Precalculus Examples

$x + y = 7$ , $x^{2} + 3 y^{2} = 79$

Step 1

Subtract $y$ from both sides of the equation.

$x = 7 - y$

$x^{2} + 3 y^{2} = 79$

Step 2

Replace all occurrences of $x$ with $7 - y$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $x^{2} + 3 y^{2} = 79$ with $7 - y$ .

${(7 - y)}^{2} + 3 y^{2} = 79$

$x = 7 - y$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify ${(7 - y)}^{2} + 3 y^{2}$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Rewrite ${(7 - y)}^{2}$ as $(7 - y) (7 - y)$ .

$(7 - y) (7 - y) + 3 y^{2} = 79$

$x = 7 - y$

Step 2.2.1.1.2

Expand $(7 - y) (7 - y)$ using the FOIL Method.

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Step 2.2.1.1.2.1

Apply the distributive property.

$7 (7 - y) - y (7 - y) + 3 y^{2} = 79$

$x = 7 - y$

Step 2.2.1.1.2.2

Apply the distributive property.

$7 \cdot 7 + 7 (- y) - y (7 - y) + 3 y^{2} = 79$

$x = 7 - y$

Step 2.2.1.1.2.3

Apply the distributive property.

$7 \cdot 7 + 7 (- y) - y \cdot 7 - y (- y) + 3 y^{2} = 79$

$x = 7 - y$

$7 \cdot 7 + 7 (- y) - y \cdot 7 - y (- y) + 3 y^{2} = 79$

$x = 7 - y$

Step 2.2.1.1.3

Simplify and combine like terms.

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Step 2.2.1.1.3.1

Simplify each term.

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Step 2.2.1.1.3.1.1

Multiply $7$ by $7$ .

$49 + 7 (- y) - y \cdot 7 - y (- y) + 3 y^{2} = 79$

$x = 7 - y$

Step 2.2.1.1.3.1.2

Multiply $- 1$ by $7$ .

$49 - 7 y - y \cdot 7 - y (- y) + 3 y^{2} = 79$

$x = 7 - y$

Step 2.2.1.1.3.1.3

Multiply $7$ by $- 1$ .

$49 - 7 y - 7 y - y (- y) + 3 y^{2} = 79$

$x = 7 - y$

Step 2.2.1.1.3.1.4

Rewrite using the commutative property of multiplication.

$49 - 7 y - 7 y - 1 \cdot (- 1 y \cdot y) + 3 y^{2} = 79$

$x = 7 - y$

Step 2.2.1.1.3.1.5

Multiply $y$ by $y$ by adding the exponents.

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Step 2.2.1.1.3.1.5.1

Move $y$ .

$49 - 7 y - 7 y - 1 \cdot (- 1 (y \cdot y)) + 3 y^{2} = 79$

$x = 7 - y$

Step 2.2.1.1.3.1.5.2

Multiply $y$ by $y$ .

$49 - 7 y - 7 y - 1 \cdot (- 1 y^{2}) + 3 y^{2} = 79$

$x = 7 - y$

$49 - 7 y - 7 y - 1 \cdot (- 1 y^{2}) + 3 y^{2} = 79$

$x = 7 - y$

Step 2.2.1.1.3.1.6

Multiply $- 1$ by $- 1$ .

$49 - 7 y - 7 y + 1 y^{2} + 3 y^{2} = 79$

$x = 7 - y$

Step 2.2.1.1.3.1.7

Multiply $y^{2}$ by $1$ .

$49 - 7 y - 7 y + y^{2} + 3 y^{2} = 79$

$x = 7 - y$

$49 - 7 y - 7 y + y^{2} + 3 y^{2} = 79$

$x = 7 - y$

Step 2.2.1.1.3.2

Subtract $7 y$ from $- 7 y$ .

$49 - 14 y + y^{2} + 3 y^{2} = 79$

$x = 7 - y$

$49 - 14 y + y^{2} + 3 y^{2} = 79$

$x = 7 - y$

$49 - 14 y + y^{2} + 3 y^{2} = 79$

$x = 7 - y$

Step 2.2.1.2

Add $y^{2}$ and $3 y^{2}$ .

$49 - 14 y + 4 y^{2} = 79$

$x = 7 - y$

$49 - 14 y + 4 y^{2} = 79$

$x = 7 - y$

$49 - 14 y + 4 y^{2} = 79$

$x = 7 - y$

$49 - 14 y + 4 y^{2} = 79$

$x = 7 - y$

Step 3

Solve for $y$ in $49 - 14 y + 4 y^{2} = 79$ .

