Algebra Examples

$5 x = 2 y + 50$ $2 y + 38 = 3 x$

Step 1

Divide each term in $5 x = 2 y + 50$ by $5$ and simplify.

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Step 1.1

Divide each term in $5 x = 2 y + 50$ by $5$ .

$\frac{5 x}{5} = \frac{2 y}{5} + \frac{50}{5}$

$2 y + 38 = 3 x$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $5$ .

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Step 1.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{2 y}{5} + \frac{50}{5}$

$2 y + 38 = 3 x$

Step 1.2.1.2

Divide $x$ by $1$ .

$x = \frac{2 y}{5} + \frac{50}{5}$

$2 y + 38 = 3 x$

$x = \frac{2 y}{5} + \frac{50}{5}$

$2 y + 38 = 3 x$

$x = \frac{2 y}{5} + \frac{50}{5}$

$2 y + 38 = 3 x$

Step 1.3

Simplify the right side.

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Step 1.3.1

Divide $50$ by $5$ .

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

$2 y + 38 = 3 x$

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

$2 y + 38 = 3 x$

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

$2 y + 38 = 3 x$

Step 2

Replace all occurrences of $x$ with $\frac{2 y}{5} + 10$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $2 y + 38 = 3 x$ with $\frac{2 y}{5} + 10$ .

$2 y + 38 = 3 (\frac{2 y}{5} + 10)$

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

Step 2.2

Simplify the right side.

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Step 2.2.1

Simplify $3 (\frac{2 y}{5} + 10)$ .

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Step 2.2.1.1

Apply the distributive property.

$2 y + 38 = 3 (\frac{2 y}{5}) + 3 \cdot 10$

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

Step 2.2.1.2

Multiply $3 \frac{2 y}{5}$ .

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Step 2.2.1.2.1

Combine $3$ and $\frac{2 y}{5}$ .

$2 y + 38 = \frac{3 (2 y)}{5} + 3 \cdot 10$

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

Step 2.2.1.2.2

Multiply $2$ by $3$ .

$2 y + 38 = \frac{6 y}{5} + 3 \cdot 10$

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

$2 y + 38 = \frac{6 y}{5} + 3 \cdot 10$

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

Step 2.2.1.3

Multiply $3$ by $10$ .

$2 y + 38 = \frac{6 y}{5} + 30$

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

$2 y + 38 = \frac{6 y}{5} + 30$

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

$2 y + 38 = \frac{6 y}{5} + 30$

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

$2 y + 38 = \frac{6 y}{5} + 30$

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

Step 3

Solve for $y$ in $2 y + 38 = \frac{6 y}{5} + 30$ .

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Step 3.1

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

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Step 3.1.1

Subtract $\frac{6 y}{5}$ from both sides of the equation.

$2 y + 38 - \frac{6 y}{5} = 30$

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

Step 3.1.2

To write $2 y$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$2 y \cdot \frac{5}{5} - \frac{6 y}{5} + 38 = 30$

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

Step 3.1.3

Combine $2 y$ and $\frac{5}{5}$ .

$\frac{2 y \cdot 5}{5} - \frac{6 y}{5} + 38 = 30$

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

Step 3.1.4

Combine the numerators over the common denominator.

$\frac{2 y \cdot 5 - 6 y}{5} + 38 = 30$

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

Step 3.1.5

Simplify the numerator.

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Step 3.1.5.1

Factor $2 y$ out of $2 y \cdot 5 - 6 y$ .

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Step 3.1.5.1.1

Factor $2 y$ out of $2 y \cdot 5$ .

$\frac{2 y (5) - 6 y}{5} + 38 = 30$

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

Step 3.1.5.1.2

Factor $2 y$ out of $- 6 y$ .

$\frac{2 y (5) + 2 y (- 3)}{5} + 38 = 30$

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

Step 3.1.5.1.3

Factor $2 y$ out of $2 y (5) + 2 y (- 3)$ .

$\frac{2 y (5 - 3)}{5} + 38 = 30$

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

$\frac{2 y (5 - 3)}{5} + 38 = 30$

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

Step 3.1.5.2

Subtract $3$ from $5$ .

$\frac{2 y \cdot 2}{5} + 38 = 30$

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

Step 3.1.5.3

Multiply $2$ by $2$ .

$\frac{4 y}{5} + 38 = 30$

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

$\frac{4 y}{5} + 38 = 30$

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

$\frac{4 y}{5} + 38 = 30$

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

Step 3.2

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

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Step 3.2.1

Subtract $38$ from both sides of the equation.

$\frac{4 y}{5} = 30 - 38$

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

Step 3.2.2

Subtract $38$ from $30$ .

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

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

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

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

Step 3.3

Multiply both sides of the equation by $\frac{5}{4}$ .

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

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

Step 3.4

Simplify both sides of the equation.

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Step 3.4.1

Simplify the left side.

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Step 3.4.1.1

Simplify $\frac{5}{4} \cdot \frac{4 y}{5}$ .

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Step 3.4.1.1.1

Cancel the common factor of $5$ .

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Step 3.4.1.1.1.1

Cancel the common factor.

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

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

Step 3.4.1.1.1.2

Rewrite the expression.

$\frac{1}{4} \cdot (4 y) = \frac{5}{4} \cdot - 8$

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

$\frac{1}{4} \cdot (4 y) = \frac{5}{4} \cdot - 8$

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

Step 3.4.1.1.2

Cancel the common factor of $4$ .

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Step 3.4.1.1.2.1

Factor $4$ out of $4 y$ .

$\frac{1}{4} \cdot (4 (y)) = \frac{5}{4} \cdot - 8$

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

Step 3.4.1.1.2.2

Cancel the common factor.

$\frac{1}{4} \cdot (4 y) = \frac{5}{4} \cdot - 8$

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

Step 3.4.1.1.2.3

Rewrite the expression.

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

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

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

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

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

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

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

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

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $\frac{5}{4} \cdot - 8$ .

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Step 3.4.2.1.1

Cancel the common factor of $4$ .

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Step 3.4.2.1.1.1

Factor $4$ out of $- 8$ .

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

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

Step 3.4.2.1.1.2

Cancel the common factor.

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

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

Step 3.4.2.1.1.3

Rewrite the expression.

$y = 5 \cdot - 2$

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

$y = 5 \cdot - 2$

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

Step 3.4.2.1.2

Multiply $5$ by $- 2$ .

$y = - 10$

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

$y = - 10$

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

$y = - 10$

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

$y = - 10$

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

$y = - 10$

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

Step 4

Replace all occurrences of $y$ with $- 10$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = \frac{2 y}{5} + 10$ with $- 10$ .

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

$y = - 10$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $\frac{2 (- 10)}{5} + 10$ .

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Step 4.2.1.1

Multiply $2$ by $- 10$ .

$x = \frac{- 20}{5} + 10$

$y = - 10$

Step 4.2.1.2

Divide $- 20$ by $5$ .

$x = - 4 + 10$

$y = - 10$

Step 4.2.1.3

Add $- 4$ and $10$ .

$x = 6$

$y = - 10$

$x = 6$

$y = - 10$

$x = 6$

$y = - 10$

$x = 6$

$y = - 10$

Step 5

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

$(6, - 10)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(6, - 10)$

Equation Form:

$x = 6, y = - 10$

Step 7