Precalculus Examples

$- 4 x - 3 y = - 15$ , $y = - \frac{4}{3} x + 5$

Step 1

Simplify each term.

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Step 1.1

Combine $x$ and $\frac{4}{3}$ .

$y = - \frac{x \cdot 4}{3} + 5$

$- 4 x - 3 y = - 15$

Step 1.2

Move $4$ to the left of $x$ .

$y = - \frac{4 x}{3} + 5$

$- 4 x - 3 y = - 15$

$y = - \frac{4 x}{3} + 5$

$- 4 x - 3 y = - 15$

Step 2

Replace all occurrences of $y$ with $- \frac{4 x}{3} + 5$ in each equation.

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Step 2.1

Replace all occurrences of $y$ in $- 4 x - 3 y = - 15$ with $- \frac{4 x}{3} + 5$ .

$- 4 x - 3 (- \frac{4 x}{3} + 5) = - 15$

$y = - \frac{4 x}{3} + 5$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $- 4 x - 3 (- \frac{4 x}{3} + 5)$ .

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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.

$- 4 x - 3 (- \frac{4 x}{3}) - 3 \cdot 5 = - 15$

$y = - \frac{4 x}{3} + 5$

Step 2.2.1.1.2

Cancel the common factor of $3$ .

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Step 2.2.1.1.2.1

Move the leading negative in $- \frac{4 x}{3}$ into the numerator.

$- 4 x - 3 \frac{- 4 x}{3} - 3 \cdot 5 = - 15$

$y = - \frac{4 x}{3} + 5$

Step 2.2.1.1.2.2

Factor $3$ out of $- 3$ .

$- 4 x + 3 (- 1) (\frac{- 4 x}{3}) - 3 \cdot 5 = - 15$

$y = - \frac{4 x}{3} + 5$

Step 2.2.1.1.2.3

Cancel the common factor.

$- 4 x + 3 \cdot (- 1 \frac{- 4 x}{3}) - 3 \cdot 5 = - 15$

$y = - \frac{4 x}{3} + 5$

Step 2.2.1.1.2.4

Rewrite the expression.

$- 4 x - 1 (- 4 x) - 3 \cdot 5 = - 15$

$y = - \frac{4 x}{3} + 5$

$- 4 x - 1 (- 4 x) - 3 \cdot 5 = - 15$

$y = - \frac{4 x}{3} + 5$

Step 2.2.1.1.3

Multiply $- 4$ by $- 1$ .

$- 4 x + 4 x - 3 \cdot 5 = - 15$

$y = - \frac{4 x}{3} + 5$

Step 2.2.1.1.4

Multiply $- 3$ by $5$ .

$- 4 x + 4 x - 15 = - 15$

$y = - \frac{4 x}{3} + 5$

$- 4 x + 4 x - 15 = - 15$

$y = - \frac{4 x}{3} + 5$

Step 2.2.1.2

Simplify by adding terms.

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Step 2.2.1.2.1

Add $- 4 x$ and $4 x$ .

$0 - 15 = - 15$

$y = - \frac{4 x}{3} + 5$

Step 2.2.1.2.2

Subtract $15$ from $0$ .

$- 15 = - 15$

$y = - \frac{4 x}{3} + 5$

$- 15 = - 15$

$y = - \frac{4 x}{3} + 5$

$- 15 = - 15$

$y = - \frac{4 x}{3} + 5$

$- 15 = - 15$

$y = - \frac{4 x}{3} + 5$

$- 15 = - 15$

$y = - \frac{4 x}{3} + 5$

Step 3

Remove any equations from the system that are always true.

$y = - \frac{4 x}{3} + 5$

Step 4