Solve the System of Equations y=3x y=x^2-4

$y = 3 x$ $y = x^{2} - 4$

Step 1

Eliminate the equal sides of each equation and combine.

$3 x = x^{2} - 4$

Step 2

Solve $3 x = x^{2} - 4$ for $x$ .

Subtract $x^{2}$ from both sides of the equation.

$3 x - x^{2} = - 4$

Add $4$ to both sides of the equation.

$3 x - x^{2} + 4 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $3 x - x^{2} + 4$ .

Reorder $3 x$ and $- x^{2}$ .

$- x^{2} + 3 x + 4 = 0$

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 3 x + 4 = 0$

Factor $- 1$ out of $3 x$ .

$- (x^{2}) - (- 3 x) + 4 = 0$

Rewrite $4$ as $- 1 (- 4)$ .

$- (x^{2}) - (- 3 x) - 1 \cdot - 4 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 3 x)$ .

$- (x^{2} - 3 x) - 1 \cdot - 4 = 0$

Factor $- 1$ out of $- (x^{2} - 3 x) - 1 (- 4)$ .

$- (x^{2} - 3 x - 4) = 0$

Factor.

Factor $x^{2} - 3 x - 4$ 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 $- 4$ and whose sum is $- 3$ .

$- 4, 1 + y = x^{2} - 4$

Write the factored form using these integers.

$- ((x - 4) (x + 1)) = 0$

Remove unnecessary parentheses.

$- (x - 4) (x + 1) = 0$

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

$x - 4 = 0$

$x + 1 = 0 + y = x^{2} - 4$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $- (x - 4) (x + 1) = 0$ true.

$x = 4, - 1$

Step 3

Evaluate $y$ when $x = 4$ .

Substitute $4$ for $x$ .

$y = {(4)}^{2} - 4$

Substitute $4$ for $x$ in $y = {(4)}^{2} - 4$ and solve for $y$ .

Remove parentheses.

$y = 4^{2} - 4$

Simplify $4^{2} - 4$ .

Raise $4$ to the power of $2$ .

$y = 16 - 4$

Subtract $4$ from $16$ .

$y = 12$

Step 4

Evaluate $y$ when $x = - 1$ .

Substitute $- 1$ for $x$ .

$y = {(- 1)}^{2} - 4$

Simplify ${(- 1)}^{2} - 4$ .

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

$y = 1 - 4$

Subtract $4$ from $1$ .

$y = - 3$

Step 5

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

$(4, 12)$

$(- 1, - 3)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 12), (- 1, - 3)$

Equation Form:

$x = 4, y = 12$

$x = - 1, y = - 3$

Step 7

Solve the System of Equations y=3x y=x^2-4

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