Algebra Examples

$2 x + y - 10 = 0$ $x - y - 4 = 0$

Step 1

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

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Step 1.1

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

$y - 10 = - 2 x$

$x - y - 4 = 0$

Step 1.2

Add $10$ to both sides of the equation.

$y = - 2 x + 10$

$x - y - 4 = 0$

$y = - 2 x + 10$

$x - y - 4 = 0$

Step 2

Replace all occurrences of $y$ with $- 2 x + 10$ in each equation.

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Step 2.1

Replace all occurrences of $y$ in $x - y - 4 = 0$ with $- 2 x + 10$ .

$x - (- 2 x + 10) - 4 = 0$

$y = - 2 x + 10$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $x - (- 2 x + 10) - 4$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Apply the distributive property.

$x - (- 2 x) - 1 \cdot 10 - 4 = 0$

$y = - 2 x + 10$

Step 2.2.1.1.2

Multiply $- 2$ by $- 1$ .

$x + 2 x - 1 \cdot 10 - 4 = 0$

$y = - 2 x + 10$

Step 2.2.1.1.3

Multiply $- 1$ by $10$ .

$x + 2 x - 10 - 4 = 0$

$y = - 2 x + 10$

$x + 2 x - 10 - 4 = 0$

$y = - 2 x + 10$

Step 2.2.1.2

Simplify by adding terms.

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Step 2.2.1.2.1

Add $x$ and $2 x$ .

$3 x - 10 - 4 = 0$

$y = - 2 x + 10$

Step 2.2.1.2.2

Subtract $4$ from $- 10$ .

$3 x - 14 = 0$

$y = - 2 x + 10$

$3 x - 14 = 0$

$y = - 2 x + 10$

$3 x - 14 = 0$

$y = - 2 x + 10$

$3 x - 14 = 0$

$y = - 2 x + 10$

$3 x - 14 = 0$

$y = - 2 x + 10$

Step 3

Solve for $x$ in $3 x - 14 = 0$ .

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Step 3.1

Add $14$ to both sides of the equation.

$3 x = 14$

$y = - 2 x + 10$

Step 3.2

Divide each term in $3 x = 14$ by $3$ and simplify.

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Step 3.2.1

Divide each term in $3 x = 14$ by $3$ .

$\frac{3 x}{3} = \frac{14}{3}$

$y = - 2 x + 10$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $3$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{14}{3}$

$y = - 2 x + 10$

Step 3.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{14}{3}$

$y = - 2 x + 10$

$x = \frac{14}{3}$

$y = - 2 x + 10$

$x = \frac{14}{3}$

$y = - 2 x + 10$

$x = \frac{14}{3}$

$y = - 2 x + 10$

$x = \frac{14}{3}$

$y = - 2 x + 10$

Step 4

Replace all occurrences of $x$ with $\frac{14}{3}$ in each equation.

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Step 4.1

Replace all occurrences of $x$ in $y = - 2 x + 10$ with $\frac{14}{3}$ .

$y = - 2 (\frac{14}{3}) + 10$

$x = \frac{14}{3}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $- 2 (\frac{14}{3}) + 10$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Multiply $- 2 (\frac{14}{3})$ .

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Step 4.2.1.1.1.1

Combine $- 2$ and $\frac{14}{3}$ .

$y = \frac{- 2 \cdot 14}{3} + 10$

$x = \frac{14}{3}$

Step 4.2.1.1.1.2

Multiply $- 2$ by $14$ .

$y = \frac{- 28}{3} + 10$

$x = \frac{14}{3}$

$y = \frac{- 28}{3} + 10$

$x = \frac{14}{3}$

Step 4.2.1.1.2

Move the negative in front of the fraction.

$y = - \frac{28}{3} + 10$

$x = \frac{14}{3}$

$y = - \frac{28}{3} + 10$

$x = \frac{14}{3}$

Step 4.2.1.2

To write $10$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = - \frac{28}{3} + 10 \cdot \frac{3}{3}$

$x = \frac{14}{3}$

Step 4.2.1.3

Combine $10$ and $\frac{3}{3}$ .

$y = - \frac{28}{3} + \frac{10 \cdot 3}{3}$

$x = \frac{14}{3}$

Step 4.2.1.4

Combine the numerators over the common denominator.

$y = \frac{- 28 + 10 \cdot 3}{3}$

$x = \frac{14}{3}$

Step 4.2.1.5

Simplify the numerator.

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Step 4.2.1.5.1

Multiply $10$ by $3$ .

$y = \frac{- 28 + 30}{3}$

$x = \frac{14}{3}$

Step 4.2.1.5.2

Add $- 28$ and $30$ .

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

$x = \frac{14}{3}$

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

$x = \frac{14}{3}$

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

$x = \frac{14}{3}$

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

$x = \frac{14}{3}$

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

$x = \frac{14}{3}$

Step 5

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

$(\frac{14}{3}, \frac{2}{3})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{14}{3}, \frac{2}{3})$

Equation Form:

$x = \frac{14}{3}, y = \frac{2}{3}$

Step 7