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Solve the system of equations: y= x^2 + 2x - 3 and y=3x+ 3.

User Alalonde
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1 Answer

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8 votes

Answer:

(-2, -3)

(3, 12)

Explanation:

To solve this, we're gonna get rid of the y's with substitution

x² + 2x - 3 = 3x + 3

Let's make this equation equal to zero

Subtract 3 from both sides

x² + 2x - 3 = 3x + 3

- 3 - 3

x² + 2x - 6 = 3x

Subtract 3x from both sides

x² + 2x - 6 = 3x

- 3x - 3x

x² - x - 6 = 0

Factor the equation

(x - 3)(x + 2) = 0

This means x can be -2 or 3

Let's solve it with -2 first, plug the new x in y = 3x + 3

y = 3(-2) + 3

y = -6 + 3 = -3

Do the same for x = 3

y = 3(3) + 3

y = 9 + 3 = 12

User DusteD
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