Algebra Examples

$17 x + 2 y = 0$

Choose a point that the parallel line will pass through.

$(1, 0)$

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.

$y = m x + b$

Subtract $17 x$ from both sides of the equation.

$2 y = - 17 x$

Divide each term by $2$ and simplify.

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Divide each term in $2 y = - 17 x$ by $2$ .

$\frac{2 y}{2} = - \frac{17 x}{2}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{2 y}{2} = - \frac{17 x}{2}$

Divide $y$ by $1$ .

$y = - \frac{17 x}{2}$

Rewrite in slope-intercept form.

$y = - \frac{17}{2} x$

Using the slope-intercept form, the slope is $- \frac{17}{2}$ .

$m = - \frac{17}{2}$

To find an equation that is parallel to $17 x + 2 y = 0$ , the slopes must be equal. Using the slope of the equation, find the parallel line using the point-slope formula.

$(1, 0)$

$m = - \frac{17}{2}$

Find the value of $b$ using the formula for the equation of a line.

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Use the formula for the equation of a line to find $b$ .

$y = m x + b$

Substitute the value of $m$ into the equation.

$y = (- \frac{17}{2}) \cdot x + b$

Substitute the value of $x$ into the equation.

$y = (- \frac{17}{2}) \cdot (1) + b$

Substitute the value of $y$ into the equation.

$(0) = (- \frac{17}{2}) \cdot (1) + b$

Find the value of $b$ .

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Rewrite the equation as $(- \frac{17}{2}) \cdot (1) + b = 0$ .

$(- \frac{17}{2}) \cdot (1) + b = 0$

Simplify each term.

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Multiply $- \frac{17}{2}$ by $1$ .

$(- \frac{17}{2}) (1) + b = 0$

Multiply $- 1$ by $1$ .

$- \frac{17}{2} + b = 0$

Add $\frac{17}{2}$ to both sides of the equation.

$b = \frac{17}{2}$

Now that the values of $m$ (slope) and $b$ (y-intercept) are known, substitute them into $y = m x + b$ to find the equation of the line.

$y = - \frac{17}{2} x + \frac{17}{2}$

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