Finite Math Examples

$\frac{2}{3} x + y = 7$ , $\frac{1}{2} x + \frac{3}{4} y = \frac{21}{4}$

Step 1

Solve for $y$ in $\frac{2}{3} x + y = 7$ .

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Step 1.1

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

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

$\frac{1}{2} x + \frac{3}{4} y = \frac{21}{4}$

Step 1.2

Subtract $\frac{2 x}{3}$ from both sides of the equation.

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

$\frac{1}{2} x + \frac{3}{4} y = \frac{21}{4}$

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

$\frac{1}{2} x + \frac{3}{4} y = \frac{21}{4}$

Step 2

Replace all occurrences of $y$ with $7 - \frac{2 x}{3}$ in each equation.

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Step 2.1

Replace all occurrences of $y$ in $\frac{1}{2} x + \frac{3}{4} y = \frac{21}{4}$ with $7 - \frac{2 x}{3}$ .

$\frac{1}{2} x + \frac{3}{4} \cdot (7 - \frac{2 x}{3}) = \frac{21}{4}$

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

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $\frac{1}{2} x + \frac{3}{4} \cdot (7 - \frac{2 x}{3})$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

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

$\frac{x}{2} + \frac{3}{4} \cdot (7 - \frac{2 x}{3}) = \frac{21}{4}$

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

Step 2.2.1.1.2

Apply the distributive property.

$\frac{x}{2} + \frac{3}{4} \cdot 7 + \frac{3}{4} \cdot (- \frac{2 x}{3}) = \frac{21}{4}$

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

Step 2.2.1.1.3

Multiply $\frac{3}{4} \cdot 7$ .

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Step 2.2.1.1.3.1

Combine $\frac{3}{4}$ and $7$ .

$\frac{x}{2} + \frac{3 \cdot 7}{4} + \frac{3}{4} \cdot (- \frac{2 x}{3}) = \frac{21}{4}$

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

Step 2.2.1.1.3.2

Multiply $3$ by $7$ .

$\frac{x}{2} + \frac{21}{4} + \frac{3}{4} \cdot (- \frac{2 x}{3}) = \frac{21}{4}$

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

$\frac{x}{2} + \frac{21}{4} + \frac{3}{4} \cdot (- \frac{2 x}{3}) = \frac{21}{4}$

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

Step 2.2.1.1.4

Cancel the common factor of $3$ .

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Step 2.2.1.1.4.1

Move the leading negative in $- \frac{2 x}{3}$ into the numerator.

$\frac{x}{2} + \frac{21}{4} + \frac{3}{4} \cdot \frac{- 2 x}{3} = \frac{21}{4}$

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

Step 2.2.1.1.4.2

Cancel the common factor.

$\frac{x}{2} + \frac{21}{4} + \frac{3}{4} \cdot \frac{- 2 x}{3} = \frac{21}{4}$

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

Step 2.2.1.1.4.3

Rewrite the expression.

$\frac{x}{2} + \frac{21}{4} + \frac{1}{4} \cdot (- 2 x) = \frac{21}{4}$

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

$\frac{x}{2} + \frac{21}{4} + \frac{1}{4} \cdot (- 2 x) = \frac{21}{4}$

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

Step 2.2.1.1.5

Cancel the common factor of $2$ .

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Step 2.2.1.1.5.1

Factor $2$ out of $4$ .

$\frac{x}{2} + \frac{21}{4} + \frac{1}{2 (2)} \cdot (- 2 x) = \frac{21}{4}$

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

Step 2.2.1.1.5.2

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

$\frac{x}{2} + \frac{21}{4} + \frac{1}{2 (2)} \cdot (2 (- x)) = \frac{21}{4}$

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

Step 2.2.1.1.5.3

Cancel the common factor.

$\frac{x}{2} + \frac{21}{4} + \frac{1}{2 \cdot 2} \cdot (2 (- x)) = \frac{21}{4}$

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

Step 2.2.1.1.5.4

Rewrite the expression.

$\frac{x}{2} + \frac{21}{4} + \frac{1}{2} \cdot (- x) = \frac{21}{4}$

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

$\frac{x}{2} + \frac{21}{4} + \frac{1}{2} \cdot (- x) = \frac{21}{4}$

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

Step 2.2.1.1.6

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

$\frac{x}{2} + \frac{21}{4} - \frac{x}{2} = \frac{21}{4}$

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

$\frac{x}{2} + \frac{21}{4} - \frac{x}{2} = \frac{21}{4}$

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

Step 2.2.1.2

Combine the opposite terms in $\frac{x}{2} + \frac{21}{4} - \frac{x}{2}$ .

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Step 2.2.1.2.1

Subtract $\frac{x}{2}$ from $\frac{x}{2}$ .

$0 + \frac{21}{4} = \frac{21}{4}$

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

Step 2.2.1.2.2

Add $0$ and $\frac{21}{4}$ .

$\frac{21}{4} = \frac{21}{4}$

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

$\frac{21}{4} = \frac{21}{4}$

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

$\frac{21}{4} = \frac{21}{4}$

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

$\frac{21}{4} = \frac{21}{4}$

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

$\frac{21}{4} = \frac{21}{4}$

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

Step 3

Remove any equations from the system that are always true.

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

Step 4