Finite Math Examples

$5 x + 2 y = 20$ , $x + 2 y = 8$

Step 1

Rewrite in slope-intercept form.

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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

Subtract $5 x$ from both sides of the equation.

$2 y = 20 - 5 x$

Step 1.3

Divide each term in $2 y = 20 - 5 x$ by $2$ and simplify.

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Step 1.3.1

Divide each term in $2 y = 20 - 5 x$ by $2$ .

$\frac{2 y}{2} = \frac{20}{2} + \frac{- 5 x}{2}$

Step 1.3.2

Simplify the left side.

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Step 1.3.2.1

Cancel the common factor of $2$ .

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Step 1.3.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{20}{2} + \frac{- 5 x}{2}$

Step 1.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{20}{2} + \frac{- 5 x}{2}$

Step 1.3.3

Simplify the right side.

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Step 1.3.3.1

Simplify each term.

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Step 1.3.3.1.1

Divide $20$ by $2$ .

$y = 10 + \frac{- 5 x}{2}$

Step 1.3.3.1.2

Move the negative in front of the fraction.

$y = 10 - \frac{5 x}{2}$

Step 1.4

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

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Step 1.4.1

Reorder $10$ and $- \frac{5 x}{2}$ .

$y = - \frac{5 x}{2} + 10$

Step 1.4.2

Reorder terms.

$y = - (\frac{5}{2} x) + 10$

Step 1.4.3

Remove parentheses.

$y = - \frac{5}{2} x + 10$

Step 2

Using the slope-intercept form, the slope is $- \frac{5}{2}$ .

$m_{1} = - \frac{5}{2}$

Step 3

Rewrite in slope-intercept form.

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Step 3.1

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 3.2

Subtract $x$ from both sides of the equation.

$2 y = 8 - x$

Step 3.3

Divide each term in $2 y = 8 - x$ by $2$ and simplify.

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Step 3.3.1

Divide each term in $2 y = 8 - x$ by $2$ .

$\frac{2 y}{2} = \frac{8}{2} + \frac{- x}{2}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $2$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{8}{2} + \frac{- x}{2}$

Step 3.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{8}{2} + \frac{- x}{2}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Simplify each term.

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Step 3.3.3.1.1

Divide $8$ by $2$ .

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

Step 3.3.3.1.2

Move the negative in front of the fraction.

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

Step 3.4

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

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Step 3.4.1

Reorder $4$ and $- \frac{x}{2}$ .

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

Step 3.4.2

Reorder terms.

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

Step 3.4.3

Remove parentheses.

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

Step 4

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

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

Step 5

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

$5 x + 2 y = 20, x + 2 y = 8$

Step 6

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

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Step 6.1

Subtract $2 y$ from both sides of the equation.

$x = 8 - 2 y$

$5 x + 2 y = 20$

Step 6.2

Replace all occurrences of $x$ with $8 - 2 y$ in each equation.

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Step 6.2.1

Replace all occurrences of $x$ in $5 x + 2 y = 20$ with $8 - 2 y$ .

$5 (8 - 2 y) + 2 y = 20$

$x = 8 - 2 y$

Step 6.2.2

Simplify the left side.

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Step 6.2.2.1

Simplify $5 (8 - 2 y) + 2 y$ .

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Step 6.2.2.1.1

Simplify each term.

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Step 6.2.2.1.1.1

Apply the distributive property.

$5 \cdot 8 + 5 (- 2 y) + 2 y = 20$

$x = 8 - 2 y$

Step 6.2.2.1.1.2

Multiply $5$ by $8$ .

$40 + 5 (- 2 y) + 2 y = 20$

$x = 8 - 2 y$

Step 6.2.2.1.1.3

Multiply $- 2$ by $5$ .

$40 - 10 y + 2 y = 20$

$x = 8 - 2 y$

$40 - 10 y + 2 y = 20$

$x = 8 - 2 y$

Step 6.2.2.1.2

Add $- 10 y$ and $2 y$ .

$40 - 8 y = 20$

$x = 8 - 2 y$

$40 - 8 y = 20$

$x = 8 - 2 y$

$40 - 8 y = 20$

$x = 8 - 2 y$

$40 - 8 y = 20$

$x = 8 - 2 y$

Step 6.3

Solve for $y$ in $40 - 8 y = 20$ .

