Precalculus Examples

$y = 4 x + 4$ , $y = 6 x$

Use the slope-intercept form to find the slope.

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

$y = m x + b$

Using the slope-intercept form, the slope is $4$ .

$m_{1} = 4$

Use the slope-intercept form to find the slope.

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

$y = m x + b$

Using the slope-intercept form, the slope is $6$ .

$m_{2} = 6$

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

$y = 4 x + 4$

$y = 6 x$

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

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Eliminate the equal sides of each equation and combine.

$4 x + 4 = 6 x$

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

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Subtract $6 x$ from both sides of the equation.

$4 x + 4 - 6 x = 0$

$y = 6 x$

Subtract $6 x$ from $4 x$ .

$- 2 x + 4 = 0$

$y = 6 x$

$- 2 x + 4 = 0$

$y = 6 x$

Subtract $4$ from both sides of the equation.

$- 2 x = - 4$

$y = 6 x$

Divide each term by $- 2$ and simplify.

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Divide each term in $- 2 x = - 4$ by $- 2$ .

$- \frac{2 x}{- 2} = - \frac{4}{- 2}$

$y = 6 x$

Simplify the left side of the $e q u a t i o n$ by cancelling the common factors.

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Reduce the expression by cancelling the common factors.

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Factor $2$ out of $2 x$ .

$- \frac{2 (x)}{- 2} = - \frac{4}{- 2}$

$y = 6 x$

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

$- (- 1 \cdot x) = - \frac{4}{- 2}$

$y = 6 x$

$- (- 1 \cdot x) = - \frac{4}{- 2}$

$y = 6 x$

Simplify the expression.

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Multiply $- 1$ by $x$ .

$- (- 1 x) = - \frac{4}{- 2}$

$y = 6 x$

Rewrite $- 1 x$ as $- x$ .

$x = - \frac{4}{- 2}$

$y = 6 x$

$x = - \frac{4}{- 2}$

$y = 6 x$

$x = - \frac{4}{- 2}$

$y = 6 x$

Divide $4$ by $- 2$ .

$x = 2$

$y = 6 x$

$x = 2$

$y = 6 x$

Solve for $x$ in the first equation.

$x = 2 y = 6 x$

Replace all occurrences of $x$ with the solution found by solving the last equation for $x$ . In this case, the value substituted is $2$ .

$x = 2$

$y = 6 (2)$

Multiply $6$ by $2$ .

$x = 2$

$y = 12$

Solve for $y$ in the second equation.

$x = 2 y = 12$

List the solutions to the system of equations.

$x = 2$

$y = 12$

The solution to the system of equations can be represented as a point.

$(2, 12)$

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

$m_{1} = 4$

$m_{2} = 6$

$(2, 12)$

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