Finite Math Examples

$a = \frac{1}{2} b + 3$ , $a - 4 b = - 11$

Step 1

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

$a = \frac{b}{2} + 3$

$a - 4 b = - 11$

Step 2

Replace all occurrences of $a$ with $\frac{b}{2} + 3$ in each equation.

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Step 2.1

Replace all occurrences of $a$ in $a - 4 b = - 11$ with $\frac{b}{2} + 3$ .

$(\frac{b}{2} + 3) - 4 b = - 11$

$a = \frac{b}{2} + 3$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $(\frac{b}{2} + 3) - 4 b$ .

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Step 2.2.1.1

To write $- 4 b$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{b}{2} - 4 b \cdot \frac{2}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

Step 2.2.1.2

Simplify terms.

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Step 2.2.1.2.1

Combine $- 4 b$ and $\frac{2}{2}$ .

$\frac{b}{2} + \frac{- 4 b \cdot 2}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

Step 2.2.1.2.2

Combine the numerators over the common denominator.

$\frac{b - 4 b \cdot 2}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

$\frac{b - 4 b \cdot 2}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

Step 2.2.1.3

Simplify each term.

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Step 2.2.1.3.1

Simplify the numerator.

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Step 2.2.1.3.1.1

Factor $b$ out of $b - 4 b \cdot 2$ .

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Step 2.2.1.3.1.1.1

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

$\frac{b - 4 b \cdot 2}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

Step 2.2.1.3.1.1.2

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

$\frac{b \cdot 1 - 4 b \cdot 2}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

Step 2.2.1.3.1.1.3

Factor $b$ out of $- 4 b \cdot 2$ .

$\frac{b \cdot 1 + b (- 4 \cdot 2)}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

Step 2.2.1.3.1.1.4

Factor $b$ out of $b \cdot 1 + b (- 4 \cdot 2)$ .

$\frac{b (1 - 4 \cdot 2)}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

$\frac{b (1 - 4 \cdot 2)}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

Step 2.2.1.3.1.2

Multiply $- 4$ by $2$ .

$\frac{b (1 - 8)}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

Step 2.2.1.3.1.3

Subtract $8$ from $1$ .

$\frac{b \cdot - 7}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

$\frac{b \cdot - 7}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

Step 2.2.1.3.2

Move $- 7$ to the left of $b$ .

$\frac{- 7 \cdot b}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

Step 2.2.1.3.3

Move the negative in front of the fraction.

$- \frac{7 b}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

$- \frac{7 b}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

$- \frac{7 b}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

$- \frac{7 b}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

$- \frac{7 b}{2} + 3 = - 11$

$a = \frac{b}{2} + 3$

Step 3

Solve for $b$ in $- \frac{7 b}{2} + 3 = - 11$ .

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Step 3.1

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

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Step 3.1.1

Subtract $3$ from both sides of the equation.

$- \frac{7 b}{2} = - 11 - 3$

$a = \frac{b}{2} + 3$

Step 3.1.2

Subtract $3$ from $- 11$ .

$- \frac{7 b}{2} = - 14$

$a = \frac{b}{2} + 3$

$- \frac{7 b}{2} = - 14$

$a = \frac{b}{2} + 3$

Step 3.2

Multiply both sides of the equation by $- \frac{2}{7}$ .

$- \frac{2}{7} \cdot (- \frac{7 b}{2}) = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

Step 3.3

Simplify both sides of the equation.

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Step 3.3.1

Simplify the left side.

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Step 3.3.1.1

Simplify $- \frac{2}{7} (- \frac{7 b}{2})$ .

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Step 3.3.1.1.1

Cancel the common factor of $2$ .

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Step 3.3.1.1.1.1

Move the leading negative in $- \frac{2}{7}$ into the numerator.

$\frac{- 2}{7} \cdot (- \frac{7 b}{2}) = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

Step 3.3.1.1.1.2

Move the leading negative in $- \frac{7 b}{2}$ into the numerator.

$\frac{- 2}{7} \cdot \frac{- 7 b}{2} = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

Step 3.3.1.1.1.3

Factor $2$ out of $- 2$ .

$\frac{2 (- 1)}{7} \cdot \frac{- 7 b}{2} = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

Step 3.3.1.1.1.4

Cancel the common factor.

$\frac{2 \cdot - 1}{7} \cdot \frac{- 7 b}{2} = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

Step 3.3.1.1.1.5

Rewrite the expression.

$\frac{- 1}{7} \cdot (- 7 b) = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

$\frac{- 1}{7} \cdot (- 7 b) = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

Step 3.3.1.1.2

Cancel the common factor of $7$ .

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Step 3.3.1.1.2.1

Factor $7$ out of $- 7 b$ .

$\frac{- 1}{7} \cdot (7 (- b)) = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

Step 3.3.1.1.2.2

Cancel the common factor.

$\frac{- 1}{7} \cdot (7 (- b)) = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

Step 3.3.1.1.2.3

Rewrite the expression.

$b = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

$b = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

Step 3.3.1.1.3

Multiply.

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Step 3.3.1.1.3.1

Multiply $- 1$ by $- 1$ .

$1 b = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

Step 3.3.1.1.3.2

Multiply $b$ by $1$ .

$b = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

$b = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

$b = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

$b = - \frac{2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

Step 3.3.2

Simplify the right side.

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Step 3.3.2.1

Simplify $- \frac{2}{7} \cdot - 14$ .

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Step 3.3.2.1.1

Cancel the common factor of $7$ .

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Step 3.3.2.1.1.1

Move the leading negative in $- \frac{2}{7}$ into the numerator.

$b = \frac{- 2}{7} \cdot - 14$

$a = \frac{b}{2} + 3$

Step 3.3.2.1.1.2

Factor $7$ out of $- 14$ .

$b = \frac{- 2}{7} \cdot (7 (- 2))$

$a = \frac{b}{2} + 3$

Step 3.3.2.1.1.3

Cancel the common factor.

$b = \frac{- 2}{7} \cdot (7 \cdot - 2)$

$a = \frac{b}{2} + 3$

Step 3.3.2.1.1.4

Rewrite the expression.

$b = - 2 \cdot - 2$

$a = \frac{b}{2} + 3$

$b = - 2 \cdot - 2$

$a = \frac{b}{2} + 3$

Step 3.3.2.1.2

Multiply $- 2$ by $- 2$ .

$b = 4$

$a = \frac{b}{2} + 3$

$b = 4$

$a = \frac{b}{2} + 3$

$b = 4$

$a = \frac{b}{2} + 3$

$b = 4$

$a = \frac{b}{2} + 3$

$b = 4$

$a = \frac{b}{2} + 3$

Step 4

Replace all occurrences of $b$ with $4$ in each equation.

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Step 4.1

Replace all occurrences of $b$ in $a = \frac{b}{2} + 3$ with $4$ .

$a = \frac{4}{2} + 3$

$b = 4$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $\frac{4}{2} + 3$ .

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Step 4.2.1.1

Divide $4$ by $2$ .

$a = 2 + 3$

$b = 4$

Step 4.2.1.2

Add $2$ and $3$ .

$a = 5$

$b = 4$

$a = 5$

$b = 4$

$a = 5$

$b = 4$

$a = 5$

$b = 4$

Step 5

List all of the solutions.

$a = 5, b = 4$