Precalculus Examples

$x^{2} + x y + 2 y^{2} = 7$ , $x - 2 y = 5$

Step 1

Add $2 y$ to both sides of the equation.

$x = 5 + 2 y$

$x^{2} + x y + 2 y^{2} = 7$

Step 2

Replace all occurrences of $x$ with $5 + 2 y$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $x^{2} + x y + 2 y^{2} = 7$ with $5 + 2 y$ .

${(5 + 2 y)}^{2} + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify ${(5 + 2 y)}^{2} + (5 + 2 y) \cdot y + 2 y^{2}$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Rewrite ${(5 + 2 y)}^{2}$ as $(5 + 2 y) (5 + 2 y)$ .

$(5 + 2 y) (5 + 2 y) + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2.1.1.2

Expand $(5 + 2 y) (5 + 2 y)$ using the FOIL Method.

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Step 2.2.1.1.2.1

Apply the distributive property.

$5 (5 + 2 y) + 2 y (5 + 2 y) + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2.1.1.2.2

Apply the distributive property.

$5 \cdot 5 + 5 (2 y) + 2 y (5 + 2 y) + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2.1.1.2.3

Apply the distributive property.

$5 \cdot 5 + 5 (2 y) + 2 y \cdot 5 + 2 y (2 y) + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

$5 \cdot 5 + 5 (2 y) + 2 y \cdot 5 + 2 y (2 y) + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

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 $5$ by $5$ .

$25 + 5 (2 y) + 2 y \cdot 5 + 2 y (2 y) + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2.1.1.3.1.2

Multiply $2$ by $5$ .

$25 + 10 y + 2 y \cdot 5 + 2 y (2 y) + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2.1.1.3.1.3

Multiply $5$ by $2$ .

$25 + 10 y + 10 y + 2 y (2 y) + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2.1.1.3.1.4

Rewrite using the commutative property of multiplication.

$25 + 10 y + 10 y + 2 \cdot (2 y \cdot y) + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

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 $y$ .

$25 + 10 y + 10 y + 2 \cdot (2 (y \cdot y)) + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2.1.1.3.1.5.2

Multiply $y$ by $y$ .

$25 + 10 y + 10 y + 2 \cdot (2 y^{2}) + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

$25 + 10 y + 10 y + 2 \cdot (2 y^{2}) + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2.1.1.3.1.6

Multiply $2$ by $2$ .

$25 + 10 y + 10 y + 4 y^{2} + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

$25 + 10 y + 10 y + 4 y^{2} + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2.1.1.3.2

Add $10 y$ and $10 y$ .

$25 + 20 y + 4 y^{2} + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

$25 + 20 y + 4 y^{2} + (5 + 2 y) \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2.1.1.4

Apply the distributive property.

$25 + 20 y + 4 y^{2} + 5 y + 2 y \cdot y + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2.1.1.5

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

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Step 2.2.1.1.5.1

Move $y$ .

$25 + 20 y + 4 y^{2} + 5 y + 2 (y \cdot y) + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2.1.1.5.2

Multiply $y$ by $y$ .

$25 + 20 y + 4 y^{2} + 5 y + 2 y^{2} + 2 y^{2} = 7$

$x = 5 + 2 y$

$25 + 20 y + 4 y^{2} + 5 y + 2 y^{2} + 2 y^{2} = 7$

$x = 5 + 2 y$

$25 + 20 y + 4 y^{2} + 5 y + 2 y^{2} + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2.1.2

Simplify by adding terms.

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Step 2.2.1.2.1

Add $20 y$ and $5 y$ .

$25 + 4 y^{2} + 25 y + 2 y^{2} + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2.1.2.2

Add $4 y^{2}$ and $2 y^{2}$ .

$25 + 6 y^{2} + 25 y + 2 y^{2} = 7$

$x = 5 + 2 y$

Step 2.2.1.2.3

Add $6 y^{2}$ and $2 y^{2}$ .

