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

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

So instead of -5x:

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)