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Elias Carpenter
Mathematics
25 September, 13:52
Y=x-4 y=x^2-4x solve solution by substitution
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Gabriel Wilcox
25 September, 14:01
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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)
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