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Step 3.1

Subtract $79$ from both sides of the equation.

$49 - 14 y + 4 y^{2} - 79 = 0$

$x = 7 - y$

Step 3.2

Subtract $79$ from $49$ .

$- 14 y + 4 y^{2} - 30 = 0$

$x = 7 - y$

Step 3.3

Factor the left side of the equation.

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Step 3.3.1

Factor $2$ out of $- 14 y + 4 y^{2} - 30$ .

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Step 3.3.1.1

Factor $2$ out of $- 14 y$ .

$2 (- 7 y) + 4 y^{2} - 30 = 0$

$x = 7 - y$

Step 3.3.1.2

Factor $2$ out of $4 y^{2}$ .

$2 (- 7 y) + 2 (2 y^{2}) - 30 = 0$

$x = 7 - y$

Step 3.3.1.3

Factor $2$ out of $- 30$ .

$2 (- 7 y) + 2 (2 y^{2}) + 2 (- 15) = 0$

$x = 7 - y$

Step 3.3.1.4

Factor $2$ out of $2 (- 7 y) + 2 (2 y^{2})$ .

$2 (- 7 y + 2 y^{2}) + 2 (- 15) = 0$

$x = 7 - y$

Step 3.3.1.5

Factor $2$ out of $2 (- 7 y + 2 y^{2}) + 2 (- 15)$ .

$2 (- 7 y + 2 y^{2} - 15) = 0$

$x = 7 - y$

$2 (- 7 y + 2 y^{2} - 15) = 0$

$x = 7 - y$

Step 3.3.2

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

$2 (- 7 u + 2 u^{2} - 15) = 0$

$x = 7 - y$

Step 3.3.3

Factor by grouping.

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Step 3.3.3.1

Reorder terms.

$2 (2 u^{2} - 7 u - 15) = 0$

$x = 7 - y$

Step 3.3.3.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 15 = - 30$ and whose sum is $b = - 7$ .

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Step 3.3.3.2.1

Factor $- 7$ out of $- 7 u$ .

$2 (2 u^{2} - 7 u - 15) = 0$

$x = 7 - y$

Step 3.3.3.2.2

Rewrite $- 7$ as $3$ plus $- 10$

$2 (2 u^{2} + (3 - 10) u - 15) = 0$

$x = 7 - y$

Step 3.3.3.2.3

Apply the distributive property.

$2 (2 u^{2} + 3 u - 10 u - 15) = 0$

$x = 7 - y$

$2 (2 u^{2} + 3 u - 10 u - 15) = 0$

$x = 7 - y$

Step 3.3.3.3

Factor out the greatest common factor from each group.

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Step 3.3.3.3.1

Group the first two terms and the last two terms.

$2 ((2 u^{2} + 3 u) - 10 u - 15) = 0$

$x = 7 - y$

Step 3.3.3.3.2

Factor out the greatest common factor (GCF) from each group.

$2 (u (2 u + 3) - 5 (2 u + 3)) = 0$

$x = 7 - y$

$2 (u (2 u + 3) - 5 (2 u + 3)) = 0$

$x = 7 - y$

Step 3.3.3.4

Factor the polynomial by factoring out the greatest common factor, $2 u + 3$ .

$2 ((2 u + 3) (u - 5)) = 0$

$x = 7 - y$

$2 ((2 u + 3) (u - 5)) = 0$

$x = 7 - y$

Step 3.3.4

Factor.

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Step 3.3.4.1

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

$2 ((2 y + 3) (y - 5)) = 0$

$x = 7 - y$

Step 3.3.4.2

Remove unnecessary parentheses.

$2 (2 y + 3) (y - 5) = 0$

$x = 7 - y$

$2 (2 y + 3) (y - 5) = 0$

$x = 7 - y$

$2 (2 y + 3) (y - 5) = 0$

$x = 7 - y$

Step 3.4

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

$2 y + 3 = 0$

$y - 5 = 0$

$x = 7 - y$

Step 3.5

Set $2 y + 3$ equal to $0$ and solve for $y$ .

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Step 3.5.1

Set $2 y + 3$ equal to $0$ .

$2 y + 3 = 0$

$x = 7 - y$

Step 3.5.2

Solve $2 y + 3 = 0$ for $y$ .

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Step 3.5.2.1

Subtract $3$ from both sides of the equation.

$2 y = - 3$

$x = 7 - y$

Step 3.5.2.2

Divide each term in $2 y = - 3$ by $2$ and simplify.

