Pre-Algebra Examples

$(4 x - 2 y) \cdot 3 + 5 y - 3 x$

Step 1

Solve for $y$ .

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Step 1.1

Subtract $9 x$ from both sides of the equation.

$- y = - 9 x$

Step 1.2

Divide each term in $- y = - 9 x$ by $- 1$ and simplify.

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Step 1.2.1

Divide each term in $- y = - 9 x$ by $- 1$ .

$\frac{- y}{- 1} = \frac{- 9 x}{- 1}$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{- 9 x}{- 1}$

Step 1.2.2.2

Divide $y$ by $1$ .

$y = \frac{- 9 x}{- 1}$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Move the negative one from the denominator of $\frac{- 9 x}{- 1}$ .

$y = - 1 \cdot (- 9 x)$

Step 1.2.3.2

Rewrite $- 1 \cdot (- 9 x)$ as $- (- 9 x)$ .

$y = - (- 9 x)$

Step 1.2.3.3

Multiply $- 9$ by $- 1$ .

$y = 9 x$

Step 2

Use the slope-intercept form to find the slope and y-intercept.

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Step 2.1

The slope-intercept form is $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept.

$y = m x + b$

Step 2.2

Find the values of $m$ and $b$ using the form $y = m x + b$ .

$m = 9$

$b = 0$

Step 2.3

The slope of the line is the value of $m$ , and the y-intercept is the value of $b$ .

Slope: $9$

y-intercept: $(0, 0)$

Slope: $9$

y-intercept: $(0, 0)$

Step 3

Any line can be graphed using two points. Select two $x$ values, and plug them into the equation to find the corresponding $y$ values.

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Step 3.1

Create a table of the $x$ and $y$ values.

$\begin{array}{c} x & y \\ 0 & 0 \\ 1 & 9 \end{array}$

Step 4

Graph the line using the slope and the y-intercept, or the points.

Slope: $9$

y-intercept: $(0, 0)$

$\begin{array}{c} x & y \\ 0 & 0 \\ 1 & 9 \end{array}$

Step 5