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

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

Step 1

Rewrite the equation as $y = 2 x - 3$ .

$y = 2 x - 3$

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

Step 2

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 ${(2 x - 3)}^{2} - x^{2}$ .

Simplify each term.

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 $- 3$ .

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

$y = 2 x - 3$

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

$y = 2 x - 3$

Subtract $6 x$ from $- 6 x$ .

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

$y = 2 x - 3$

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

$y = 2 x - 3$

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

$y = 2 x - 3$

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

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

$y = 2 x - 3$

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

$y = 2 x - 3$

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

$y = 2 x - 3$

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

$y = 2 x - 3$

Step 3

Solve for $x$ in $3 x^{2} - 12 x + 9 = 9$ .

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 $0$ .

$x - 4 = 0$

$y = 2 x - 3$

Add $4$ to 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.

$x = 0, 4$

$y = 2 x - 3$

$x = 0, 4$

$y = 2 x - 3$

Step 4

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 $0$ .

$y = 0 - 3$

$x = 0$

Subtract $3$ from $0$ .

$y = - 3$

$x = 0$

$y = - 3$

$x = 0$

$y = - 3$

$x = 0$

$y = - 3$

$x = 0$

Step 5

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 $4$ .

$y = 8 - 3$

$x = 4$

Subtract $3$ from $8$ .

$y = 5$

$x = 4$

$y = 5$

$x = 4$

$y = 5$

$x = 4$

$y = 5$

$x = 4$

Step 6

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

$(0, - 3)$

$(4, 5)$

Step 7

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

$x = 0, y = - 3$

$x = 4, y = 5$

Step 8

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

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