Algebra Examples

$4 x - \frac{1}{3} y = 8$ $\frac{2}{3} x + \frac{1}{3} y = \frac{11}{3}$

Step 1

Multiply each term in $4 x - \frac{1}{3} y = 8$ by $3$ to eliminate the fractions.

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Step 1.1

Multiply each term in $4 x - \frac{1}{3} y = 8$ by $3$ .

$4 x \cdot 3 - \frac{1}{3} y \cdot 3 = 8 \cdot 3$

$\frac{2}{3} x + \frac{1}{3} y = \frac{11}{3}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Simplify each term.

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Step 1.2.1.1

Multiply $3$ by $4$ .

$12 x - \frac{1}{3} y \cdot 3 = 8 \cdot 3$

$\frac{2}{3} x + \frac{1}{3} y = \frac{11}{3}$

Step 1.2.1.2

Combine $y$ and $\frac{1}{3}$ .

$12 x - \frac{y}{3} \cdot 3 = 8 \cdot 3$

$\frac{2}{3} x + \frac{1}{3} y = \frac{11}{3}$

Step 1.2.1.3

Cancel the common factor of $3$ .

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Step 1.2.1.3.1

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

$12 x + \frac{- y}{3} \cdot 3 = 8 \cdot 3$

$\frac{2}{3} x + \frac{1}{3} y = \frac{11}{3}$

Step 1.2.1.3.2

Cancel the common factor.

$12 x + \frac{- y}{3} \cdot 3 = 8 \cdot 3$

$\frac{2}{3} x + \frac{1}{3} y = \frac{11}{3}$

Step 1.2.1.3.3

Rewrite the expression.

$12 x - y = 8 \cdot 3$

$\frac{2}{3} x + \frac{1}{3} y = \frac{11}{3}$

$12 x - y = 8 \cdot 3$

$\frac{2}{3} x + \frac{1}{3} y = \frac{11}{3}$

$12 x - y = 8 \cdot 3$

$\frac{2}{3} x + \frac{1}{3} y = \frac{11}{3}$

$12 x - y = 8 \cdot 3$

$\frac{2}{3} x + \frac{1}{3} y = \frac{11}{3}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Multiply $8$ by $3$ .

$12 x - y = 24$

$\frac{2}{3} x + \frac{1}{3} y = \frac{11}{3}$

$12 x - y = 24$

$\frac{2}{3} x + \frac{1}{3} y = \frac{11}{3}$

$12 x - y = 24$

$\frac{2}{3} x + \frac{1}{3} y = \frac{11}{3}$

Step 2

Multiply each term in $\frac{2}{3} x + \frac{1}{3} y = \frac{11}{3}$ by $3$ to eliminate the fractions.

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Step 2.1

Multiply each term in $\frac{2}{3} x + \frac{1}{3} y = \frac{11}{3}$ by $3$ .

$\frac{2}{3} x \cdot 3 + \frac{1}{3} y \cdot 3 = \frac{11}{3} \cdot 3$

$12 x - y = 24$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

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

$\frac{2 x}{3} \cdot 3 + \frac{1}{3} y \cdot 3 = \frac{11}{3} \cdot 3$

$12 x - y = 24$

Step 2.2.1.2

Cancel the common factor of $3$ .

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Step 2.2.1.2.1

Cancel the common factor.

$\frac{2 x}{3} \cdot 3 + \frac{1}{3} y \cdot 3 = \frac{11}{3} \cdot 3$

$12 x - y = 24$

Step 2.2.1.2.2

Rewrite the expression.

$2 x + \frac{1}{3} y \cdot 3 = \frac{11}{3} \cdot 3$

$12 x - y = 24$

$2 x + \frac{1}{3} y \cdot 3 = \frac{11}{3} \cdot 3$

$12 x - y = 24$

Step 2.2.1.3

Combine $\frac{1}{3}$ and $y$ .

$2 x + \frac{y}{3} \cdot 3 = \frac{11}{3} \cdot 3$

$12 x - y = 24$

Step 2.2.1.4

Cancel the common factor of $3$ .

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Step 2.2.1.4.1

Cancel the common factor.

$2 x + \frac{y}{3} \cdot 3 = \frac{11}{3} \cdot 3$

$12 x - y = 24$

Step 2.2.1.4.2

Rewrite the expression.

