25 September, 13:52

# Y=x-4 y=x^2-4x solve solution by substitution

+3
1. 25 September, 14:01
0
y = x - 4

y = x² - 4x

Substitute the first equation into the second equation:

y = x² - 4x

x - 4 = x² - 4x [since y = x - 4, you can substitute (x - 4) for y] Add 4 on both sides

x = x² - 4x + 4 Subtract x on both sides to get the equation equal to 0

0 = x² - 5x + 4 Now you need to factor the equation. Find the factors of 4 that add or subtract to - 5

Factors of 4: 1, 2, 4 [1 * 4, 2 * 2]

1 and 4 add up to 5

0 = x² - 5x + 4

You can replace it with:

0 = x² - 1x - 4x + 4 [the sum of - 1x and - 4x is - 5x] Now factor the equation separately. Factor out x from (x² - 1x), and factor out - 4 from (-4x + 4)

0 = x (x - 1) - 4 (x - 1) Factor out (x - 1)

0 = (x - 1) (x - 4) Now set (x-1) and (x-4) equal to 0

x - 1 = 0

x = 1

x - 4 = 0

x = 4

x = 4, x = 1 Now that you found x, plug it into one of the equations (does not matter which equation) to find y

y = x - 4

y = 4 - 4

y = 0 (4, 0)

y = x - 4

y = 1 - 4

y = - 3 (1, - 3)