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
Answers
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