$25 + 8 y^{2} + 25 y = 7$

$x = 5 + 2 y$

$25 + 8 y^{2} + 25 y = 7$

$x = 5 + 2 y$

$25 + 8 y^{2} + 25 y = 7$

$x = 5 + 2 y$

$25 + 8 y^{2} + 25 y = 7$

$x = 5 + 2 y$

$25 + 8 y^{2} + 25 y = 7$

$x = 5 + 2 y$

Step 3

Solve for $y$ in $25 + 8 y^{2} + 25 y = 7$ .

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Step 3.1

Subtract $7$ from both sides of the equation.

$25 + 8 y^{2} + 25 y - 7 = 0$

$x = 5 + 2 y$

Step 3.2

Subtract $7$ from $25$ .

$8 y^{2} + 25 y + 18 = 0$

$x = 5 + 2 y$

Step 3.3

Factor by grouping.

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Step 3.3.1

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 $a \cdot c = 8 \cdot 18 = 144$ and whose sum is $b = 25$ .

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Step 3.3.1.1

Factor $25$ out of $25 y$ .

$8 y^{2} + 25 (y) + 18 = 0$

$x = 5 + 2 y$

Step 3.3.1.2

Rewrite $25$ as $9$ plus $16$

$8 y^{2} + (9 + 16) y + 18 = 0$

$x = 5 + 2 y$

Step 3.3.1.3

Apply the distributive property.

$8 y^{2} + 9 y + 16 y + 18 = 0$

$x = 5 + 2 y$

$8 y^{2} + 9 y + 16 y + 18 = 0$

$x = 5 + 2 y$

Step 3.3.2

Factor out the greatest common factor from each group.

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Step 3.3.2.1

Group the first two terms and the last two terms.

$(8 y^{2} + 9 y) + 16 y + 18 = 0$

$x = 5 + 2 y$

Step 3.3.2.2

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

$y (8 y + 9) + 2 (8 y + 9) = 0$

$x = 5 + 2 y$

$y (8 y + 9) + 2 (8 y + 9) = 0$

$x = 5 + 2 y$

Step 3.3.3

Factor the polynomial by factoring out the greatest common factor, $8 y + 9$ .

$(8 y + 9) (y + 2) = 0$

$x = 5 + 2 y$

$(8 y + 9) (y + 2) = 0$

$x = 5 + 2 y$

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 $0$ .

$8 y + 9 = 0$

$y + 2 = 0$

$x = 5 + 2 y$

Step 3.5

Set $8 y + 9$ equal to $0$ and solve for $y$ .

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Step 3.5.1

Set $8 y + 9$ equal to $0$ .

$8 y + 9 = 0$

$x = 5 + 2 y$

Step 3.5.2

Solve $8 y + 9 = 0$ for $y$ .

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Step 3.5.2.1

Subtract $9$ from both sides of the equation.

$8 y = - 9$

$x = 5 + 2 y$

Step 3.5.2.2

Divide each term in $8 y = - 9$ by $8$ and simplify.

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Step 3.5.2.2.1

Divide each term in $8 y = - 9$ by $8$ .

$\frac{8 y}{8} = \frac{- 9}{8}$

$x = 5 + 2 y$

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 $8$ .

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Step 3.5.2.2.2.1.1

Cancel the common factor.

$\frac{8 y}{8} = \frac{- 9}{8}$

$x = 5 + 2 y$

Step 3.5.2.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 9}{8}$

$x = 5 + 2 y$

$y = \frac{- 9}{8}$

$x = 5 + 2 y$

$y = \frac{- 9}{8}$

$x = 5 + 2 y$

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.

$y = - \frac{9}{8}$

$x = 5 + 2 y$

$y = - \frac{9}{8}$

$x = 5 + 2 y$

$y = - \frac{9}{8}$

$x = 5 + 2 y$

$y = - \frac{9}{8}$

$x = 5 + 2 y$

$y = - \frac{9}{8}$

$x = 5 + 2 y$

Step 3.6

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

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Step 3.6.1

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

$y + 2 = 0$

$x = 5 + 2 y$

Step 3.6.2

Subtract $2$ from both sides of the equation.

