Finite Math Examples

$y = - 2 x + 1$ , $y = \frac{1}{2} x + 4$

Step 1

Use the slope-intercept form to find the slope.

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Step 1.1

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 1.2

Using the slope-intercept form, the slope is $- 2$ .

$m_{1} = - 2$

Step 2

Rewrite in slope-intercept form.

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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

Simplify the right side.

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Step 2.2.1

Combine $\frac{1}{2}$ and $x$ .

$y = \frac{x}{2} + 4$

Step 2.3

Reorder terms.

$y = \frac{1}{2} x + 4$

Step 3

Using the slope-intercept form, the slope is $\frac{1}{2}$ .

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

Step 4

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

$y = - 2 x + 1, y = \frac{1}{2} x + 4$

Step 5

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

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Step 5.1

Eliminate the equal sides of each equation and combine.

$- 2 x + 1 = \frac{1}{2} x + 4$

Step 5.2

Solve $- 2 x + 1 = \frac{1}{2} x + 4$ for $x$ .

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Step 5.2.1

Combine $\frac{1}{2}$ and $x$ .

$- 2 x + 1 = \frac{x}{2} + 4$

Step 5.2.2

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

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Step 5.2.2.1

Subtract $\frac{x}{2}$ from both sides of the equation.

$- 2 x + 1 - \frac{x}{2} = 4$

Step 5.2.2.2

To write $- 2 x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 5.2.2.3

Combine $- 2 x$ and $\frac{2}{2}$ .

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

Step 5.2.2.4

Combine the numerators over the common denominator.

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

Step 5.2.2.5

Simplify each term.

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Step 5.2.2.5.1

Simplify the numerator.

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Step 5.2.2.5.1.1

Factor $x$ out of $- 2 x \cdot 2 - x$ .

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Step 5.2.2.5.1.1.1

Factor $x$ out of $- 2 x \cdot 2$ .

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

Step 5.2.2.5.1.1.2

Factor $x$ out of $- x$ .

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

Step 5.2.2.5.1.1.3

Factor $x$ out of $x (- 2 \cdot 2) + x \cdot - 1$ .

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

Step 5.2.2.5.1.2

Multiply $- 2$ by $2$ .

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

Step 5.2.2.5.1.3

Subtract $1$ from $- 4$ .

$\frac{x \cdot - 5}{2} + 1 = 4$

Step 5.2.2.5.2

Move $- 5$ to the left of $x$ .

$\frac{- 5 \cdot x}{2} + 1 = 4$

Step 5.2.2.5.3

Move the negative in front of the fraction.

$- \frac{5 x}{2} + 1 = 4$

Step 5.2.3

Move all terms not containing $x$ to the right side of the equation.

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Step 5.2.3.1

Subtract $1$ from both sides of the equation.

$- \frac{5 x}{2} = 4 - 1$

Step 5.2.3.2

Subtract $1$ from $4$ .

$- \frac{5 x}{2} = 3$

Step 5.2.4

Multiply both sides of the equation by $- \frac{2}{5}$ .

$- \frac{2}{5} (- \frac{5 x}{2}) = - \frac{2}{5} \cdot 3$

Step 5.2.5

Simplify both sides of the equation.

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Step 5.2.5.1

Simplify the left side.

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Step 5.2.5.1.1

Simplify $- \frac{2}{5} (- \frac{5 x}{2})$ .

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Step 5.2.5.1.1.1

Cancel the common factor of $2$ .

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Step 5.2.5.1.1.1.1

Move the leading negative in $- \frac{2}{5}$ into the numerator.

$\frac{- 2}{5} (- \frac{5 x}{2}) = - \frac{2}{5} \cdot 3$

Step 5.2.5.1.1.1.2

Move the leading negative in $- \frac{5 x}{2}$ into the numerator.

$\frac{- 2}{5} \cdot \frac{- 5 x}{2} = - \frac{2}{5} \cdot 3$

Step 5.2.5.1.1.1.3

Factor $2$ out of $- 2$ .

$\frac{2 (- 1)}{5} \cdot \frac{- 5 x}{2} = - \frac{2}{5} \cdot 3$

Step 5.2.5.1.1.1.4

Cancel the common factor.

$\frac{2 \cdot - 1}{5} \cdot \frac{- 5 x}{2} = - \frac{2}{5} \cdot 3$

Step 5.2.5.1.1.1.5

Rewrite the expression.

