Question

23. Write Four solutions for each of the following equations
(ii) y= 2x +1​

Answer 1

Step-by-step explanation:

1. y = 2x +1
2. if x=0
3. y= 2×0+1
4. y= 1
5. (0,1)
6. if x=1
7. y= 2×1 +1
8. y=3
9. so (1,3)

Answer 2

Answer:

1)2x+y=9⟹y=9−2x

1)2x+y=9⟹y=9−2x x=1 then y=9−2(1)=7

1)2x+y=9⟹y=9−2x x=1 then y=9−2(1)=7 x=0 then y=9−2(0)=9

1)2x+y=9⟹y=9−2x x=1 then y=9−2(1)=7 x=0 then y=9−2(0)=9 x=2 then y=9−2(2)=5

1)2x+y=9⟹y=9−2x x=1 then y=9−2(1)=7 x=0 then y=9−2(0)=9 x=2 then y=9−2(2)=5 x=3 then y=9−2(3)=3

1)2x+y=9⟹y=9−2x x=1 then y=9−2(1)=7 x=0 then y=9−2(0)=9 x=2 then y=9−2(2)=5 x=3 then y=9−2(3)=3 (0,9),(1,7),(2,5),(3,3) are four solutions

$2) \: xy+y=9⟹y= \: \frac{9}{x + 1}$

$x=0 then y= \: \frac{9}{0 + 1} \: = \: 9$

$x=1  then y = \: \frac{9}{1 + 1} \: = \: 4.5$

$x=8  then y= \: \frac{9}{1 + 8} \: = \: 1$

(0,9),(1,4.5),(2,3),(8,1) are four solutions

3)x=4y

3)x=4y y=0 then x=0

3)x=4y y=0 then x=0 y=1 then x=4

3)x=4y y=0 then x=0 y=1 then x=4 y=2 then x=8

3)x=4y y=0 then x=0 y=1 then x=4 y=2 then x=8 y=3 then x=12

3)x=4y y=0 then x=0 y=1 then x=4 y=2 then x=8 y=3 then x=12(0,0),(4,1),(8,2),(12,3) are four solutions