Solve the System of Equations y=2x^2-5x -5x+y=-12

$y = 2 x^{2} - 5 x$ $- 5 x + y = - 12$

Replace all occurrences of $y$ with $2 x^{2} - 5 x$ in each equation.

Replace all occurrences of $y$ in $- 5 x + y = - 12$ with $2 x^{2} - 5 x$ .

$- 5 x + 2 x^{2} - 5 x = - 12$

$y = 2 x^{2} - 5 x$

Simplify the left side.

Simplify $- 5 x + 2 x^{2} - 5 x$ .

Remove parentheses.

$- 5 x + 2 x^{2} - 5 x = - 12$

$y = 2 x^{2} - 5 x$

Subtract $5 x$ from $- 5 x$ .

$2 x^{2} - 10 x = - 12$

$y = 2 x^{2} - 5 x$

$2 x^{2} - 10 x = - 12$

$y = 2 x^{2} - 5 x$

$2 x^{2} - 10 x = - 12$

$y = 2 x^{2} - 5 x$

$2 x^{2} - 10 x = - 12$

$y = 2 x^{2} - 5 x$

Solve for $x$ in $2 x^{2} - 10 x = - 12$ .

Add $12$ to both sides of the equation.

$2 x^{2} - 10 x + 12 = 0$

$y = 2 x^{2} - 5 x$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 10 x + 12$ .

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

$2 (x^{2}) - 10 x + 12 = 0$

$y = 2 x^{2} - 5 x$

Factor $2$ out of $- 10 x$ .

$2 (x^{2}) + 2 (- 5 x) + 12 = 0$

$y = 2 x^{2} - 5 x$

Factor $2$ out of $12$ .

$2 x^{2} + 2 (- 5 x) + 2 \cdot 6 = 0$

$y = 2 x^{2} - 5 x$

Factor $2$ out of $2 x^{2} + 2 (- 5 x)$ .

$2 (x^{2} - 5 x) + 2 \cdot 6 = 0$

$y = 2 x^{2} - 5 x$

Factor $2$ out of $2 (x^{2} - 5 x) + 2 \cdot 6$ .

$2 (x^{2} - 5 x + 6) = 0$

$y = 2 x^{2} - 5 x$

$2 (x^{2} - 5 x + 6) = 0$

$y = 2 x^{2} - 5 x$

Factor.

Factor $x^{2} - 5 x + 6$ 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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

$y = 2 x^{2} - 5 x$

Write the factored form using these integers.

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

$y = 2 x^{2} - 5 x$

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

$y = 2 x^{2} - 5 x$

Remove unnecessary parentheses.

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

$y = 2 x^{2} - 5 x$

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

$y = 2 x^{2} - 5 x$

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

$y = 2 x^{2} - 5 x$

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

$x - 3 = 0$

$x - 2 = 0$

$y = 2 x^{2} - 5 x$

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

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

$x - 3 = 0$

$y = 2 x^{2} - 5 x$

Add $3$ to both sides of the equation.

$x = 3$

$y = 2 x^{2} - 5 x$

$x = 3$

$y = 2 x^{2} - 5 x$

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

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

$x - 2 = 0$

$y = 2 x^{2} - 5 x$

Add $2$ to both sides of the equation.

$x = 2$

$y = 2 x^{2} - 5 x$

$x = 2$

$y = 2 x^{2} - 5 x$

The final solution is all the values that make $2 (x - 3) (x - 2) = 0$ true.

$x = 3, 2$

$y = 2 x^{2} - 5 x$

$x = 3, 2$

$y = 2 x^{2} - 5 x$

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

Replace all occurrences of $x$ in $y = 2 x^{2} - 5 x$ with $3$ .

$y = 2 {(3)}^{2} - 5 \cdot 3$

$x = 3$

Simplify the right side.

Simplify $2 {(3)}^{2} - 5 \cdot 3$ .

Simplify each term.

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

$y = 2 \cdot 9 - 5 \cdot 3$

$x = 3$

Multiply $2$ by $9$ .

$y = 18 - 5 \cdot 3$

$x = 3$

Multiply $- 5$ by $3$ .

$y = 18 - 15$

$x = 3$

$y = 18 - 15$

$x = 3$

Subtract $15$ from $18$ .

$y = 3$

$x = 3$

$y = 3$

$x = 3$

$y = 3$

$x = 3$

$y = 3$

$x = 3$

Replace all occurrences of $x$ with $2$ in each equation.

Replace all occurrences of $x$ in $y = 2 x^{2} - 5 x$ with $2$ .

$y = 2 {(2)}^{2} - 5 \cdot 2$

$x = 2$

Simplify the right side.

Simplify $2 {(2)}^{2} - 5 \cdot 2$ .

Simplify each term.

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

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

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

$y = 2 {(2)}^{2} - 5 \cdot 2$

$x = 2$

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

$y = 2^{1 + 2} - 5 \cdot 2$

$x = 2$

$y = 2^{1 + 2} - 5 \cdot 2$

$x = 2$

Add $1$ and $2$ .

$y = 2^{3} - 5 \cdot 2$

$x = 2$

$y = 2^{3} - 5 \cdot 2$

$x = 2$

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

$y = 8 - 5 \cdot 2$

$x = 2$

Multiply $- 5$ by $2$ .

$y = 8 - 10$

$x = 2$

$y = 8 - 10$

$x = 2$

Subtract $10$ from $8$ .

$y = - 2$

$x = 2$

$y = - 2$

$x = 2$

$y = - 2$

$x = 2$

$y = - 2$

$x = 2$

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

$(3, 3)$

$(2, - 2)$

The result can be shown in multiple forms.

Point Form:

$(3, 3), (2, - 2)$

Equation Form:

$x = 3, y = 3$

$x = 2, y = - 2$

Solve the System of Equations y=2x^2-5x -5x+y=-12

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