Precalculus Examples

$y = \frac{1}{7} x + 3$ , $x - 7 y = - 21$

Step 1

Combine $\frac{1}{7}$ and $x$ .

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

$x - 7 y = - 21$

Step 2

Subtract $x$ from both sides of the equation.

$- 7 y = - 21 - x$

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

Step 3

Divide each term in $- 7 y = - 21 - x$ by $- 7$ and simplify.

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Step 3.1

Divide each term in $- 7 y = - 21 - x$ by $- 7$ .

$\frac{- 7 y}{- 7} = \frac{- 21}{- 7} + \frac{- x}{- 7}$

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

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $- 7$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{- 7 y}{- 7} = \frac{- 21}{- 7} + \frac{- x}{- 7}$

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

Step 3.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 21}{- 7} + \frac{- x}{- 7}$

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

$y = \frac{- 21}{- 7} + \frac{- x}{- 7}$

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

$y = \frac{- 21}{- 7} + \frac{- x}{- 7}$

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

Step 3.3

Simplify the right side.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Divide $- 21$ by $- 7$ .

$y = 3 + \frac{- x}{- 7}$

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

Step 3.3.1.2

Dividing two negative values results in a positive value.

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

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

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

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

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

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

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

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

Step 4

Create a graph to locate the intersection of the equations. The intersection of the system of equations is the solution.

Always true

Step 5