Solve the System of Equations y=3x^2-3x-20 11=3x-y

$y = 3 x^{2} - 3 x - 20$ $11 = 3 x - y$

Step 1

Replace all occurrences of $y$ with $3 x^{2} - 3 x - 20$ in each equation.

Replace all occurrences of $y$ in $11 = 3 x - y$ with $3 x^{2} - 3 x - 20$ .

$11 = 3 x - (3 x^{2} - 3 x - 20)$

$y = 3 x^{2} - 3 x - 20$

Simplify the right side.

Simplify $3 x - (3 x^{2} - 3 x - 20)$ .

Simplify each term.

Apply the distributive property.

$11 = 3 x - (3 x^{2}) - (- 3 x) + 20$

$y = 3 x^{2} - 3 x - 20$

Simplify.

Multiply $3$ by $- 1$ .

$11 = 3 x - 3 x^{2} - (- 3 x) + 20$

$y = 3 x^{2} - 3 x - 20$

Multiply $- 3$ by $- 1$ .

$11 = 3 x - 3 x^{2} + 3 x + 20$

$y = 3 x^{2} - 3 x - 20$

Multiply $- 1$ by $- 20$ .

$11 = 3 x - 3 x^{2} + 3 x + 20$

$y = 3 x^{2} - 3 x - 20$

$11 = 3 x - 3 x^{2} + 3 x + 20$

$y = 3 x^{2} - 3 x - 20$

$11 = 3 x - 3 x^{2} + 3 x + 20$

$y = 3 x^{2} - 3 x - 20$

Add $3 x$ and $3 x$ .

$11 = - 3 x^{2} + 6 x + 20$

$y = 3 x^{2} - 3 x - 20$

$11 = - 3 x^{2} + 6 x + 20$

$y = 3 x^{2} - 3 x - 20$

$11 = - 3 x^{2} + 6 x + 20$

$y = 3 x^{2} - 3 x - 20$

$11 = - 3 x^{2} + 6 x + 20$

$y = 3 x^{2} - 3 x - 20$

Step 2

Solve for $x$ in $11 = - 3 x^{2} + 6 x + 20$ .

Rewrite the equation as $- 3 x^{2} + 6 x + 20 = 11$ .

$- 3 x^{2} + 6 x + 20 = 11$

$y = 3 x^{2} - 3 x - 20$

Subtract $11$ from both sides of the equation.

$- 3 x^{2} + 6 x + 20 - 11 = 0$

$y = 3 x^{2} - 3 x - 20$

Subtract $11$ from $20$ .

$- 3 x^{2} + 6 x + 9 = 0$

$y = 3 x^{2} - 3 x - 20$

Factor the left side of the equation.

Factor $- 3$ out of $- 3 x^{2} + 6 x + 9$ .

Factor $- 3$ out of $- 3 x^{2}$ .

$- 3 x^{2} + 6 x + 9 = 0$

$y = 3 x^{2} - 3 x - 20$

Factor $- 3$ out of $6 x$ .

$- 3 x^{2} - 3 (- 2 x) + 9 = 0$

$y = 3 x^{2} - 3 x - 20$

Factor $- 3$ out of $9$ .

$- 3 x^{2} - 3 (- 2 x) - 3 \cdot - 3 = 0$

$y = 3 x^{2} - 3 x - 20$

Factor $- 3$ out of $- 3 (x^{2}) - 3 (- 2 x)$ .

$- 3 (x^{2} - 2 x) - 3 \cdot - 3 = 0$

$y = 3 x^{2} - 3 x - 20$

Factor $- 3$ out of $- 3 (x^{2} - 2 x) - 3 (- 3)$ .

$- 3 (x^{2} - 2 x - 3) = 0$

$y = 3 x^{2} - 3 x - 20$

$- 3 (x^{2} - 2 x - 3) = 0$

$y = 3 x^{2} - 3 x - 20$

Factor.

Factor $x^{2} - 2 x - 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

$y = 3 x^{2} - 3 x - 20$

Write the factored form using these integers.

$- 3 ((x - 3) (x + 1)) = 0$

$y = 3 x^{2} - 3 x - 20$

$- 3 ((x - 3) (x + 1)) = 0$

$y = 3 x^{2} - 3 x - 20$

Remove unnecessary parentheses.

