrm{y}\text{'}\mathrm{term},\mathrm{we}\text{have}:\phantom{\rule{0ex}{0ex}}a=1,b=0,c=-\frac{5}{3}\phantom{\rule{0ex}{0ex}}\text{U}\mathrm{sin}\text{g}\text{thesevalueswe}\mathrm{get}\mathrm{the}\mathrm{following}\mathrm{equation}:\phantom{\rule{0ex}{0ex}}x+0.y-\frac{5}{3}=0\phantom{\rule{0ex}{0ex}}\mathrm{or},3x+0.y-5=0\phantom{\rule{0ex}{0ex}}$

(ii) $5y=4$ or $y=\frac{4}{5}$

$\Rightarrow y-\frac{4}{5}=0\phantom{\rule{0ex}{0ex}}\mathrm{Since}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{x}\mathrm{term},\mathrm{we}\text{have}:\phantom{\rule{0ex}{0ex}}a=0,b=1,c=-\frac{4}{5}\phantom{\rule{0ex}{0ex}}\text{U}\mathrm{sin}\text{gtheabovevalueswe}\mathrm{get}\mathrm{the}\mathrm{following}\mathrm{equation}:\phantom{\rule{0ex}{0ex}}0.x+1.y-\frac{4}{5}=0\phantom{\rule{0ex}{0ex}}\mathrm{or},0.x+5y-4=0\phantom{\rule{0ex}{0ex}}$

#### Page No 272:

#### Question 27:

The total runs scored by two batsmen in a one-day cricket match is 215. Express this information in the form of a linear equation in two variables.

#### Answer:

Let $\text{x}$ be the number of runs scored by the first batsman and $y$ be the number of runs scored by the second batsman.

Then, total runs = 215 (Given)

So, the linear equation in two variable is $x+y=215$.

#### Page No 273:

#### Question 28:

The weight of a book is three times the weight of a note book. Express this fact in the form of an equation in two variables.

#### Answer:

Let the weight of notebook be $\text{x}$ gm.

Let the weight of book be $y$ gm.

According to the given condition, we have:

$y=3x$

Hence, the equation in two variables is $y=3x$.

#### Page No 273:

#### Question 29:

Check which of the following are solutions of the equation 2*x* − 3*y* = 6.

(i) (3, 0)

(ii) (0, 2)

(iii) (2, 6)

(iv) (6, 2)

#### Answer:

(i) $(3,0)$

Given equation: $2x-3y=6$

Putting $x=3,y=0$ in the given equation, we get:

$LHS=2\times 3-3\times 0=6=RHS$

So, it satisfies the given equation.

(ii) $(0,2)$

Given equation: $2x-3y=6$

Putting $x=0,y=2$ in the given equation, we get:

$LHS\hspace{0.17em}=2\times 0-3\times 2=-6$ ≠ $RHS$

So, it does not satisfy the given equation.

(iii) $(2,6)$

Given equation: $2x-3y=6$

Putting $x=2,y=6$ in the given equation, we get:

$LHS=2\times 2-3\times 6=-14$ ≠ $RHS$

So, it does not satisfy the given equation.

(iv) $(6,2)$

Given equation: $2x-3y=6$

Putting $x=6,y=2$ in the given equation, we get:

$LHS=2\times 6-3\times 2=12-6=6=RHS$

So, it satisfies the given equation.

Hence, $(3,0),(6,2)$ are the solutions of the given equation.

#### Page No 273:

#### Question 30:

Find the value of *k* if *x* = 3, *y* = 1 is a solution of the equation 2*x* + 5*y* = *k*.

#### Answer:

Given equation: $2x+5y=k$; $x=3,y=1$ are the solutions of the given equation.

So, they must satisfy the given equation.

Putting $x=3,y=1$ in the equation, we get:

$\Rightarrow 2\times 3+5\times 1=k\phantom{\rule{0ex}{0ex}}\Rightarrow 6+5=k\phantom{\rule{0ex}{0ex}}$

$\Rightarrow k=11$

Hence, the value of $k=11$.

#### Page No 273:

#### Question 31:

Find four different solutions of 2*x* + *y* = 6.

#### Answer:

Given: $2x+y=6$

Or, $y=6-2x$

When $x=1$, $y=6-2=4$.

When $x=2$, $y=6-4=2$.

When $x=3$, $y=6-6=0$.

