Algebra Examples

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

Rewrite in slope-intercept form.

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

$y = m x + b$

Reorder terms.

$y = \frac{2}{3} x$

Using the slope-intercept form, the slope is $\frac{2}{3}$ .

$m_{1} = \frac{2}{3}$

Rewrite in slope-intercept form.

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

$y = m x + b$

Write in $y = m x + b$ form.

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Reorder terms.

$y = - (\frac{2}{3} x)$

Remove parentheses.

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

Using the slope-intercept form, the slope is $- \frac{2}{3}$ .

$m_{2} = - \frac{2}{3}$

Set up the system of equations to find any points of intersection.

$y = \frac{2 x}{3}$

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

Solve the system of equations to find the point of intersection.

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Eliminate the equal sides of each equation and combine.

$\frac{2 x}{3} = - \frac{2 x}{3}$

Solve $\frac{2 x}{3} = - \frac{2 x}{3}$ for $x$ .

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Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$2 x = - 2 x$

Move all terms containing $x$ to the left side of the equation.

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

$2 x + 2 x = 0$

Add $2 x$ and $2 x$ .

$4 x = 0$

Divide each term in $4 x = 0$ by $4$ and simplify.

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Divide each term in $4 x = 0$ by $4$ .

$\frac{4 x}{4} = \frac{0}{4}$

Simplify the left side.

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Cancel the common factor of $4$ .

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Cancel the common factor.

$\frac{4 x}{4} = \frac{0}{4}$

Divide $x$ by $1$ .

$x = \frac{0}{4}$

Simplify the right side.

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Divide $0$ by $4$ .

$x = 0$

Evaluate $y$ when $x = 0$ .

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Substitute $0$ for $x$ .

$y = - \frac{2 (0)}{3}$

Simplify $- \frac{2 (0)}{3}$ .

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Cancel the common factor of $0$ and $3$ .

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Factor $3$ out of $2 (0)$ .

$y = - \frac{3 (2 \cdot (0))}{3}$

Cancel the common factors.

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Factor $3$ out of $3$ .

$y = - \frac{3 (2 \cdot (0))}{3 (1)}$

Cancel the common factor.

$y = - \frac{3 (2 \cdot (0))}{3 \cdot 1}$

Rewrite the expression.

$y = - \frac{2 \cdot (0)}{1}$

Divide $2 \cdot (0)$ by $1$ .

$y = - (2 \cdot (0))$

Multiply by zero.

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Multiply $2$ by $0$ .

$y = - 0$

Multiply $- 1$ by $0$ .

$y = 0$

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

$(0, 0)$

Since the slopes are different, the lines will have exactly one intersection point.

$m_{1} = \frac{2}{3}$

$m_{2} = - \frac{2}{3}$

$(0, 0)$