$- 3 (x - 3) (x + 1) = 0$

$y = 3 x^{2} - 3 x - 20$

$- 3 (x - 3) (x + 1) = 0$

$y = 3 x^{2} - 3 x - 20$

$- 3 (x - 3) (x + 1) = 0$

$y = 3 x^{2} - 3 x - 20$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x + 1 = 0$

$y = 3 x^{2} - 3 x - 20$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

$y = 3 x^{2} - 3 x - 20$

Add $3$ to both sides of the equation.

$x = 3$

$y = 3 x^{2} - 3 x - 20$

$x = 3$

$y = 3 x^{2} - 3 x - 20$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

$y = 3 x^{2} - 3 x - 20$

Subtract $1$ from both sides of the equation.

$x = - 1$

$y = 3 x^{2} - 3 x - 20$

$x = - 1$

$y = 3 x^{2} - 3 x - 20$

The final solution is all the values that make $- 3 (x - 3) (x + 1) = 0$ true.

$x = 3, - 1$

$y = 3 x^{2} - 3 x - 20$

$x = 3, - 1$

$y = 3 x^{2} - 3 x - 20$

Step 3

Replace all occurrences of $x$ with $3$ in each equation.

Replace all occurrences of $x$ in $y = 3 x^{2} - 3 x - 20$ with $3$ .

$y = 3 {(3)}^{2} - 3 \cdot 3 - 20$

$x = 3$

Simplify the right side.

Simplify $3 {(3)}^{2} - 3 \cdot 3 - 20$ .

Simplify each term.

Multiply $3$ by ${(3)}^{2}$ by adding the exponents.

Multiply $3$ by ${(3)}^{2}$ .

Raise $3$ to the power of $1$ .

$y = 3 {(3)}^{2} - 3 \cdot 3 - 20$

$x = 3$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = 3^{1 + 2} - 3 \cdot 3 - 20$

$x = 3$

$y = 3^{1 + 2} - 3 \cdot 3 - 20$

$x = 3$

Add $1$ and $2$ .

$y = 3^{3} - 3 \cdot 3 - 20$

$x = 3$

$y = 3^{3} - 3 \cdot 3 - 20$

$x = 3$

Raise $3$ to the power of $3$ .

$y = 27 - 3 \cdot 3 - 20$

$x = 3$

Multiply $- 3$ by $3$ .

$y = 27 - 9 - 20$

$x = 3$

$y = 27 - 9 - 20$

$x = 3$

Simplify by subtracting numbers.

Subtract $9$ from $27$ .

$y = 18 - 20$

$x = 3$

Subtract $20$ from $18$ .

$y = - 2$

$x = 3$

$y = - 2$

$x = 3$

$y = - 2$

$x = 3$

$y = - 2$

$x = 3$

$y = - 2$

$x = 3$

Step 4

Replace all occurrences of $x$ with $- 1$ in each equation.

Replace all occurrences of $x$ in $y = 3 x^{2} - 3 x - 20$ with $- 1$ .

$y = 3 {(- 1)}^{2} - 3 \cdot - 1 - 20$

$x = - 1$

Simplify the right side.

Simplify $3 {(- 1)}^{2} - 3 \cdot - 1 - 20$ .

Simplify each term.

Raise $- 1$ to the power of $2$ .

$y = 3 \cdot 1 - 3 \cdot - 1 - 20$

$x = - 1$

Multiply $3$ by $1$ .

$y = 3 - 3 \cdot - 1 - 20$

$x = - 1$

Multiply $- 3$ by $- 1$ .

$y = 3 + 3 - 20$

$x = - 1$

$y = 3 + 3 - 20$

$x = - 1$

Simplify by adding and subtracting.

Add $3$ and $3$ .

$y = 6 - 20$

$x = - 1$

Subtract $20$ from $6$ .

$y = - 14$

$x = - 1$

$y = - 14$

$x = - 1$

$y = - 14$

$x = - 1$

$y = - 14$

$x = - 1$

$y = - 14$

$x = - 1$

Step 5

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(3, - 2)$

$(- 1, - 14)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(3, - 2), (- 1, - 14)$

Equation Form:

$x = 3, y = - 2$

$x = - 1, y = - 14$

Step 7

Solve the System of Equations y=3x^2-3x-20 11=3x-y

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