Precalculus Examples

$2 y^{2} + x y = 5$ , $4 y + x = 7$

Step 1

Subtract $4 y$ from both sides of the equation.

$x = 7 - 4 y$

$2 y^{2} + x y = 5$

Step 2

Replace all occurrences of $x$ with $7 - 4 y$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $2 y^{2} + x y = 5$ with $7 - 4 y$ .

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

$x = 7 - 4 y$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $2 y^{2} + (7 - 4 y) \cdot y$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Apply the distributive property.

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

$x = 7 - 4 y$

Step 2.2.1.1.2

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

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Step 2.2.1.1.2.1

Move $y$ .

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

$x = 7 - 4 y$

Step 2.2.1.1.2.2

Multiply $y$ by $y$ .

$2 y^{2} + 7 y - 4 y^{2} = 5$

$x = 7 - 4 y$

$2 y^{2} + 7 y - 4 y^{2} = 5$

$x = 7 - 4 y$

$2 y^{2} + 7 y - 4 y^{2} = 5$

$x = 7 - 4 y$

Step 2.2.1.2

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

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

$x = 7 - 4 y$

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

$x = 7 - 4 y$

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

$x = 7 - 4 y$

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

$x = 7 - 4 y$

Step 3

Solve for $y$ in $- 2 y^{2} + 7 y = 5$ .

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Step 3.1

Subtract $5$ from both sides of the equation.

$- 2 y^{2} + 7 y - 5 = 0$

$x = 7 - 4 y$

Step 3.2

Factor the left side of the equation.

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Step 3.2.1

Factor $- 1$ out of $- 2 y^{2} + 7 y - 5$ .

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Step 3.2.1.1

Factor $- 1$ out of $- 2 y^{2}$ .

$- (2 y^{2}) + 7 y - 5 = 0$

$x = 7 - 4 y$

Step 3.2.1.2

Factor $- 1$ out of $7 y$ .

$- (2 y^{2}) - (- 7 y) - 5 = 0$

$x = 7 - 4 y$

Step 3.2.1.3

Rewrite $- 5$ as $- 1 (5)$ .

$- (2 y^{2}) - (- 7 y) - 1 \cdot 5 = 0$

$x = 7 - 4 y$

Step 3.2.1.4

Factor $- 1$ out of $- (2 y^{2}) - (- 7 y)$ .

$- (2 y^{2} - 7 y) - 1 \cdot 5 = 0$

$x = 7 - 4 y$

Step 3.2.1.5

Factor $- 1$ out of $- (2 y^{2} - 7 y) - 1 (5)$ .

$- (2 y^{2} - 7 y + 5) = 0$

$x = 7 - 4 y$

$- (2 y^{2} - 7 y + 5) = 0$

$x = 7 - 4 y$

Step 3.2.2

Factor.

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Step 3.2.2.1

Factor by grouping.

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Step 3.2.2.1.1

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 $a \cdot c = 2 \cdot 5 = 10$ and whose sum is $b = - 7$ .

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Step 3.2.2.1.1.1

Factor $- 7$ out of $- 7 y$ .

$- (2 y^{2} - 7 y + 5) = 0$

$x = 7 - 4 y$

Step 3.2.2.1.1.2

Rewrite $- 7$ as $- 2$ plus $- 5$

$- (2 y^{2} + (- 2 - 5) y + 5) = 0$

$x = 7 - 4 y$

Step 3.2.2.1.1.3

Apply the distributive property.

$- (2 y^{2} - 2 y - 5 y + 5) = 0$

$x = 7 - 4 y$

$- (2 y^{2} - 2 y - 5 y + 5) = 0$

$x = 7 - 4 y$

Step 3.2.2.1.2

Factor out the greatest common factor from each group.

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Step 3.2.2.1.2.1

Group the first two terms and the last two terms.

$- ((2 y^{2} - 2 y) - 5 y + 5) = 0$

$x = 7 - 4 y$

Step 3.2.2.1.2.2

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

$- (2 y (y - 1) - 5 (y - 1)) = 0$

$x = 7 - 4 y$

$- (2 y (y - 1) - 5 (y - 1)) = 0$

$x = 7 - 4 y$

Step 3.2.2.1.3

Factor the polynomial by factoring out the greatest common factor, $y - 1$ .

$- ((y - 1) (2 y - 5)) = 0$

$x = 7 - 4 y$

$- ((y - 1) (2 y - 5)) = 0$

$x = 7 - 4 y$

Step 3.2.2.2

Remove unnecessary parentheses.

