Solve the System of Equations y=3x^2+2 9x+y=-4

$y = 3 x^{2} + 2$ $9 x + y = - 4$

Replace all occurrences of $y$ with $3 x^{2} + 2$ in each equation.

Replace all occurrences of $y$ in $9 x + y = - 4$ with $3 x^{2} + 2$ .

$9 x + 3 x^{2} + 2 = - 4$

$y = 3 x^{2} + 2$

Simplify the left side.

Remove parentheses.

$9 x + 3 x^{2} + 2 = - 4$

$y = 3 x^{2} + 2$

$9 x + 3 x^{2} + 2 = - 4$

$y = 3 x^{2} + 2$

$9 x + 3 x^{2} + 2 = - 4$

$y = 3 x^{2} + 2$

Solve for $x$ in $9 x + 3 x^{2} + 2 = - 4$ .

Add $4$ to both sides of the equation.

$9 x + 3 x^{2} + 2 + 4 = 0$

$y = 3 x^{2} + 2$

Add $2$ and $4$ .

$9 x + 3 x^{2} + 6 = 0$

$y = 3 x^{2} + 2$

Factor the left side of the equation.

Factor $3$ out of $9 x + 3 x^{2} + 6$ .

Factor $3$ out of $9 x$ .

$3 (3 x) + 3 x^{2} + 6 = 0$

$y = 3 x^{2} + 2$

Factor $3$ out of $3 x^{2}$ .

$3 (3 x) + 3 (x^{2}) + 6 = 0$

$y = 3 x^{2} + 2$

Factor $3$ out of $6$ .

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

$y = 3 x^{2} + 2$

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

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

$y = 3 x^{2} + 2$

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

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

$y = 3 x^{2} + 2$

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

$y = 3 x^{2} + 2$

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

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

$y = 3 x^{2} + 2$

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

$1, 2$

$y = 3 x^{2} + 2$

Write the factored form using these integers.

$3 ((u + 1) (u + 2)) = 0$

$y = 3 x^{2} + 2$

$3 ((u + 1) (u + 2)) = 0$

$y = 3 x^{2} + 2$

Factor.

Replace all occurrences of $u$ with $x$ .

$3 ((x + 1) (x + 2)) = 0$

$y = 3 x^{2} + 2$

Remove unnecessary parentheses.

$3 (x + 1) (x + 2) = 0$

$y = 3 x^{2} + 2$

$3 (x + 1) (x + 2) = 0$

$y = 3 x^{2} + 2$

$3 (x + 1) (x + 2) = 0$

$y = 3 x^{2} + 2$

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

$x + 1 = 0$

$x + 2 = 0$

$y = 3 x^{2} + 2$

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

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

$x + 1 = 0$

$y = 3 x^{2} + 2$

Subtract $1$ from both sides of the equation.

$x = - 1$

$y = 3 x^{2} + 2$

$x = - 1$

$y = 3 x^{2} + 2$

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

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

$x + 2 = 0$

$y = 3 x^{2} + 2$

Subtract $2$ from both sides of the equation.

$x = - 2$

$y = 3 x^{2} + 2$

$x = - 2$

$y = 3 x^{2} + 2$

The final solution is all the values that make $3 (x + 1) (x + 2) = 0$ true.

$x = - 1, - 2$

$y = 3 x^{2} + 2$

$x = - 1, - 2$

$y = 3 x^{2} + 2$

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

Replace all occurrences of $x$ in $y = 3 x^{2} + 2$ with $- 1$ .

$y = 3 {(- 1)}^{2} + 2$

$x = - 1$

Simplify the right side.

Simplify $3 {(- 1)}^{2} + 2$ .

Simplify each term.

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

$y = 3 \cdot 1 + 2$

$x = - 1$

Multiply $3$ by $1$ .

$y = 3 + 2$

$x = - 1$

$y = 3 + 2$

$x = - 1$

Add $3$ and $2$ .

$y = 5$

$x = - 1$

$y = 5$

$x = - 1$

$y = 5$

$x = - 1$

$y = 5$

$x = - 1$

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

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

$y = 3 {(- 2)}^{2} + 2$

$x = - 2$

Simplify the right side.

Simplify $3 {(- 2)}^{2} + 2$ .

Simplify each term.

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

$y = 3 \cdot 4 + 2$

$x = - 2$

Multiply $3$ by $4$ .

$y = 12 + 2$

$x = - 2$

$y = 12 + 2$

$x = - 2$

Add $12$ and $2$ .

$y = 14$

$x = - 2$

$y = 14$

$x = - 2$

$y = 14$

$x = - 2$

$y = 14$

$x = - 2$

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

$(- 1, 5)$

$(- 2, 14)$

The result can be shown in multiple forms.

Point Form:

$(- 1, 5), (- 2, 14)$

Equation Form:

$x = - 1, y = 5$

$x = - 2, y = 14$

Solve the System of Equations y=3x^2+2 9x+y=-4

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