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

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

Add $3$ to both sides of the equation.

$y = 2 x + 3$

$y^{2} = x^{2} + 9$

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

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

${(2 x + 3)}^{2} = x^{2} + 9$

$y = 2 x + 3$

Simplify the left side.

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

Rewrite ${(2 x + 3)}^{2}$ as $(2 x + 3) (2 x + 3)$ .

$(2 x + 3) (2 x + 3) = x^{2} + 9$

$y = 2 x + 3$

Expand $(2 x + 3) (2 x + 3)$ using the FOIL Method.

Apply the distributive property.

$2 x (2 x + 3) + 3 (2 x + 3) = x^{2} + 9$

$y = 2 x + 3$

Apply the distributive property.

$2 x (2 x) + 2 x \cdot 3 + 3 (2 x + 3) = x^{2} + 9$

$y = 2 x + 3$

Apply the distributive property.

$2 x (2 x) + 2 x \cdot 3 + 3 (2 x) + 3 \cdot 3 = x^{2} + 9$

$y = 2 x + 3$

$2 x (2 x) + 2 x \cdot 3 + 3 (2 x) + 3 \cdot 3 = x^{2} + 9$

$y = 2 x + 3$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$2 \cdot (2 x \cdot x) + 2 x \cdot 3 + 3 (2 x) + 3 \cdot 3 = x^{2} + 9$

$y = 2 x + 3$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$2 \cdot (2 (x \cdot x)) + 2 x \cdot 3 + 3 (2 x) + 3 \cdot 3 = x^{2} + 9$

$y = 2 x + 3$

Multiply $x$ by $x$ .

$2 \cdot (2 x^{2}) + 2 x \cdot 3 + 3 (2 x) + 3 \cdot 3 = x^{2} + 9$

$y = 2 x + 3$

$2 \cdot (2 x^{2}) + 2 x \cdot 3 + 3 (2 x) + 3 \cdot 3 = x^{2} + 9$

$y = 2 x + 3$

Multiply $2$ by $2$ .

$4 x^{2} + 2 x \cdot 3 + 3 (2 x) + 3 \cdot 3 = x^{2} + 9$

$y = 2 x + 3$

Multiply $3$ by $2$ .

$4 x^{2} + 6 x + 3 (2 x) + 3 \cdot 3 = x^{2} + 9$

$y = 2 x + 3$

Multiply $2$ by $3$ .

$4 x^{2} + 6 x + 6 x + 3 \cdot 3 = x^{2} + 9$

$y = 2 x + 3$

Multiply $3$ by $3$ .

$4 x^{2} + 6 x + 6 x + 9 = x^{2} + 9$

$y = 2 x + 3$

$4 x^{2} + 6 x + 6 x + 9 = x^{2} + 9$

$y = 2 x + 3$

Add $6 x$ and $6 x$ .

$4 x^{2} + 12 x + 9 = x^{2} + 9$

$y = 2 x + 3$

$4 x^{2} + 12 x + 9 = x^{2} + 9$

$y = 2 x + 3$

$4 x^{2} + 12 x + 9 = x^{2} + 9$

$y = 2 x + 3$

$4 x^{2} + 12 x + 9 = x^{2} + 9$

$y = 2 x + 3$

$4 x^{2} + 12 x + 9 = x^{2} + 9$

$y = 2 x + 3$

Solve for $x$ in $4 x^{2} + 12 x + 9 = x^{2} + 9$ .

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

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

$4 x^{2} + 12 x + 9 - x^{2} = 9$

$y = 2 x + 3$

Subtract $x^{2}$ from $4 x^{2}$ .

$3 x^{2} + 12 x + 9 = 9$

$y = 2 x + 3$

$3 x^{2} + 12 x + 9 = 9$

$y = 2 x + 3$

Subtract $9$ from both sides of the equation.

$3 x^{2} + 12 x + 9 - 9 = 0$

$y = 2 x + 3$

Combine the opposite terms in $3 x^{2} + 12 x + 9 - 9$ .

Subtract $9$ from $9$ .

$3 x^{2} + 12 x + 0 = 0$

$y = 2 x + 3$

Add $3 x^{2} + 12 x$ and $0$ .

$3 x^{2} + 12 x = 0$

$y = 2 x + 3$

$3 x^{2} + 12 x = 0$

$y = 2 x + 3$

Factor $3 x$ out of $3 x^{2} + 12 x$ .

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

$3 x (x) + 12 x = 0$

$y = 2 x + 3$

Factor $3 x$ out of $12 x$ .

$3 x (x) + 3 x (4) = 0$

$y = 2 x + 3$

Factor $3 x$ out of $3 x (x) + 3 x (4)$ .

$3 x (x + 4) = 0$

$y = 2 x + 3$

$3 x (x + 4) = 0$

$y = 2 x + 3$

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

$x = 0$

$x + 4 = 0$

$y = 2 x + 3$

Set $x$ equal to $0$ .

$x = 0$

$y = 2 x + 3$

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

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

$x + 4 = 0$

$y = 2 x + 3$

Subtract $4$ from both sides of the equation.

$x = - 4$

$y = 2 x + 3$

$x = - 4$

$y = 2 x + 3$

The final solution is all the values that make $3 x (x + 4) = 0$ true.

$x = 0, - 4$

$y = 2 x + 3$

$x = 0, - 4$

$y = 2 x + 3$

Replace all occurrences of $x$ with $0$ in each equation.

Replace all occurrences of $x$ in $y = 2 x + 3$ with $0$ .

$y = 2 (0) + 3$

$x = 0$

Simplify the right side.

Simplify $2 (0) + 3$ .

Multiply $2$ by $0$ .

$y = 0 + 3$

$x = 0$

Add $0$ and $3$ .

$y = 3$

$x = 0$

$y = 3$

$x = 0$

$y = 3$

$x = 0$

$y = 3$

$x = 0$

Replace all occurrences of $x$ with $- 4$ in each equation.

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

$y = 2 (- 4) + 3$

$x = - 4$

Simplify the right side.

Simplify $2 (- 4) + 3$ .

Multiply $2$ by $- 4$ .

$y = - 8 + 3$

$x = - 4$

Add $- 8$ and $3$ .

$y = - 5$

$x = - 4$

$y = - 5$

$x = - 4$

$y = - 5$

$x = - 4$

$y = - 5$

$x = - 4$

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

$(0, 3)$

$(- 4, - 5)$

The result can be shown in multiple forms.

Point Form:

$(0, 3), (- 4, - 5)$

Equation Form:

$x = 0, y = 3$

$x = - 4, y = - 5$

Related Questions