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Step 6.3.1

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

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Step 6.3.1.1

Subtract $40$ from both sides of the equation.

$- 8 y = 20 - 40$

$x = 8 - 2 y$

Step 6.3.1.2

Subtract $40$ from $20$ .

$- 8 y = - 20$

$x = 8 - 2 y$

$- 8 y = - 20$

$x = 8 - 2 y$

Step 6.3.2

Divide each term in $- 8 y = - 20$ by $- 8$ and simplify.

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Step 6.3.2.1

Divide each term in $- 8 y = - 20$ by $- 8$ .

$\frac{- 8 y}{- 8} = \frac{- 20}{- 8}$

$x = 8 - 2 y$

Step 6.3.2.2

Simplify the left side.

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Step 6.3.2.2.1

Cancel the common factor of $- 8$ .

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Step 6.3.2.2.1.1

Cancel the common factor.

$\frac{- 8 y}{- 8} = \frac{- 20}{- 8}$

$x = 8 - 2 y$

Step 6.3.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 20}{- 8}$

$x = 8 - 2 y$

$y = \frac{- 20}{- 8}$

$x = 8 - 2 y$

$y = \frac{- 20}{- 8}$

$x = 8 - 2 y$

Step 6.3.2.3

Simplify the right side.

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Step 6.3.2.3.1

Cancel the common factor of $- 20$ and $- 8$ .

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Step 6.3.2.3.1.1

Factor $- 4$ out of $- 20$ .

$y = \frac{- 4 \cdot 5}{- 8}$

$x = 8 - 2 y$

Step 6.3.2.3.1.2

Cancel the common factors.

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Step 6.3.2.3.1.2.1

Factor $- 4$ out of $- 8$ .

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

$x = 8 - 2 y$

Step 6.3.2.3.1.2.2

Cancel the common factor.

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

$x = 8 - 2 y$

Step 6.3.2.3.1.2.3

Rewrite the expression.

$y = \frac{5}{2}$

$x = 8 - 2 y$

$y = \frac{5}{2}$

$x = 8 - 2 y$

$y = \frac{5}{2}$

$x = 8 - 2 y$

$y = \frac{5}{2}$

$x = 8 - 2 y$

$y = \frac{5}{2}$

$x = 8 - 2 y$

$y = \frac{5}{2}$

$x = 8 - 2 y$

Step 6.4

Replace all occurrences of $y$ with $\frac{5}{2}$ in each equation.

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Step 6.4.1

Replace all occurrences of $y$ in $x = 8 - 2 y$ with $\frac{5}{2}$ .

$x = 8 - 2 (\frac{5}{2})$

$y = \frac{5}{2}$

Step 6.4.2

Simplify the right side.

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Step 6.4.2.1

Simplify $8 - 2 (\frac{5}{2})$ .

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Step 6.4.2.1.1

Simplify each term.

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Step 6.4.2.1.1.1

Cancel the common factor of $2$ .

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Step 6.4.2.1.1.1.1

Factor $2$ out of $- 2$ .

$x = 8 + 2 (- 1) (\frac{5}{2})$

$y = \frac{5}{2}$

Step 6.4.2.1.1.1.2

Cancel the common factor.

$x = 8 + 2 \cdot (- 1 (\frac{5}{2}))$

$y = \frac{5}{2}$

Step 6.4.2.1.1.1.3

Rewrite the expression.

$x = 8 - 1 \cdot 5$

$y = \frac{5}{2}$

$x = 8 - 1 \cdot 5$

$y = \frac{5}{2}$

Step 6.4.2.1.1.2

Multiply $- 1$ by $5$ .

$x = 8 - 5$

$y = \frac{5}{2}$

$x = 8 - 5$

$y = \frac{5}{2}$

Step 6.4.2.1.2

Subtract $5$ from $8$ .

$x = 3$

$y = \frac{5}{2}$

$x = 3$

$y = \frac{5}{2}$

$x = 3$

$y = \frac{5}{2}$

$x = 3$

$y = \frac{5}{2}$

Step 6.5

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

$(3, \frac{5}{2})$

Step 7

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

$m_{1} = - \frac{5}{2}$

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

$(3, \frac{5}{2})$

Step 8