Solve the System of Equations 10x+y=-20 y=2x^2-4x-16

$10 x + y = - 20$ $y = 2 x^{2} - 4 x - 16$

Replace all occurrences of $y$ with $2 x^{2} - 4 x - 16$ in each equation.

Replace all occurrences of $y$ in $10 x + y = - 20$ with $2 x^{2} - 4 x - 16$ .

$10 x + 2 x^{2} - 4 x - 16 = - 20$

$y = 2 x^{2} - 4 x - 16$

Simplify the left side.

Simplify $10 x + 2 x^{2} - 4 x - 16$ .

Remove parentheses.

$10 x + 2 x^{2} - 4 x - 16 = - 20$

$y = 2 x^{2} - 4 x - 16$

Subtract $4 x$ from $10 x$ .

$2 x^{2} + 6 x - 16 = - 20$

$y = 2 x^{2} - 4 x - 16$

$2 x^{2} + 6 x - 16 = - 20$

$y = 2 x^{2} - 4 x - 16$

$2 x^{2} + 6 x - 16 = - 20$

$y = 2 x^{2} - 4 x - 16$

$2 x^{2} + 6 x - 16 = - 20$

$y = 2 x^{2} - 4 x - 16$

Solve for $x$ in $2 x^{2} + 6 x - 16 = - 20$ .

Add $20$ to both sides of the equation.

$2 x^{2} + 6 x - 16 + 20 = 0$

$y = 2 x^{2} - 4 x - 16$

Add $- 16$ and $20$ .

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

$y = 2 x^{2} - 4 x - 16$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} + 6 x + 4$ .

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

$2 (x^{2}) + 6 x + 4 = 0$

$y = 2 x^{2} - 4 x - 16$

Factor $2$ out of $6 x$ .

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

$y = 2 x^{2} - 4 x - 16$

Factor $2$ out of $4$ .

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

$y = 2 x^{2} - 4 x - 16$

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

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

$y = 2 x^{2} - 4 x - 16$

Factor $2$ out of $2 (x^{2} + 3 x) + 2 \cdot 2$ .

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

$y = 2 x^{2} - 4 x - 16$

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

$y = 2 x^{2} - 4 x - 16$

Factor.

Factor $x^{2} + 3 x + 2$ 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 $3$ .

$1, 2$

$y = 2 x^{2} - 4 x - 16$

Write the factored form using these integers.

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

$y = 2 x^{2} - 4 x - 16$

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

$y = 2 x^{2} - 4 x - 16$

Remove unnecessary parentheses.

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

$y = 2 x^{2} - 4 x - 16$

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

$y = 2 x^{2} - 4 x - 16$

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

$y = 2 x^{2} - 4 x - 16$

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

$x + 1 = 0$

$x + 2 = 0$

$y = 2 x^{2} - 4 x - 16$

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

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

$x + 1 = 0$

$y = 2 x^{2} - 4 x - 16$

Subtract $1$ from both sides of the equation.

$x = - 1$

$y = 2 x^{2} - 4 x - 16$

$x = - 1$

$y = 2 x^{2} - 4 x - 16$

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

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

$x + 2 = 0$

$y = 2 x^{2} - 4 x - 16$

Subtract $2$ from both sides of the equation.

$x = - 2$

$y = 2 x^{2} - 4 x - 16$

$x = - 2$

$y = 2 x^{2} - 4 x - 16$

The final solution is all the values that make $2 (x + 1) (x + 2) = 0$ true.

$x = - 1, - 2$

$y = 2 x^{2} - 4 x - 16$

$x = - 1, - 2$

$y = 2 x^{2} - 4 x - 16$

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

Replace all occurrences of $x$ in $y = 2 x^{2} - 4 x - 16$ with $- 1$ .

$y = 2 {(- 1)}^{2} - 4 \cdot - 1 - 16$

$x = - 1$

Simplify the right side.

Simplify $2 {(- 1)}^{2} - 4 \cdot - 1 - 16$ .

Simplify each term.

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

$y = 2 \cdot 1 - 4 \cdot - 1 - 16$

$x = - 1$

Multiply $2$ by $1$ .

$y = 2 - 4 \cdot - 1 - 16$

$x = - 1$

Multiply $- 4$ by $- 1$ .

$y = 2 + 4 - 16$

$x = - 1$

$y = 2 + 4 - 16$

$x = - 1$

Simplify by adding and subtracting.

Add $2$ and $4$ .

$y = 6 - 16$

$x = - 1$

Subtract $16$ from $6$ .

$y = - 10$

$x = - 1$

$y = - 10$

$x = - 1$

$y = - 10$

$x = - 1$

$y = - 10$

$x = - 1$

$y = - 10$

$x = - 1$

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

Replace all occurrences of $x$ in $y = 2 x^{2} - 4 x - 16$ with $- 2$ .

$y = 2 {(- 2)}^{2} - 4 \cdot - 2 - 16$

$x = - 2$

Simplify the right side.

Simplify $2 {(- 2)}^{2} - 4 \cdot - 2 - 16$ .

Simplify each term.

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

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

$x = - 2$

Multiply $2$ by $4$ .

$y = 8 - 4 \cdot - 2 - 16$

$x = - 2$

Multiply $- 4$ by $- 2$ .

$y = 8 + 8 - 16$

$x = - 2$

$y = 8 + 8 - 16$

$x = - 2$

Simplify by adding and subtracting.

Add $8$ and $8$ .

$y = 16 - 16$

$x = - 2$

Subtract $16$ from $16$ .

$y = 0$

$x = - 2$

$y = 0$

$x = - 2$

$y = 0$

$x = - 2$

$y = 0$

$x = - 2$

$y = 0$

$x = - 2$

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

$(- 1, - 10)$

$(- 2, 0)$

The result can be shown in multiple forms.

Point Form:

$(- 1, - 10), (- 2, 0)$

Equation Form:

$x = - 1, y = - 10$

$x = - 2, y = 0$

Solve the System of Equations 10x+y=-20 y=2x^2-4x-16

Related Questions

Support LCMGCF.com by disabling your adblocker.