$- (y - 1) (2 y - 5) = 0$

$x = 7 - 4 y$

$- (y - 1) (2 y - 5) = 0$

$x = 7 - 4 y$

$- (y - 1) (2 y - 5) = 0$

$x = 7 - 4 y$

Step 3.3

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

$y - 1 = 0$

$2 y - 5 = 0$

$x = 7 - 4 y$

Step 3.4

Set $y - 1$ equal to $0$ and solve for $y$ .

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Step 3.4.1

Set $y - 1$ equal to $0$ .

$y - 1 = 0$

$x = 7 - 4 y$

Step 3.4.2

Add $1$ to both sides of the equation.

$y = 1$

$x = 7 - 4 y$

$y = 1$

$x = 7 - 4 y$

Step 3.5

Set $2 y - 5$ equal to $0$ and solve for $y$ .

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Step 3.5.1

Set $2 y - 5$ equal to $0$ .

$2 y - 5 = 0$

$x = 7 - 4 y$

Step 3.5.2

Solve $2 y - 5 = 0$ for $y$ .

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Step 3.5.2.1

Add $5$ to both sides of the equation.

$2 y = 5$

$x = 7 - 4 y$

Step 3.5.2.2

Divide each term in $2 y = 5$ by $2$ and simplify.

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Step 3.5.2.2.1

Divide each term in $2 y = 5$ by $2$ .

$\frac{2 y}{2} = \frac{5}{2}$

$x = 7 - 4 y$

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.

$\frac{2 y}{2} = \frac{5}{2}$

$x = 7 - 4 y$

Step 3.5.2.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{5}{2}$

$x = 7 - 4 y$

$y = \frac{5}{2}$

$x = 7 - 4 y$

$y = \frac{5}{2}$

$x = 7 - 4 y$

$y = \frac{5}{2}$

$x = 7 - 4 y$

$y = \frac{5}{2}$

$x = 7 - 4 y$

$y = \frac{5}{2}$

$x = 7 - 4 y$

Step 3.6

The final solution is all the values that make $- (y - 1) (2 y - 5) = 0$ true.

$y = 1, \frac{5}{2}$

$x = 7 - 4 y$

$y = 1, \frac{5}{2}$

$x = 7 - 4 y$

Step 4

Replace all occurrences of $y$ with $1$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = 7 - 4 y$ with $1$ .

$x = 7 - 4 \cdot 1$

$y = 1$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $7 - 4 \cdot 1$ .

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Step 4.2.1.1

Multiply $- 4$ by $1$ .

$x = 7 - 4$

$y = 1$

Step 4.2.1.2

Subtract $4$ from $7$ .

$x = 3$

$y = 1$

$x = 3$

$y = 1$

$x = 3$

$y = 1$

$x = 3$

$y = 1$

Step 5

Replace all occurrences of $y$ with $\frac{5}{2}$ in each equation.

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Step 5.1

Replace all occurrences of $y$ in $x = 7 - 4 y$ with $\frac{5}{2}$ .

$x = 7 - 4 (\frac{5}{2})$

$y = \frac{5}{2}$

Step 5.2

Simplify the right side.

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Step 5.2.1

Simplify $7 - 4 (\frac{5}{2})$ .

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Step 5.2.1.1

Simplify each term.

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Step 5.2.1.1.1

Cancel the common factor of $2$ .

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Step 5.2.1.1.1.1

Factor $2$ out of $- 4$ .

$x = 7 + 2 (- 2) (\frac{5}{2})$

$y = \frac{5}{2}$

Step 5.2.1.1.1.2

Cancel the common factor.

$x = 7 + 2 \cdot (- 2 (\frac{5}{2}))$

$y = \frac{5}{2}$

Step 5.2.1.1.1.3

Rewrite the expression.

$x = 7 - 2 \cdot 5$

$y = \frac{5}{2}$

$x = 7 - 2 \cdot 5$

$y = \frac{5}{2}$

Step 5.2.1.1.2

Multiply $- 2$ by $5$ .

$x = 7 - 10$

$y = \frac{5}{2}$

$x = 7 - 10$

$y = \frac{5}{2}$

Step 5.2.1.2

Subtract $10$ from $7$ .

$x = - 3$

$y = \frac{5}{2}$

$x = - 3$

$y = \frac{5}{2}$

$x = - 3$

$y = \frac{5}{2}$

$x = - 3$

$y = \frac{5}{2}$

Step 6

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

$(3, 1)$

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

Step 7

The result can be shown in multiple forms.

Point Form:

$(3, 1), (- 3, \frac{5}{2})$

Equation Form:

$x = 3, y = 1$

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

Step 8