Solve the System of Equations y=x^2-x+2 y=2x^2+x-6

$y = x^{2} - x + 2$ $y = 2 x^{2} + x - 6$

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} - x + 2 = 2 x^{2} + x - 6$

Step 2

Solve $x^{2} - x + 2 = 2 x^{2} + x - 6$ for $x$ .

Move all terms containing $x$ to the left side of the equation.

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

$x^{2} - x + 2 - 2 x^{2} = x - 6$

Subtract $x$ from both sides of the equation.

$x^{2} - x + 2 - 2 x^{2} - x = - 6$

Subtract $2 x^{2}$ from $x^{2}$ .

$- x^{2} - x + 2 - x = - 6$

Subtract $x$ from $- x$ .

$- x^{2} - 2 x + 2 = - 6$

Add $6$ to both sides of the equation.

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

Add $2$ and $6$ .

$- x^{2} - 2 x + 8 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} - 2 x + 8$ .

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

$- (x^{2}) - 2 x + 8 = 0$

Factor $- 1$ out of $- 2 x$ .

$- (x^{2}) - (2 x) + 8 = 0$

Rewrite $8$ as $- 1 (- 8)$ .

$- (x^{2}) - (2 x) - 1 \cdot - 8 = 0$

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

$- (x^{2} + 2 x) - 1 \cdot - 8 = 0$

Factor $- 1$ out of $- (x^{2} + 2 x) - 1 (- 8)$ .

$- (x^{2} + 2 x - 8) = 0$

Factor.

Factor $x^{2} + 2 x - 8$ 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 $- 8$ and whose sum is $2$ .

$- 2, 4 + y = 2 x^{2} + x - 6$

Write the factored form using these integers.

$- ((x - 2) (x + 4)) = 0$

Remove unnecessary parentheses.

$- (x - 2) (x + 4) = 0$

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

$x - 2 = 0$

$x + 4 = 0 + y = 2 x^{2} + x - 6$

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 $x + 4$ equal to $0$ and solve for $x$ .

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $- (x - 2) (x + 4) = 0$ true.

$x = 2, - 4$

Step 3

Evaluate $y$ when $x = 2$ .

Substitute $2$ for $x$ .

$y = 2 {(2)}^{2} + 2 - 6$

Substitute $2$ for $x$ in $y = 2 {(2)}^{2} + 2 - 6$ and solve for $y$ .

Remove parentheses.

$y = 2 {(2)}^{2} + 2 - 6$

Remove parentheses.

$y = 2 \cdot 2^{2} + 2 - 6$

Remove parentheses.

$y = 2 {(2)}^{2} + 2 - 6$

Simplify $2 {(2)}^{2} + 2 - 6$ .

Simplify each term.

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

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

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

$y = 2^{1} {(2)}^{2} + 2 - 6$

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

$y = 2^{1 + 2} + 2 - 6$

Add $1$ and $2$ .

$y = 2^{3} + 2 - 6$

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

$y = 8 + 2 - 6$

Simplify by adding and subtracting.

Add $8$ and $2$ .

$y = 10 - 6$

Subtract $6$ from $10$ .

$y = 4$

Step 4

Evaluate $y$ when $x = - 4$ .

Substitute $- 4$ for $x$ .

$y = 2 {(- 4)}^{2} - 4 - 6$

Substitute $- 4$ for $x$ in $y = 2 {(- 4)}^{2} - 4 - 6$ and solve for $y$ .

Remove parentheses.

$y = 2 {(- 4)}^{2} - 4 - 6$

Simplify $2 {(- 4)}^{2} - 4 - 6$ .

Simplify each term.

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

$y = 2 \cdot 16 - 4 - 6$

Multiply $2$ by $16$ .

$y = 32 - 4 - 6$

Simplify by subtracting numbers.

Subtract $4$ from $32$ .

$y = 28 - 6$

Subtract $6$ from $28$ .

$y = 22$

Step 5

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

$(2, 4)$

$(- 4, 22)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(2, 4), (- 4, 22)$

Equation Form:

$x = 2, y = 4$

$x = - 4, y = 22$

Step 7

Solve the System of Equations y=x^2-x+2 y=2x^2+x-6

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