When $\text{x}=4$, $y=6-8=-2$.

Hence, four different solutions are $(1,4),(2,2),(3,0)$ and $(4,-2)$.

#### Page No 273:

#### Question 32:

Express *y* in terms of *x*, given that $\frac{x}{5}+2y=3$. Check whether (−5, 2) is a solution of the given equation.

#### Answer:

Given: $\frac{x}{5}+2y=3$

$\Rightarrow x+10y=15$

$\Rightarrow y=\frac{15-x}{10}$

We have to check whether $(-5,2)$ is a solution of the given equation.

So,

$LHS=\frac{x}{5}+2y=\frac{-5}{5}+2\times 2=-1+4=3=RHS$

Hence, $(-5,2)$ is a solution of the given equation.

#### Page No 273:

#### Question 33:

Show that (3, 1) as well as (2, −2) are the solutions of the equation 3*x* − *y* = 8. Find two more solutions. How many solutions can we find?

#### Answer:

Given equation: $3x-y=8$

To show: $(3,1),(2,-2)$ are the solutions of the given equation.

Putting $x=3,y=1$ in the equation, we get:

$LHS=3\times 3-1=9-1=8=RHS$

Now, putting $x=2,y=-2$ in the equation, we get:

$LHS=3\times 2-(-2)=6+2=8=RHS$

Hence $(3,1),(2,-2)$ are the solutions of $3x-y=8$.

Further solutions:

The equation can be rewritten as $y=3x-8$

Putting $x=1$, we get: $y=3-8=-5$

Putting $x=4$, we get: $y=3\times 4-8=4$

Putting $x=5$, we get: $y=3\times 5-8=7$.....and so on

Two more solutions are $(1,-5)\text{and}(4,4).$

Hence, we can find infinite solutions for the given equation.

#### Page No 273:

#### Question 34:

For the equation 6*x* − 5*y* = 8, verify that

(i) (3, 2) is a solution

(ii) (2, 3) is not a solution

#### Answer:

Given equation: $6x-5y=8$

(i) $(3,2)$ is a solution.

Putting $x=3,y=2$ in the equation, we get:

$LHS=6\times 3-5\times 2=18-10=8=RHS$.

Hence, $(3,2)$ is a solution of the given equation.

(ii) $(2,3)$ is not a solution.

Putting $x=2,y=3$ in the equation, we get:

$LHS=6\times 2-5\times 3=12-15=-3$ ≠ $RHS$.

Hence, $(2,3)$ is not a solution of the given equation.

#### Page No 273:

#### Question 35:

If the point (3, 4) lies on the graph of the equation 3*y* = *ax* + 7, find the value of *a*.

#### Answer:

Given equation: $3y=ax+7$.

Also, $(3,4)$ lies on the graph of the equation.

Putting $x=3,y=4$ in the equation, we get:

$3\times 4=3a+7$

⇒ $12=3a+7$

⇒ $3a=12-7=5$

⇒ $a=\frac{5}{3}$

#### Page No 273:

#### Question 36:

Find two solutions for each of the following:

(i) 3*x* + 4*y* = 12

(ii) 3*x* + 5*y* = 0

(iii) 4*y* + 5 = 0

#### Answer:

(i) $3x+4y=12$

Or, $y=\frac{12-3x}{4}$

When $x=0$, $y=\frac{12}{4}=3$.

When $x=4$, $y=\frac{0}{4}=0$.

Hence , the two solutions are $(0,3)$and $(4,0)$.

(ii) $3x+5y=0$

Or, $y=-\frac{3}{5}x$

When $x=5$, $y=-\frac{3}{5}\times 5=-3$.

When $x=10$, $y=-\frac{3}{5}\times 10=-6$.

Hence, the two solution are $(5,-3)$ and $(10,-6)$.

(iii) $4y+5=0$

Or, $y=-\frac{5}{4}$

It is independent of $\text{x}$.

When $x=1$ , $y=-\frac{5}{4}$.

When $x=2$, $y=-\frac{5}{4}$.

Hence, the two solutions are $(1,-\frac{5}{4})$ and $(2,-\frac{5}{4})$.

#### Page No 273:

#### Question 37:

Study the graph given below. Choose the equation whose graph is given:

(i) *y* = *x*

(ii) *y* = 2*x*

(iii) *y* = 2*x* + 1

(iv) *x* + *y* = 0

#### Answer:

(iii) *y* = 2*x* + 1

The two given points are $A(1,3),B(-1,-1)$.