$2 x + y = \frac{11}{3} \cdot 3$

$12 x - y = 24$

$2 x + y = \frac{11}{3} \cdot 3$

$12 x - y = 24$

$2 x + y = \frac{11}{3} \cdot 3$

$12 x - y = 24$

$2 x + y = \frac{11}{3} \cdot 3$

$12 x - y = 24$

Step 2.3

Simplify the right side.

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Step 2.3.1

Cancel the common factor of $3$ .

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Step 2.3.1.1

Cancel the common factor.

$2 x + y = \frac{11}{3} \cdot 3$

$12 x - y = 24$

Step 2.3.1.2

Rewrite the expression.

$2 x + y = 11$

$12 x - y = 24$

$2 x + y = 11$

$12 x - y = 24$

$2 x + y = 11$

$12 x - y = 24$

$2 x + y = 11$

$12 x - y = 24$

Step 3

Add the two equations together to eliminate $y$ from the system.

	$1$	$2$	$x$	$-$	$y$	$=$	$2$	$4$
$+$		$2$	$x$	$+$	$y$	$=$	$1$	$1$
	$1$	$4$	$x$			$=$	$3$	$5$

Step 4

Divide each term in $14 x = 35$ by $14$ and simplify.

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Step 4.1

Divide each term in $14 x = 35$ by $14$ .

$\frac{14 x}{14} = \frac{35}{14}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $14$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{14 x}{14} = \frac{35}{14}$

Step 4.2.1.2

Divide $x$ by $1$ .

$x = \frac{35}{14}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Cancel the common factor of $35$ and $14$ .

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Step 4.3.1.1

Factor $7$ out of $35$ .

$x = \frac{7 (5)}{14}$

Step 4.3.1.2

Cancel the common factors.

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Step 4.3.1.2.1

Factor $7$ out of $14$ .

$x = \frac{7 \cdot 5}{7 \cdot 2}$

Step 4.3.1.2.2

Cancel the common factor.

$x = \frac{7 \cdot 5}{7 \cdot 2}$

Step 4.3.1.2.3

Rewrite the expression.

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

Step 5

Substitute the value found for $x$ into one of the original equations, then solve for $y$ .

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Step 5.1

Substitute the value found for $x$ into one of the original equations to solve for $y$ .

$12 (\frac{5}{2}) - y = 24$

Step 5.2

Simplify each term.

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Step 5.2.1

Cancel the common factor of $2$ .

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Step 5.2.1.1

Factor $2$ out of $12$ .

$2 (6) \frac{5}{2} - y = 24$

Step 5.2.1.2

Cancel the common factor.

$2 \cdot 6 \frac{5}{2} - y = 24$

Step 5.2.1.3

Rewrite the expression.

$6 \cdot 5 - y = 24$

Step 5.2.2

Multiply $6$ by $5$ .

$30 - y = 24$

Step 5.3

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

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Step 5.3.1

Subtract $30$ from both sides of the equation.

$- y = 24 - 30$

Step 5.3.2

Subtract $30$ from $24$ .

$- y = - 6$

Step 5.4

Divide each term in $- y = - 6$ by $- 1$ and simplify.

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Step 5.4.1

Divide each term in $- y = - 6$ by $- 1$ .

$\frac{- y}{- 1} = \frac{- 6}{- 1}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{- 6}{- 1}$

Step 5.4.2.2

Divide $y$ by $1$ .

$y = \frac{- 6}{- 1}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Divide $- 6$ by $- 1$ .

$y = 6$

Step 6

The solution to the independent system of equations can be represented as a point.

$(\frac{5}{2}, 6)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(\frac{5}{2}, 6)$

Equation Form:

$x = \frac{5}{2}, y = 6$

Step 8

	$1$	$2$	$x$	$-$	$y$	$=$	$2$	$4$
$+$		$2$	$x$	$+$	$y$	$=$	$1$	$1$
	$1$	$4$	$x$			$=$	$3$	$5$

	$1$	$2$	$x$	$-$	$y$	$=$	$2$	$4$
$+$		$2$	$x$	$+$	$y$	$=$	$1$	$1$
	$1$	$4$	$x$			$=$	$3$	$5$

	$1$	$2$	$x$	$-$	$y$	$=$	$2$	$4$
$+$		$2$	$x$	$+$	$y$	$=$	$1$	$1$
	$1$	$4$	$x$			$=$	$3$	$5$