Solve by Substitution x^2-y^2=39 , x-y=3

$x^{2} - y^{2} = 39$ , $x - y = 3$

Step 1

Add $y$ to both sides of the equation.

$x = 3 + y$

$x^{2} - y^{2} = 39$

Step 2

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

Replace all occurrences of $x$ in $x^{2} - y^{2} = 39$ with $3 + y$ .

${(3 + y)}^{2} - y^{2} = 39$

$x = 3 + y$

Simplify the left side.

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

Simplify each term.

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

$(3 + y) (3 + y) - y^{2} = 39$

$x = 3 + y$

Expand $(3 + y) (3 + y)$ using the FOIL Method.

Apply the distributive property.

$3 (3 + y) + y (3 + y) - y^{2} = 39$

$x = 3 + y$

Apply the distributive property.

$3 \cdot 3 + 3 y + y (3 + y) - y^{2} = 39$

$x = 3 + y$

Apply the distributive property.

$3 \cdot 3 + 3 y + y \cdot 3 + y \cdot y - y^{2} = 39$

$x = 3 + y$

$3 \cdot 3 + 3 y + y \cdot 3 + y \cdot y - y^{2} = 39$

$x = 3 + y$

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $3$ .

$9 + 3 y + y \cdot 3 + y \cdot y - y^{2} = 39$

$x = 3 + y$

Move $3$ to the left of $y$ .

$9 + 3 y + 3 \cdot y + y \cdot y - y^{2} = 39$

$x = 3 + y$

Multiply $y$ by $y$ .

$9 + 3 y + 3 y + y^{2} - y^{2} = 39$

$x = 3 + y$

$9 + 3 y + 3 y + y^{2} - y^{2} = 39$

$x = 3 + y$

Add $3 y$ and $3 y$ .

$9 + 6 y + y^{2} - y^{2} = 39$

$x = 3 + y$

$9 + 6 y + y^{2} - y^{2} = 39$

$x = 3 + y$

$9 + 6 y + y^{2} - y^{2} = 39$

$x = 3 + y$

Combine the opposite terms in $9 + 6 y + y^{2} - y^{2}$ .

Subtract $y^{2}$ from $y^{2}$ .

$9 + 6 y + 0 = 39$

$x = 3 + y$

Add $9 + 6 y$ and $0$ .

$9 + 6 y = 39$

$x = 3 + y$

$9 + 6 y = 39$

$x = 3 + y$

$9 + 6 y = 39$

$x = 3 + y$

$9 + 6 y = 39$

$x = 3 + y$

$9 + 6 y = 39$

$x = 3 + y$

Step 3

Solve for $y$ in $9 + 6 y = 39$ .

Move all terms not containing $y$ to the right side of the equation.

Subtract $9$ from both sides of the equation.

$6 y = 39 - 9$

$x = 3 + y$

Subtract $9$ from $39$ .

$6 y = 30$

$x = 3 + y$

$6 y = 30$

$x = 3 + y$

Divide each term in $6 y = 30$ by $6$ and simplify.

Divide each term in $6 y = 30$ by $6$ .

$\frac{6 y}{6} = \frac{30}{6}$

$x = 3 + y$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 y}{6} = \frac{30}{6}$

$x = 3 + y$

Divide $y$ by $1$ .

$y = \frac{30}{6}$

$x = 3 + y$

$y = \frac{30}{6}$

$x = 3 + y$

$y = \frac{30}{6}$

$x = 3 + y$

Simplify the right side.

Divide $30$ by $6$ .

$y = 5$

$x = 3 + y$

$y = 5$

$x = 3 + y$

$y = 5$

$x = 3 + y$

$y = 5$

$x = 3 + y$

Step 4

Replace all occurrences of $y$ with $5$ in each equation.

Replace all occurrences of $y$ in $x = 3 + y$ with $5$ .

$x = 3 + 5$

$y = 5$

Simplify $x = 3 + 5$ .

Simplify the left side.

Remove parentheses.

$x = 3 + 5$

$y = 5$

$x = 3 + 5$

$y = 5$

Simplify the right side.

Add $3$ and $5$ .

$x = 8$

$y = 5$

$x = 8$

$y = 5$

$x = 8$

$y = 5$

$x = 8$

$y = 5$

Step 5

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

$(8, 5)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(8, 5)$

Equation Form:

$x = 8, y = 5$

Step 7

Solve by Substitution x^2-y^2=39 , x-y=3

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