These points must satisfy the equation in order to form the graph.

Let $y=2x+1$ be the equation of this graph.

At $A(1,3)$, $LHS=3,RHS=2x+1=2\times 1+1=3$

So, $LHS=RHS$

Now, at $B(-1,-1)$, $LHS=-1,RHS=2x+1=2\times (-1)+1=-1$

So, $LHS=RHS$

Hence, the given points satisfy this equation.

Hence, $y=2x+1$ is the equation of the graph.

#### Page No 273:

#### Question 38:

Draw the graph of the equation 3*x* + 5*y* − 15 = 0 and show that *x* = 1, *y* = 2 is not a solution of the given equation.

#### Answer:

Given equation: $3x+5y-15=0$

Or, $y=\frac{15-3x}{5}$

When $x=0$, then $y=\frac{15}{5}=3$.

When $x=5$, then $y=\frac{0}{5}=0$.

Thus, we get the following table:

$\text{x}$ | 0 | 5 |

$y$ | 3 | 0 |

Plot these points $A(0,3),B(5,0)$ on the graph paper. Join the points and extend them in both the directions.

Now, from the graph we can see that $x=1,y=2$ does not lie on the line.

Moreover, we can check that $x=1,y=2$ does not satisfy the equation as well.

$LHS=3x+5y-15=3\times 1+5\times 2-15=13-15=-2$ ≠ $RHS$.

Hence, $x=1,y=2$ is not a solution of the given equation.

#### Page No 274:

#### Question 39:

Draw the graph of the equation 3*x* + 2*y* = 12. At what points does the graph cut the *x*-axis and the *y*-axis?

#### Answer:

Given equation: $3x+2y=12$

or, $y=\frac{12-3x}{2}$

When $x=0$, then $y=\frac{12}{2}=6$.

When $x=2$, then $y=\frac{12-6}{2}=\frac{6}{2}=3$.

When $\text{x}=4$, then $y=\frac{12-12}{2}=0$.

Thus, we have the following table:

$\text{x}$ | 0 | 2 | 4 |

$y$ | 6 | 3 | 0 |

Plot the points $A(0,6),B(4,0)$ and $C(2,3)$ on the graph paper.

Join the points and extend them in both the direction.

We see the graph cuts the $\text{x}$-axis at $(4,0)$ and the $y$-axis at $(0,6)$.

#### Page No 274:

#### Question 40:

Draw the graph of the equation *x* − 2*y* = 6. Verify that each of the points *P*(2, −2), *Q*(4, −1) and *R*(−2, −4) lies on the straight line.

#### Answer:

Given, $x-2y=6$

(or) $y=\frac{x-6}{2}$

When $x=0$, then $y=-\frac{6}{2}=-3$.

When $x=2$, then $y=-\frac{4}{2}=-2$.

When $\text{x}=4$, then $y=-\frac{2}{2}=-1$.

When $x=6$, then $y=\frac{0}{2}=0$.

Thus, we get the following table:

$\text{x}$ | 0 | 2 | 6 |

$y$ | - 3 | -2 | 0 |

Plot the points $(0,3),(2,-2),(6,0)$ on the graph paper. Join the points and extend them in both the directions.

From the graph, we can see that all the points lie on the straight line.

#### Page No 274:

#### Question 41:

There are two scales of measuring temperature, namely, Fahrenheit (F) and Celsius (C).

The relation between the two scales is

$F=\frac{9}{5}C+32$

(i) Draw the graph of the given linear equation taking *C* along the *x*-axis and *F* along the *y*-axis.

Fill in the blanks given below:

(ii) 0°C = (......)°F

(iii) 95°F = (......)°C

(iv) 0°F = (......)°C

(v) Find the temperature which is numerically the same in both (F) and (C).

#### Answer:

The given equation is $F=\frac{9}{5}C+32$.

On a graph paper, take $C$ along the $\text{x}$-axis and $F$ along the $y$-axis.

Along the $\text{x}$-axis, take 10 small divisions = $1\xb0C$

Along the $y$-axis, take 10 small divisions = $10\xb0F$

Putting $C=0$ in the given equation, we get: $F=32$

Putting $C=5$ in the given equation, we get: $F=41$

Thus, we get the following table:

$\text{x}$ | 0 | 5 |

$y$ | 32 | 41 |

Plot the points $A(0,32),B(5,41)$ on the graph paper. Join these points and extend them in both the directions.

