Finite Math Examples

$y = 2 x + 2$ , $y = (- \frac{1}{3}) x + \frac{14}{3}$

Step 1

Eliminate the equal sides of each equation and combine.

$2 x + 2 = - \frac{x}{3} + \frac{14}{3}$

Step 2

Solve $2 x + 2 = - \frac{x}{3} + \frac{14}{3}$ for $x$ .

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Step 2.1

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

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Step 2.1.1

Add $\frac{x}{3}$ to both sides of the equation.

$2 x + 2 + \frac{x}{3} = \frac{14}{3}$

Step 2.1.2

To write $2 x$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$2 x \cdot \frac{3}{3} + \frac{x}{3} + 2 = \frac{14}{3}$

Step 2.1.3

Combine $2 x$ and $\frac{3}{3}$ .

$\frac{2 x \cdot 3}{3} + \frac{x}{3} + 2 = \frac{14}{3}$

Step 2.1.4

Combine the numerators over the common denominator.

$\frac{2 x \cdot 3 + x}{3} + 2 = \frac{14}{3}$

Step 2.1.5

Simplify each term.

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Step 2.1.5.1

Simplify the numerator.

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Step 2.1.5.1.1

Factor $x$ out of $2 x \cdot 3 + x$ .

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Step 2.1.5.1.1.1

Factor $x$ out of $2 x \cdot 3$ .

$\frac{x (2 \cdot 3) + x}{3} + 2 = \frac{14}{3}$

Step 2.1.5.1.1.2

Raise $x$ to the power of $1$ .

$\frac{x (2 \cdot 3) + x^{1}}{3} + 2 = \frac{14}{3}$

Step 2.1.5.1.1.3

Factor $x$ out of $x^{1}$ .

$\frac{x (2 \cdot 3) + x \cdot 1}{3} + 2 = \frac{14}{3}$

Step 2.1.5.1.1.4

Factor $x$ out of $x (2 \cdot 3) + x \cdot 1$ .

$\frac{x (2 \cdot 3 + 1)}{3} + 2 = \frac{14}{3}$

Step 2.1.5.1.2

Multiply $2$ by $3$ .

$\frac{x (6 + 1)}{3} + 2 = \frac{14}{3}$

Step 2.1.5.1.3

Add $6$ and $1$ .

$\frac{x \cdot 7}{3} + 2 = \frac{14}{3}$

Step 2.1.5.2

Move $7$ to the left of $x$ .

$\frac{7 x}{3} + 2 = \frac{14}{3}$

Step 2.2

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

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Step 2.2.1

Subtract $2$ from both sides of the equation.

$\frac{7 x}{3} = \frac{14}{3} - 2$

Step 2.2.2

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{7 x}{3} = \frac{14}{3} - 2 \cdot \frac{3}{3}$

Step 2.2.3

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

$\frac{7 x}{3} = \frac{14}{3} + \frac{- 2 \cdot 3}{3}$

Step 2.2.4

Combine the numerators over the common denominator.

$\frac{7 x}{3} = \frac{14 - 2 \cdot 3}{3}$

Step 2.2.5

Simplify the numerator.

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Step 2.2.5.1

Multiply $- 2$ by $3$ .

$\frac{7 x}{3} = \frac{14 - 6}{3}$

Step 2.2.5.2

Subtract $6$ from $14$ .

$\frac{7 x}{3} = \frac{8}{3}$

Step 2.3

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

$7 x = 8$

Step 2.4

Divide each term in $7 x = 8$ by $7$ and simplify.

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Step 2.4.1

Divide each term in $7 x = 8$ by $7$ .

$\frac{7 x}{7} = \frac{8}{7}$

Step 2.4.2

Simplify the left side.

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Step 2.4.2.1

Cancel the common factor of $7$ .

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Step 2.4.2.1.1

Cancel the common factor.

$\frac{7 x}{7} = \frac{8}{7}$

Step 2.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{8}{7}$

Step 3

Evaluate $y$ when $x = \frac{8}{7}$ .

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Step 3.1

Substitute $\frac{8}{7}$ for $x$ .

$y = - \frac{\frac{8}{7}}{3} + \frac{14}{3}$

Step 3.2

Simplify $- \frac{\frac{8}{7}}{3} + \frac{14}{3}$ .

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Step 3.2.1

Combine the numerators over the common denominator.

$y = \frac{- \frac{8}{7} + 14}{3}$

Step 3.2.2

To write $14$ as a fraction with a common denominator, multiply by $\frac{7}{7}$ .

$y = \frac{- \frac{8}{7} + 14 \cdot \frac{7}{7}}{3}$

Step 3.2.3

Combine $14$ and $\frac{7}{7}$ .

$y = \frac{- \frac{8}{7} + \frac{14 \cdot 7}{7}}{3}$

Step 3.2.4

Combine the numerators over the common denominator.

$y = \frac{\frac{- 8 + 14 \cdot 7}{7}}{3}$

Step 3.2.5

Simplify the numerator.

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Step 3.2.5.1

Multiply $14$ by $7$ .

$y = \frac{\frac{- 8 + 98}{7}}{3}$

Step 3.2.5.2

Add $- 8$ and $98$ .

$y = \frac{\frac{90}{7}}{3}$

Step 3.2.6

Multiply the numerator by the reciprocal of the denominator.

$y = \frac{90}{7} \cdot \frac{1}{3}$

Step 3.2.7

Cancel the common factor of $3$ .

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Step 3.2.7.1

Factor $3$ out of $90$ .

$y = \frac{3 (30)}{7} \cdot \frac{1}{3}$

Step 3.2.7.2

Cancel the common factor.

$y = \frac{3 \cdot 30}{7} \cdot \frac{1}{3}$

Step 3.2.7.3

Rewrite the expression.

$y = \frac{30}{7}$

Step 4

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

$(\frac{8}{7}, \frac{30}{7})$

Step 5

The result can be shown in multiple forms.

Point Form:

$(\frac{8}{7}, \frac{30}{7})$

Equation Form:

$x = \frac{8}{7}, y = \frac{30}{7}$

Step 6