Algebra Examples

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

Step 1

Subtract $2 y$ from both sides of the equation.

$x = - 2 - 2 y$

$y = - \frac{1}{3} x$

Step 2

Replace all occurrences of $x$ with $- 2 - 2 y$ in each equation.

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Step 2.1

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

$y = - \frac{1}{3} \cdot (- 2 - 2 y)$

$x = - 2 - 2 y$

Step 2.2

Simplify the right side.

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Step 2.2.1

Simplify $- \frac{1}{3} \cdot (- 2 - 2 y)$ .

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Step 2.2.1.1

Apply the distributive property.

$y = - \frac{1}{3} \cdot - 2 - \frac{1}{3} \cdot (- 2 y)$

$x = - 2 - 2 y$

Step 2.2.1.2

Multiply $- \frac{1}{3} \cdot - 2$ .

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Step 2.2.1.2.1

Multiply $- 2$ by $- 1$ .

$y = 2 (\frac{1}{3}) - \frac{1}{3} \cdot (- 2 y)$

$x = - 2 - 2 y$

Step 2.2.1.2.2

Combine $2$ and $\frac{1}{3}$ .

$y = \frac{2}{3} - \frac{1}{3} \cdot (- 2 y)$

$x = - 2 - 2 y$

$y = \frac{2}{3} - \frac{1}{3} \cdot (- 2 y)$

$x = - 2 - 2 y$

Step 2.2.1.3

Multiply $- \frac{1}{3} (- 2 y)$ .

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Step 2.2.1.3.1

Multiply $- 2$ by $- 1$ .

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

$x = - 2 - 2 y$

Step 2.2.1.3.2

Combine $2$ and $\frac{1}{3}$ .

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

$x = - 2 - 2 y$

Step 2.2.1.3.3

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

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

$x = - 2 - 2 y$

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

$x = - 2 - 2 y$

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

$x = - 2 - 2 y$

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

$x = - 2 - 2 y$

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

$x = - 2 - 2 y$

Step 3

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

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

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

$x = - 2 - 2 y$

Step 3.1.2

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

$y \cdot \frac{3}{3} - \frac{2 y}{3} = \frac{2}{3}$

$x = - 2 - 2 y$

Step 3.1.3

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

$\frac{y \cdot 3}{3} - \frac{2 y}{3} = \frac{2}{3}$

$x = - 2 - 2 y$

Step 3.1.4

Combine the numerators over the common denominator.

$\frac{y \cdot 3 - 2 y}{3} = \frac{2}{3}$

$x = - 2 - 2 y$

Step 3.1.5

Simplify the numerator.

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Step 3.1.5.1

Move $3$ to the left of $y$ .

$\frac{3 \cdot y - 2 y}{3} = \frac{2}{3}$

$x = - 2 - 2 y$

Step 3.1.5.2

Subtract $2 y$ from $3 y$ .

$\frac{y}{3} = \frac{2}{3}$

$x = - 2 - 2 y$

$\frac{y}{3} = \frac{2}{3}$

$x = - 2 - 2 y$

$\frac{y}{3} = \frac{2}{3}$

$x = - 2 - 2 y$

Step 3.2

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

$y = 2$

$x = - 2 - 2 y$

$y = 2$

$x = - 2 - 2 y$

Step 4

Replace all occurrences of $y$ with $2$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = - 2 - 2 y$ with $2$ .

$x = - 2 - 2 \cdot 2$

$y = 2$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $- 2 - 2 \cdot 2$ .

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Step 4.2.1.1

Multiply $- 2$ by $2$ .

$x = - 2 - 4$

$y = 2$

Step 4.2.1.2

Subtract $4$ from $- 2$ .

$x = - 6$

$y = 2$

$x = - 6$

$y = 2$

$x = - 6$

$y = 2$

$x = - 6$

$y = 2$

Step 5

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

$(- 6, 2)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 6, 2)$

Equation Form:

$x = - 6, y = 2$

Step 7

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$