Question

Solve by substitution X^2= 2y + 10 3x-y = 9

Answer 1

Answer:

$x=-4\\ x=-2\\ y=-21\\ y=-15$

Step-by-step explanation:

$x^{2} =2y+10$

$y=-9+3x$

$x^{2} =2(-9+3x)+10$

$x^{2} =-18+6x+10$

$x^{2} =6x-8$

$x^{2} -6x+8$

$x=-4, -2$

$y=-9+3(-4)$

$y=-9-12$

$y=-21$

$y=-9+3(-2)$

$y=-9-6$

$y=-15$

Answer 2

Answer:

(4,3) (2,−3)

Step-by-step explanation:

x^2 =2(−9+3x)+10

2(−9+3x)+10

x2 = −18+6x+10

x2=6x−8

x2−6x+8=0

We now have to find integers that find product is 8 and whose sum is −6

= -4 -2 we find this is set to 4 2 as x−4=0

Add 4 to both sides of the equation. x=4

Set the next factor equal to 0

x−2=0 Add 2 to both sides of the equation.x=2

Confirms and proves x = 4,  x = 2

We solve for y

2y+10=(4)2

2y+10=16

2y=16−10

2y = 6

y = 3

We rewrite and solve for y again

(2)^2=2y+10

2y+10=(2)*2

2y+10=4  

2y = 4(-10)

2y = -6

y= -3