Use substitution to solve the system

Question

y=2x^2-6x-8

y=2x-y=16

Help!!

Answer 1

1. y = 2x² - 6x - 8 equation 1
2. 2x - y = 16 equation 2

From the second equation; we need to isolate y :

2x-y=16

subtract 2x on both sides:

1. 2x-2x-y=16-2x
2. -y=16-2x
3. divide both sides by -1

y=2x-16

We will substitute y by 2x-16 on our first equation to find x

2x-16= 2x² - 6x - 8, set your equation equal to 0

2x² - 6x - 8-2x+16=0

2x² - 8x +8=0 factor 2 from the equation

2(x² - 4x + 4)=0

You can either use the box method , factoring or quadratic equation to solve for x

2(x-2)²₌₀

_x=2

_{then solve for y}

_{y=2(2)-16= 4-16=-12}

_{(x,y)= (2,-12)}

_{you always check your answers if its correct by substituting the value of x and y on both equations to check if the statement is correct.}

Answer 2

y = 2x² - 6x - 8 and 2x - y = 16

From the second equation, we have y = 2x - 16.

Replace y by 2x - 16 in the first equation to get 2x - 16 = 2x² - 6x - 8.

So, 2x² - 8x + 8 = 0

Divide by 2 to obtain x² - 4x + 4 = 0

(x - 2)² = 0

Therefore, x = 2 and y = 2(2) - 16 = -12.