(ii) Putting $C=0$ in the given equation, we get $F=32$

$0\xb0C=32\xb0F$

(iii) Putting $F=95\xb0$ in the given equation, we get:

$\frac{9C}{5}=95-32$

⇒ $\frac{9}{5}C=63$

⇒ $C=\frac{63\times 5}{9}=35$

∴ $95\xb0F=35\xb0C$

(iv) Putting $F=0$ in the given equation, we get:

$\frac{9}{5}C=-32$

⇒ $C=-\frac{5\times 32}{9}=-\frac{160}{9}=-17.8$

$0\xb0F=-17.8\xb0C$

(v) Putting $C=F$ in the given equation, we get:

$F-\frac{9}{5}F=32$

⇒ $4F=-160\phantom{\rule{0ex}{0ex}}$

⇒ $F=-\frac{160}{4}==-40$

∴ $-40\xb0F=-40\xb0C$

#### Page No 274:

#### Question 42:

A taxi charges Rs 20 for the first km and Rs 12 per km for subsequent distance covered. Taking the distance covered as *x* km and total fare as Rs *y*, write a linear equation, depicting the relation in *x* and *y*.

Draw the graph between *x* and *y*.

From the graph, find the taxi charges for covering 16 km.

#### Answer:

Taxi charges for the first km = Rs 20

Taxi charges for the subsequent distance covered = Rs 12 per km

Distance covered = x km (given)

Total fare = Rs y (given)

Fare charged for the distance covered leaving the first kilometre = Rs 12(x - 1)

So, taxi charge = 20 + 12(x - 1)

But total fare = Rs y (given)

∴ y = 20 + 12x - 12 = 12x + 8

Hence, the required relation between x and y is y = 12x + 8

For, $x=0$ $\Rightarrow y=8$

For, $x=3$ $\Rightarrow y=44$

Plot these points $A(0,8),B(3,44)$ on the graph paper.

Join them and extend in both the directions.

Then, $x=16$ gives $y=200$.

So, for 16 km taxi charge = Rs 200.

#### Page No 274:

#### Question 43:

If the work done by a body on applying a constant force is directly proportional to the distance travelled by the body, then express this in the form of an equation in two variables by taking the constant force as 4 units. From the graph, find the work done when the distance travelled is

(i) 2 units

(ii) 0 unit

(iii) 5 units

#### Answer:

Let work done by the body = $W$

Constant force = 4 units

Let distance travelled by the body = $d$

According to the given condition, we have:

$W=4d$ (∵ Work = Distance $\times $ Force)

So,

when $d=0$, then $W=0$

when $d=2$, then $W=8$

Plot these points on the graph paper. Join the points and extend them in both the directions.

Then, we have:

(i) If $d=2$, then $W=8$ units

(ii) If $d=0$, then $W=0$ units

(iii) If $d=5$, then $W=20$ units

#### Page No 280:

#### Question 1:

For the equation 5*x* + 8*y* = 50, if *y* = 10, the value of *x* is

(a) 6

(b) −6

(c) 12

(d) −12

#### Answer:

(b) −6

Given equation: $5x+8y=50$

If $y=10$, then

$5x+80=50$

⇒ $5x=-30$.

⇒ $x=-\frac{30}{5}=-6$

Hence, $x=-6$.

#### Page No 280:

#### Question 2:

The linear equation 2*x* + 5*y* = 16 has

(a) a unique solution

(b) two solutions

(c) no solution

(d) infinitely many solutions

#### Answer:

(d) infinitely many solutions

The linear equation is $2x+5y=16$ or, $y=\frac{16-2x}{5}$

When $x=3$, then $y=\frac{10}{5}=2$

When $x=8$, then $y=\frac{0}{5}=0$

When $x=13$, then $y=\frac{-10}{5}=-2$

Plotting these points $(3,2),(8,0)$ and $(13,-2)$ on the graph paper, we see that the equation has infinite solutions.

#### Page No 280:

#### Question 3:

Express $\frac{2x}{3}+\frac{y}{6}-5=0$ in the form *ax* + by + *c* = 0.