$y = - 2$

$x = 5 + 2 y$

$y = - 2$

$x = 5 + 2 y$

Step 3.7

The final solution is all the values that make $(8 y + 9) (y + 2) = 0$ true.

$y = - \frac{9}{8}, - 2$

$x = 5 + 2 y$

$y = - \frac{9}{8}, - 2$

$x = 5 + 2 y$

Step 4

Replace all occurrences of $y$ with $- \frac{9}{8}$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = 5 + 2 y$ with $- \frac{9}{8}$ .

$x = 5 + 2 (- \frac{9}{8})$

$y = - \frac{9}{8}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $5 + 2 (- \frac{9}{8})$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Cancel the common factor of $2$ .

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Step 4.2.1.1.1.1

Move the leading negative in $- \frac{9}{8}$ into the numerator.

$x = 5 + 2 (\frac{- 9}{8})$

$y = - \frac{9}{8}$

Step 4.2.1.1.1.2

Factor $2$ out of $8$ .

$x = 5 + 2 (\frac{- 9}{2 (4)})$

$y = - \frac{9}{8}$

Step 4.2.1.1.1.3

Cancel the common factor.

$x = 5 + 2 (\frac{- 9}{2 \cdot 4})$

$y = - \frac{9}{8}$

Step 4.2.1.1.1.4

Rewrite the expression.

$x = 5 + \frac{- 9}{4}$

$y = - \frac{9}{8}$

$x = 5 + \frac{- 9}{4}$

$y = - \frac{9}{8}$

Step 4.2.1.1.2

Move the negative in front of the fraction.

$x = 5 - \frac{9}{4}$

$y = - \frac{9}{8}$

$x = 5 - \frac{9}{4}$

$y = - \frac{9}{8}$

Step 4.2.1.2

To write $5$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x = 5 \cdot \frac{4}{4} - \frac{9}{4}$

$y = - \frac{9}{8}$

Step 4.2.1.3

Combine $5$ and $\frac{4}{4}$ .

$x = \frac{5 \cdot 4}{4} - \frac{9}{4}$

$y = - \frac{9}{8}$

Step 4.2.1.4

Combine the numerators over the common denominator.

$x = \frac{5 \cdot 4 - 9}{4}$

$y = - \frac{9}{8}$

Step 4.2.1.5

Simplify the numerator.

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Step 4.2.1.5.1

Multiply $5$ by $4$ .

$x = \frac{20 - 9}{4}$

$y = - \frac{9}{8}$

Step 4.2.1.5.2

Subtract $9$ from $20$ .

$x = \frac{11}{4}$

$y = - \frac{9}{8}$

$x = \frac{11}{4}$

$y = - \frac{9}{8}$

$x = \frac{11}{4}$

$y = - \frac{9}{8}$

$x = \frac{11}{4}$

$y = - \frac{9}{8}$

$x = \frac{11}{4}$

$y = - \frac{9}{8}$

Step 5

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

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Step 5.1

Replace all occurrences of $y$ in $x = 5 + 2 y$ with $- 2$ .

$x = 5 + 2 (- 2)$

$y = - 2$

Step 5.2

Simplify the right side.

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Step 5.2.1

Simplify $5 + 2 (- 2)$ .

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Step 5.2.1.1

Multiply $2$ by $- 2$ .

$x = 5 - 4$

$y = - 2$

Step 5.2.1.2

Subtract $4$ from $5$ .

$x = 1$

$y = - 2$

$x = 1$

$y = - 2$

$x = 1$

$y = - 2$

$x = 1$

$y = - 2$

Step 6

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

$(\frac{11}{4}, - \frac{9}{8})$

$(1, - 2)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(\frac{11}{4}, - \frac{9}{8}), (1, - 2)$

Equation Form:

$x = \frac{11}{4}, y = - \frac{9}{8}$

$x = 1, y = - 2$

Step 8