Solve the System of Equations 3x^2+y=10 3x-y=-4

$3 x^{2} + y = 10$ $3 x - y = - 4$

Subtract $3 x^{2}$ from both sides of the equation.

$y = 10 - 3 x^{2}$

$3 x - y = - 4$

Replace all occurrences of $y$ with $10 - 3 x^{2}$ in each equation.

Replace all occurrences of $y$ in $3 x - y = - 4$ with $10 - 3 x^{2}$ .

$3 x - (10 - 3 x^{2}) = - 4$

$y = 10 - 3 x^{2}$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$3 x - 1 \cdot 10 - (- 3 x^{2}) = - 4$

$y = 10 - 3 x^{2}$

Multiply $- 1$ by $10$ .

$3 x - 10 - (- 3 x^{2}) = - 4$

$y = 10 - 3 x^{2}$

Multiply $- 3$ by $- 1$ .

$3 x - 10 + 3 x^{2} = - 4$

$y = 10 - 3 x^{2}$

$3 x - 10 + 3 x^{2} = - 4$

$y = 10 - 3 x^{2}$

$3 x - 10 + 3 x^{2} = - 4$

$y = 10 - 3 x^{2}$

$3 x - 10 + 3 x^{2} = - 4$

$y = 10 - 3 x^{2}$

Solve for $x$ in $3 x - 10 + 3 x^{2} = - 4$ .

Add $4$ to both sides of the equation.

$3 x - 10 + 3 x^{2} + 4 = 0$

$y = 10 - 3 x^{2}$

Add $- 10$ and $4$ .

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

$y = 10 - 3 x^{2}$

Factor the left side of the equation.

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

Factor $3$ out of $3 x$ .

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

$y = 10 - 3 x^{2}$

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

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

$y = 10 - 3 x^{2}$

Factor $3$ out of $- 6$ .

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

$y = 10 - 3 x^{2}$

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

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

$y = 10 - 3 x^{2}$

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

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

$y = 10 - 3 x^{2}$

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

$y = 10 - 3 x^{2}$

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

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

$y = 10 - 3 x^{2}$

Factor $u + u^{2} - 2$ 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 $1$ .

$- 1, 2$

$y = 10 - 3 x^{2}$

Write the factored form using these integers.

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

$y = 10 - 3 x^{2}$

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

$y = 10 - 3 x^{2}$

Factor.

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

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

$y = 10 - 3 x^{2}$

Remove unnecessary parentheses.

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

$y = 10 - 3 x^{2}$

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

$y = 10 - 3 x^{2}$

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

$y = 10 - 3 x^{2}$

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

$x - 1 = 0$

$x + 2 = 0$

$y = 10 - 3 x^{2}$

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

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

$x - 1 = 0$

$y = 10 - 3 x^{2}$

Add $1$ to both sides of the equation.

$x = 1$

$y = 10 - 3 x^{2}$

$x = 1$

$y = 10 - 3 x^{2}$

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

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

$x + 2 = 0$

$y = 10 - 3 x^{2}$

Subtract $2$ from both sides of the equation.

$x = - 2$

$y = 10 - 3 x^{2}$

$x = - 2$

$y = 10 - 3 x^{2}$

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

$x = 1, - 2$

$y = 10 - 3 x^{2}$

$x = 1, - 2$

$y = 10 - 3 x^{2}$

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

Replace all occurrences of $x$ in $y = 10 - 3 x^{2}$ with $1$ .

$y = 10 - 3 {(1)}^{2}$

$x = 1$

Simplify the right side.

Simplify $10 - 3 {(1)}^{2}$ .

Simplify each term.

One to any power is one.

$y = 10 - 3 \cdot 1$

$x = 1$

Multiply $- 3$ by $1$ .

$y = 10 - 3$

$x = 1$

$y = 10 - 3$

$x = 1$

Subtract $3$ from $10$ .

$y = 7$

$x = 1$

$y = 7$

$x = 1$

$y = 7$

$x = 1$

$y = 7$

$x = 1$

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

Replace all occurrences of $x$ in $y = 10 - 3 x^{2}$ with $- 2$ .

$y = 10 - 3 {(- 2)}^{2}$

$x = - 2$

Simplify the right side.

Simplify $10 - 3 {(- 2)}^{2}$ .

Simplify each term.

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

$y = 10 - 3 \cdot 4$

$x = - 2$

Multiply $- 3$ by $4$ .

$y = 10 - 12$

$x = - 2$

$y = 10 - 12$

$x = - 2$

Subtract $12$ from $10$ .

$y = - 2$

$x = - 2$

$y = - 2$

$x = - 2$

$y = - 2$

$x = - 2$

$y = - 2$

$x = - 2$

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

$(1, 7)$

$(- 2, - 2)$

The result can be shown in multiple forms.

Point Form:

$(1, 7), (- 2, - 2)$

Equation Form:

$x = 1, y = 7$

$x = - 2, y = - 2$

Solve the System of Equations 3x^2+y=10 3x-y=-4

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