#### Answer:

Given, $\frac{2x}{3}+\frac{y}{6}-5=0$

Taking $LCM$ of the denominator, we get:

$\frac{4x+y-30}{6}=0$

⇒ $4x+y-30=0$,

which is the required form.

#### Page No 280:

#### Question 4:

If 5*y* − 3*x* + 15 = 0, then express *y* in terms of *x*.

#### Answer:

Given, $5y-3x+15=0$.

$\Rightarrow 5y=3x-15\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{(3x-15)}{5}$

which is the expression in $\text{x}$.

#### Page No 280:

#### Question 5:

For what value of *k* does the point (*k*, −3) lie on the line 3*x* − *y* = 6?

#### Answer:

Given line: $3x-y=6$

Also $(k,-3)$ lies on the line.

Therefore, putting $x=k,y=-3$ in the equation of the line, we get:

$3k-(-3)=6\phantom{\rule{0ex}{0ex}}\Rightarrow 3k+3=6\phantom{\rule{0ex}{0ex}}\Rightarrow 3k=3\phantom{\rule{0ex}{0ex}}\Rightarrow k=1$

Hence, $k=1$.

#### Page No 280:

#### Question 6:

If *x* = 3, *y* = −2 satisfy 2*x* − 3*y* = *k*, then find the value of *k*.

#### Answer:

Given: $x=3,y=-2$ satisfy the equation $2x-3y=k$. Then,

$2\times 3-3\times (-2)=k$

$\Rightarrow 6+6=k\phantom{\rule{0ex}{0ex}}\Rightarrow k=12$

Hence, $k=12$.

#### Page No 280:

#### Question 7:

Find the points where the graph of the equation 3*x* + 4*y* = 12 cuts the *x*-axis and the *y*-axis.

#### Answer:

Given equation: $3x+4y=12$

or, $y=\frac{12-3x}{4}$

Now,

When $x=0$, then $y=\frac{12}{4}=3$

When $\text{x}=4$, then $y=\frac{0}{4}=0$

Thus, we have the following table:

$\text{x}$ | 0 | 4 |

$y$ | 3 | 0 |

Thus, the graph of the equation cuts the $\text{x}$-axis at (4, 0) and the $y$-axis at (0,3).

#### Page No 280:

#### Question 8:

The area of the triangle formed by the line *x* + 3*y* = 12 and the coordinate axes is

(a) 12 sq units

(b) 18 sq units

(c) 24 sq units

(d) 30 sq units

#### Answer:

Given: $x+3y=12$ or, $y=\frac{12-x}{3}$

When $x=0$, then $y=\frac{12}{3}=4$

When $x=3$, then $y=\frac{9}{3}=3$

When $x=6$, then $y=\frac{6}{3}=2$

When $x=9$, then $y=\frac{3}{3}=1$

When $x=12$, then $y=\frac{0}{3}=0$

Thus, we get the following table:

$\text{x}$ | 0 | 3 | 6 | 9 | 12 |

$y$ | 4 | 3 | 2 | 1 | 0 |

Plot these points on the graph paper. Join them to get a right-angled triangle $AOE$, such that

$AO=4$ unit and $OE=12$ unit.

Therefore,

$areaof\u25b3AOE=\frac{1}{2}(base\times height)=\frac{1}{2}(OE\times AO)=\frac{1}{2}\times 12\times 4=24$ sq units

#### Page No 281:

#### Question 9:

**Assertion:** *y* = *mx* represents a line passing through the origin.

**Reason:** Any line parallel to the *x*-axis is *y* = *k*.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

(c) Assertion is true and Reason is false.

(d) Assertion is false and Reason is true.

#### Answer:

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

We know that $y=mx$ represents a line passing through the origin.

So, Assertion (A) is true.

Moreover, we know that equation of a line parallel to the $\text{x}$-axis is $y=k$.

So, Reason (R) is also true.

Hence both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

#### Page No 281:

#### Question 10:

**Assertion:** *x* = 3 is a line parallel to the *y*-axis.

**Reason:** Any line parallel to the *y*-axis is *y* = *k*, where *k* *∈* *R*.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

(c) Assertion is true and Reason is false.

(d) Assertion is false and Reason is true.

#### Answer:

(c) Assertion is true and Reason is false.

We know that $x=3$ is a line parallel to $y$-axis.

Hence, Assertion (A) is true.