$\frac{- 1}{5} (- 5 x) = - \frac{2}{5} \cdot 3$

Step 5.2.5.1.1.2

Cancel the common factor of $5$ .

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Step 5.2.5.1.1.2.1

Factor $5$ out of $- 5 x$ .

$\frac{- 1}{5} (5 (- x)) = - \frac{2}{5} \cdot 3$

Step 5.2.5.1.1.2.2

Cancel the common factor.

$\frac{- 1}{5} (5 (- x)) = - \frac{2}{5} \cdot 3$

Step 5.2.5.1.1.2.3

Rewrite the expression.

$- - x = - \frac{2}{5} \cdot 3$

Step 5.2.5.1.1.3

Multiply.

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Step 5.2.5.1.1.3.1

Multiply $- 1$ by $- 1$ .

$1 x = - \frac{2}{5} \cdot 3$

Step 5.2.5.1.1.3.2

Multiply $x$ by $1$ .

$x = - \frac{2}{5} \cdot 3$

Step 5.2.5.2

Simplify the right side.

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Step 5.2.5.2.1

Simplify $- \frac{2}{5} \cdot 3$ .

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Step 5.2.5.2.1.1

Multiply $- \frac{2}{5} \cdot 3$ .

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Step 5.2.5.2.1.1.1

Multiply $3$ by $- 1$ .

$x = - 3 (\frac{2}{5})$

Step 5.2.5.2.1.1.2

Combine $- 3$ and $\frac{2}{5}$ .

$x = \frac{- 3 \cdot 2}{5}$

Step 5.2.5.2.1.1.3

Multiply $- 3$ by $2$ .

$x = \frac{- 6}{5}$

Step 5.2.5.2.1.2

Move the negative in front of the fraction.

$x = - \frac{6}{5}$

Step 5.3

Evaluate $y$ when $x = - \frac{6}{5}$ .

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Step 5.3.1

Substitute $- \frac{6}{5}$ for $x$ .

$y = \frac{1}{2} \cdot (- \frac{6}{5}) + 4$

Step 5.3.2

Substitute $- \frac{6}{5}$ for $x$ in $y = \frac{1}{2} \cdot (- \frac{6}{5}) + 4$ and solve for $y$ .

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Step 5.3.2.1

Remove parentheses.

$y = \frac{1}{2} \cdot (- 1 (\frac{6}{5})) + 4$

Step 5.3.2.2

Simplify $\frac{1}{2} \cdot (- 1 (\frac{6}{5})) + 4$ .

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Step 5.3.2.2.1

Simplify each term.

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Step 5.3.2.2.1.1

Rewrite $- 1 (\frac{6}{5})$ as $- (\frac{6}{5})$ .

$y = \frac{1}{2} \cdot (- (\frac{6}{5})) + 4$

Step 5.3.2.2.1.2

Cancel the common factor of $2$ .

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Step 5.3.2.2.1.2.1

Move the leading negative in $- (\frac{6}{5})$ into the numerator.

$y = \frac{1}{2} \cdot \frac{- 6}{5} + 4$

Step 5.3.2.2.1.2.2

Factor $2$ out of $- 6$ .

$y = \frac{1}{2} \cdot \frac{2 (- 3)}{5} + 4$

Step 5.3.2.2.1.2.3

Cancel the common factor.

$y = \frac{1}{2} \cdot \frac{2 \cdot - 3}{5} + 4$

Step 5.3.2.2.1.2.4

Rewrite the expression.

$y = \frac{- 3}{5} + 4$

Step 5.3.2.2.1.3

Move the negative in front of the fraction.

$y = - \frac{3}{5} + 4$

Step 5.3.2.2.2

To write $4$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$y = - \frac{3}{5} + 4 \cdot \frac{5}{5}$

Step 5.3.2.2.3

Combine $4$ and $\frac{5}{5}$ .

$y = - \frac{3}{5} + \frac{4 \cdot 5}{5}$

Step 5.3.2.2.4

Combine the numerators over the common denominator.

$y = \frac{- 3 + 4 \cdot 5}{5}$

Step 5.3.2.2.5

Simplify the numerator.

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Step 5.3.2.2.5.1

Multiply $4$ by $5$ .

$y = \frac{- 3 + 20}{5}$

Step 5.3.2.2.5.2

Add $- 3$ and $20$ .

$y = \frac{17}{5}$

Step 5.4

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

$(- \frac{6}{5}, \frac{17}{5})$

Step 6

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

$m_{1} = - 2$

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

$(- \frac{6}{5}, \frac{17}{5})$

Step 7