Question

draw the graph of the following equation, from the graph find the value of y when x=-2
(I) x+2y=6
(II) y-x=3
(III) y-4x​

Answer 1

Answer:

Q1. x + y = 3; 2x + 5y = 12

Sol:

2x + 5y = 12

We have,

x + y = 3

When y = 0 we have x =3

When x= 0 we have y =3

Thus we have the following table giving points on the line x+y =3

X 0 3

Y 3 0

Now, 2+5y= 12

= y = 12−2x5

When x= 1, we have

= y = 12−2(1)5 = 4

Thus we have the following table giving points on the line 2x+5y=12

X 1 -4

Y 2 4

Graph of the equation x+y =3 and 2x+5y=12 is

1

Clearly two lines intersect at a point P (1,2)

Hence x= 1 and y = 2

Q2: x- 2y=5, 2x+3y=10

Sol:

We have, x- 2y=5 and 2x+3y=10

Now, x- 2y=5

= x = 5+2y

When y=0 then, x= 5

When y=-2 then, x=1

Thus, we have the following table giving points on the line x-2y =5

X 5 -1

Y 0 -2

Now , 2x+3y=10 =x= 10−3y2

When y=0, then x=5

When y=2 , then x= 2Thus, we have the following table giving points on the line 2x+3y =10

X 5 2

Y 0 2

Graph of the equation x-2y =5 and 2x+3y =10

2

Clearly, two lines intersect at a point P (5,0)

Hence x= 5 and y = 0

Q3: 3x+y+1=0, 2x-3y+8=0

Sol:

We have, 3x+y+1=0 and 2x-3y+8=0

Now 3x+y+1=0

= y = -1-3x

When x=0 then, x= -1

When y=-1 then, x=2

Thus, we have the following table giving points on the line x-2y =5

X -1 0

Y 2 -1

Now, 2x-3y+8=0

=x= 3y−82

When y=0, then x=-4

When y=2 , then x= 1Thus, we have the following table giving points on the line 2x+3y =10

X -4 -1

Y 0 -2

Graph of the equation 3x+y+1=0 and 2x-3y+8=0

3

Clearly two lines intersect at a point P (-1, 2)

Hence x= -1 and y = 2

Q4: 2x+y-3 = 0, 2x-3y-7 =0

Sol:

We have, 2x+y-3 = 0 and 2x-3y-7 =0

Now 2x+y-3 = 0

= y = 3-2x

When x=0 then, x= 3

When x=1 then, x=1

Thus, we have the following table giving points on the line 2x+y-3 = 0

X 0 1

Y 3 1

Now, 2x-3y-7 =0

=y= 2x−73

When x=0, then y=1

When x=2, then y= -1Thus, we have the following table giving points on the line 2x+3y =10

X 2 5

Y -1 1

Graph of the equation 2x+y-3 = 0 and 2x-3y-7 =0