Any line parallel to the $y$-axis is of the form $x=k$, where $k\in R$.

Hence Reason (R) is not true.

Hence, Assertion (A) is true and Reason (R) is false.

#### Page No 281:

#### Question 11:

Match the following columns:

Column I |
Column II |

(a) Any line parallel to the x-axis is |
(p) 3 |

(b) Any line parallel to the y-axis is |
(q) y = mx |

(c) Any line passing through the origin is | (r) x = k |

(d) If the point (−2, 2) lies on the line ax + 4y = 2, then a = |
(s) y = k |

(b) ......,

(c) ......,

(d) ......,

#### Answer:

(a) Any line parallel to the $\text{x}$-axis is $y=k$.

(b) Any line parallel to the $y$-axis is $x=k$.

(c) Any line passing through the origin is $y=mx$.

(d) The point (-2, 2) lies on the line $ax+4y=2$.

Putting $x=-2,y=2$ in the equation, we get:

$-2a+8=2\phantom{\rule{0ex}{0ex}}\Rightarrow -2a=2-8=-6\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{-6}{-2}=3$

Column I |
Column II |

(a) Any line parallel to the $\text{x}$-axis is | (s) $y=k$ |

(b) Any line parallel to the $y$-axis is | (r) $x=k$ |

(c) Any line passing through the origin is | (q) $y=mx$ |

(d) If the point (-2, 2) lies on the line $ax+4y=2$, then $a=$ | (p) 3 |

#### Page No 281:

#### Question 12:

Give the geometrical representation of *x* = 3 as an equation in

(i) one variable

(ii) two variables

#### Answer:

(i) Geometrical representation of $x=3$ in one variable is $x=3$.

(ii) Geometrical representation of $x=3$ in two variable is $x+0.y=3$.

#### Page No 281:

#### Question 13:

For the line 2*x* + 3*y* = 6

(i) *x*-intercept = ......

(ii) *y*-intercept = ......

#### Answer:

Line: $2x+3y=6$

(i) Putting $y=0$ in the equation , we get:

$2x=6$

$\Rightarrow x=\frac{6}{2}=3$

Hence, $\text{x}$-intercept = 3

(ii) Putting $x=0$ in the equation, we get:

$3y=6$

$\Rightarrow y=\frac{6}{3}=2$

Hence, $y$-intercept = 2

#### Page No 281:

#### Question 14:

Draw the graph of the line *y* = *x* and show that the point (2, 3) does not lie on it.

#### Answer:

Equation of line : $y=x$.

Then,

$Whenx=1,y=1\phantom{\rule{0ex}{0ex}}Whenx=2,y=2\phantom{\rule{0ex}{0ex}}Whenx=3,y=3\phantom{\rule{0ex}{0ex}}Whenx=4,y=4\phantom{\rule{0ex}{0ex}}.\phantom{\rule{0ex}{0ex}}.\phantom{\rule{0ex}{0ex}}.$

and so on.

Plot the points $(1,1),(2,2),(3,3),(4,4)......$ .

Join the points and extend them in both the directions.

We see that $(2,3)$ does not lie on the line $y=x$.

#### Page No 281:

#### Question 15:

Draw the graph of 2*x* − 3*y* = 4. From the graph, find whether *x* = −1, *y* = −2 is a solution or not.

#### Answer:

Given: $2x-3y=4$ or, $y=\frac{2x-4}{3}$

When $x=2$, then $y=\frac{0}{3}=0$

When $x=5$, then $y=\frac{6}{3}=2$

When $x=-1$, then $y=-\frac{6}{3}=-2$

Thus, we get the following table:

$\text{x}$ | 2 | 5 | -1 |

$y$ | 0 | 2 | -2 |

Plot the points $(2,0),(5,2)$ and $(-1,-2)$ on the graph paper. Join them and extend them in both the directions.

Thus, we see that $x=-1,y=-2$ is a solution of the equation.

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#### Question 16:

The runs scored by two batsmen in a cricket match are 164. Write a linear equation in two variables *x* and *y*. Also write a solution of this equation.

#### Answer:

Let one batsman score $\text{x}$ runs.

Then, the other batsman scores $y$ runs.

Given, total runs = 164

Then, according the given condition, we have:

$x+y=164$

Solution:

Putting $x=0$ in the equation, we get:

$y=164$

Again, putting $y=0$ in the equation, we get:

$x=164$

Hence, $(0,164),(164,0)$ are the two solutions of the equation.