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Step 3.5.2.2.1

Divide each term in $2 y = - 3$ by $2$ .

$\frac{2 y}{2} = \frac{- 3}{2}$

$x = 7 - y$

Step 3.5.2.2.2

Simplify the left side.

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Step 3.5.2.2.2.1

Cancel the common factor of $2$ .

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Step 3.5.2.2.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{- 3}{2}$

$x = 7 - y$

Step 3.5.2.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 3}{2}$

$x = 7 - y$

$y = \frac{- 3}{2}$

$x = 7 - y$

$y = \frac{- 3}{2}$

$x = 7 - y$

Step 3.5.2.2.3

Simplify the right side.

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Step 3.5.2.2.3.1

Move the negative in front of the fraction.

$y = - \frac{3}{2}$

$x = 7 - y$

$y = - \frac{3}{2}$

$x = 7 - y$

$y = - \frac{3}{2}$

$x = 7 - y$

$y = - \frac{3}{2}$

$x = 7 - y$

$y = - \frac{3}{2}$

$x = 7 - y$

Step 3.6

Set $y - 5$ equal to $0$ and solve for $y$ .

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Step 3.6.1

Set $y - 5$ equal to $0$ .

$y - 5 = 0$

$x = 7 - y$

Step 3.6.2

Add $5$ to both sides of the equation.

$y = 5$

$x = 7 - y$

$y = 5$

$x = 7 - y$

Step 3.7

The final solution is all the values that make $2 (2 y + 3) (y - 5) = 0$ true.

$y = - \frac{3}{2}, 5$

$x = 7 - y$

$y = - \frac{3}{2}, 5$

$x = 7 - y$

Step 4

Replace all occurrences of $y$ with $- \frac{3}{2}$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = 7 - y$ with $- \frac{3}{2}$ .

$x = 7 - (- \frac{3}{2})$

$y = - \frac{3}{2}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $7 - (- \frac{3}{2})$ .

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Step 4.2.1.1

Multiply $- (- \frac{3}{2})$ .

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Step 4.2.1.1.1

Multiply $- 1$ by $- 1$ .

$x = 7 + 1 (\frac{3}{2})$

$y = - \frac{3}{2}$

Step 4.2.1.1.2

Multiply $\frac{3}{2}$ by $1$ .

$x = 7 + \frac{3}{2}$

$y = - \frac{3}{2}$

$x = 7 + \frac{3}{2}$

$y = - \frac{3}{2}$

Step 4.2.1.2

To write $7$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = 7 \cdot \frac{2}{2} + \frac{3}{2}$

$y = - \frac{3}{2}$

Step 4.2.1.3

Combine $7$ and $\frac{2}{2}$ .

$x = \frac{7 \cdot 2}{2} + \frac{3}{2}$

$y = - \frac{3}{2}$

Step 4.2.1.4

Combine the numerators over the common denominator.

$x = \frac{7 \cdot 2 + 3}{2}$

$y = - \frac{3}{2}$

Step 4.2.1.5

Simplify the numerator.

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Step 4.2.1.5.1

Multiply $7$ by $2$ .

$x = \frac{14 + 3}{2}$

$y = - \frac{3}{2}$

Step 4.2.1.5.2

Add $14$ and $3$ .

$x = \frac{17}{2}$

$y = - \frac{3}{2}$

$x = \frac{17}{2}$

$y = - \frac{3}{2}$

$x = \frac{17}{2}$

$y = - \frac{3}{2}$

$x = \frac{17}{2}$

$y = - \frac{3}{2}$

$x = \frac{17}{2}$

$y = - \frac{3}{2}$

Step 5

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

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Step 5.1

Replace all occurrences of $y$ in $x = 7 - y$ with $5$ .

$x = 7 - (5)$

$y = 5$

Step 5.2

Simplify the right side.

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Step 5.2.1

Simplify $7 - (5)$ .

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Step 5.2.1.1

Multiply $- 1$ by $5$ .

$x = 7 - 5$

$y = 5$

Step 5.2.1.2

Subtract $5$ from $7$ .

$x = 2$

$y = 5$

$x = 2$

$y = 5$

$x = 2$

$y = 5$

$x = 2$

$y = 5$

Step 6

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

$(\frac{17}{2}, - \frac{3}{2})$

$(2, 5)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(\frac{17}{2}, - \frac{3}{2}), (2, 5)$

Equation Form:

$x = \frac{17}{2}, y = - \frac{3}{2}$

$x = 2, y = 5$

Step 8