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if the graph of the equations 3x+4y=12 and (m+n)x+2(m-n)y=5m-1 is a coincident line,then

1.m=-1,n=-5

2.m=1,n=5

3.m=5,n=1

4.m=-5,n=-1

 Since the pair of linear equations have infinite solutions they obey the criterion- (a1/a2)=(b1/b2)=(c1/c2)

Here a1=3 ,b1=4, c1=12

a2=(m+n), b2=2(m-n), c2=5m -1

applying these values we get,

{3/(m+n)} = {4/2(m-n)} = {12/(5m-1)}

equating the first and second part, we get

{3/(m+n)} = {4/2(m-n)}

3/(m+n) = 2/(m-n)

3(m-n) = 2(m+n)

3m - 3n = 2m + 2n

3m - 2m = 3n + 2n

M = 5n ........... (1)

Equating the first and third part

3/(m+n) = 12/(5m-1)

1/(m+n) = 4/(5m-1)

5m – 1 = 4(m+n)

5m – 1 = 4m + 4n

5m – 4m -4n = 1

m – 4n = 1

Substituting m = 5n from eqn(1)

5n -4n = 1

n = 1

m = 5n

m = 5(1)

m = 5

Therefore m = 5 and n = 1 are the required values.

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Draw the graph of 4x-3y+12=0. From the graph find the value of : (1) y when x= -3,3 (2) x when y= 4,-4

In the given line i.e.,

4x – 3y + 12 = 0

When

x = 0, y = 4

When

y = 0, x = –3

So by joining the points (0, 4) and (–3, 0) we obtain the required graph

Clearly

when x = – 3, y = 0

when x = 3, y = 8

when y = 4, x = 0

when y = – 4, x = – 6

  • -1

 Given system of equations is: 3x+4y=12 and (a+b)x = 5a - 1.

They can be rewritten as:

3x + 4y - 12 = 0 and

(a+b)x + 0y - (5a-1) = 0

We know that, for the given system of equations to have infinitely many solutions,  ==  , i.e.,

 = 

⇒  = 

⇒ 15a - 3 = 12a + 12b

⇒ 3a - 12b = 3

⇒ a - 4b = 1

⇒ a = 1+ 4b,is the required for the given equations of line to have infinitely many solutions.

 

2)

Given system of equations is: ax+3y=a-2 , 12x+ay=a.

They can be rewritten as:

ax + 3y - (a-2) = 0 and

12x + ay - a = 0

We know that, for the given system of equations to have infinitely many solutions,  =≠  , i.e.,

 = 

⇒ a2 = 36

⇒ a = ± 6 ,is the required condition for the given system of equations to have no solution.

  • -1
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