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#### Question 17:

Find whether the given statement is true or false:

(i) *x* = 2, *y* = 3 is a solution of the equation 5*x* − 3*y* = 1.

(ii) *y* = 2*x* + 5 is a straight line passing through the point (1, 5).

(iii) The area bounded by the line *x* + *y* = 6, the *x*-axis and the *y*-axis is 18 sq units.

#### Answer:

(i) Given equation: $5x-3y=1$.

To check wether $x=2,y=3$ is the solution of the equation.

Putting $x=2,y=3$ in the equation, we get:

$LHS=5x-3y=5\times 2-3\times 3=10-9=1=RHS$

Hence, $x=2,y=3$ is the solution of the equation.

Hence, the statement is true.

(ii) Given equation: $y=2x+5$

When $x=-3$, then $y=-1$.

When $x=-2$, then $y=1$.

When $x=-1$, then $y=3$.

When $x=0$, then $y=5$.

When $x=1$, then $y=7$.

Plot these points on a graph paper and join them in both the directions.

From here, we see that for $x=1$ we have $y=7$≠ 5. Moreover, from the graph we see that $(1,5)$ does not lie on the line $y=2x+5$.

Hence, the statement is false.

(iii) Given : $x+y=6$.

When $x=0$, then $y=6$.

When $y=0$, then $x=6$.

Plotting these points on the graph paper, we get the right-angled triangle$AOB$, such that $AO=6$ unit and $OB=6$ unit.

Hence, $areaof\u25b3AOB=\frac{1}{2}\times base\times height=\frac{1}{2}\times 6\times 6=18$ sq unit.

Hence, the statement is true.

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#### Question 18:

Two men start from points *A* and *B* respectively, 42 km apart. One walks from *A* to *B* at 4 km/hr and another walks from *B* to *A* at a certain uniform speed. They meet each other after 6 hours. Find the speed of the second man.

#### Answer:

Given: Two men start from $A,B$ respectively, which are 42 km apart.

Speed of man at $A$ = 4 km/hr.

Let the speed of man at $B$ = x km/hr.

They meet after 6 hours.

Then, relative speed of the two men = $\frac{Dis\mathrm{tan}ce}{time}=\frac{42}{6}=7$ km/hr

Therefore, speed of man at $B$ = Relative speed of the two - Speed of man at $A$ = 7 - 4 = 3 km/hr

Hence, speed of the second man = 3 km/hr.

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#### Question 19:

The taxi fare in a city is such that Rs 50 is the fixed amount and Rs 16 per km is charged. Taking the distance covered as *x* km and total fare as Rs *y*, write a linear equation in *x* and *y*. What is the total fare for 20 km?

#### Answer:

Given: Taxi fare in the city (fixed amount) = Rs 50

Taxi fare per km = Rs 16

Distance covered = x km

Total fare = Rs y

Therefore, fare charged per km = Rs 16x

Hence,

Total fare = y = 50 + 16x

Hence, the linear equation: y = 50 + 16x

Total fare for 20 km:

Putting x = 20 in the linear equation, we get:

y = 50 + 16$\times $20 = 50 + 320 = Rs 370

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#### Question 20:

Draw the graphs for the equations *x* + *y* = 6 and *x* − *y* = 2 on the same graph paper and find the coordinates of the point where the two straight lines intersect.

#### Answer:

The equations are $x+y=6$ and $x-y=2$.

Now, consider $x+y=6$.

When $x=0$, then $y=6$.

When $x=2$, then $y=4$.

When $x=6$, then $y=0$.

Thus, we get the following table:

$\text{x}$ | 0 | 2 | 6 |

$y$ | 6 | 4 | 0 |

Plot the points $(0,6),(2,4)$ and $(6,0)$ on the graph paper. Then, join the points and extend them in both the directions.

Now, consider $x-y=2$.

When $x=0$, then $y=-2$.

When $x=1,$ then $y=-1$.

When $x=2$, then $y=0$.

Thus, we get the following table:

$\text{x}$ | 0 | 1 | 2 |

$y$ | -2 | -1 | 0 |

Plot these points on the graph paper. Then, join the points and extend them in both the directions.

From the graph, we can see that the two lines intersect at (4,2) .

View NCERT Solutions for all chapters of Class 9