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

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

Step 1

Eliminate the equal sides of each equation and combine.

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

Step 2

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

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

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

Add $4$ to both sides of the equation.

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

Factor the left side of the equation.

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

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

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

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

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

Factor $- 2$ out of $2 x$ .

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

Factor $- 2$ out of $4$ .

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

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

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

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

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

Factor.

Factor $x^{2} - x - 2$ 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 $- 2$ and whose sum is $- 1$ .

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

Write the factored form using these integers.

$- 2 ((x - 2) (x + 1)) = 0$

Remove unnecessary parentheses.

$- 2 (x - 2) (x + 1) = 0$

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

$x - 2 = 0$

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

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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 $- 2 (x - 2) (x + 1) = 0$ true.

$x = 2, - 1$

Step 3

Evaluate $y$ when $x = 2$ .

Substitute $2$ for $x$ .

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

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

Remove parentheses.

$y = 2 \cdot 2^{2} - 4$

Simplify $2 \cdot 2^{2} - 4$ .

Simplify each term.

Multiply $2$ by $2^{2}$ by adding the exponents.

Multiply $2$ by $2^{2}$ .

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

$y = 2^{1} \cdot 2^{2} - 4$

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

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

Add $1$ and $2$ .

$y = 2^{3} - 4$

Raise $2$ to the power of $3$ .

$y = 8 - 4$

Subtract $4$ from $8$ .

$y = 4$

Step 4

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

Substitute $- 1$ for $x$ .

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

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

Simplify each term.

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

$y = 2 \cdot 1 - 4$

Multiply $2$ by $1$ .

$y = 2 - 4$

Subtract $4$ from $2$ .

$y = - 2$

Step 5

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

$(2, 4)$

$(- 1, - 2)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(2, 4), (- 1, - 2)$

Equation Form:

$x = 2, y = 4$

$x = - 1, y = - 2$

Step 7

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

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