Algebra Examples

$y = - 2 x$ , $y = 4 x + 1$

Step 1

Eliminate the equal sides of each equation and combine.

$- 2 x = 4 x + 1$

Step 2

Solve $- 2 x = 4 x + 1$ for $x$ .

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Step 2.1

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

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Step 2.1.1

Subtract $4 x$ from both sides of the equation.

$- 2 x - 4 x = 1$

Step 2.1.2

Subtract $4 x$ from $- 2 x$ .

$- 6 x = 1$

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $- 6$ .

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Step 2.2.2.1.1

Cancel the common factor.

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

Step 2.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Move the negative in front of the fraction.

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

Step 3

Evaluate $y$ when $x = - \frac{1}{6}$ .

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Step 3.1

Substitute $- \frac{1}{6}$ for $x$ .

$y = 4 (- \frac{1}{6}) + 1$

Step 3.2

Simplify $4 (- \frac{1}{6}) + 1$ .

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Cancel the common factor of $2$ .

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Step 3.2.1.1.1

Move the leading negative in $- \frac{1}{6}$ into the numerator.

$y = 4 (\frac{- 1}{6}) + 1$

Step 3.2.1.1.2

Factor $2$ out of $4$ .

$y = 2 (2) \frac{- 1}{6} + 1$

Step 3.2.1.1.3

Factor $2$ out of $6$ .

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

Step 3.2.1.1.4

Cancel the common factor.

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

Step 3.2.1.1.5

Rewrite the expression.

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

Step 3.2.1.2

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

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

Step 3.2.1.3

Multiply $2$ by $- 1$ .

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

Step 3.2.1.4

Move the negative in front of the fraction.

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

Step 3.2.2

Simplify the expression.

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Step 3.2.2.1

Write $1$ as a fraction with a common denominator.

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

Step 3.2.2.2

Combine the numerators over the common denominator.

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

Step 3.2.2.3

Add $- 2$ and $3$ .

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

Step 4

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

$(- \frac{1}{6}, \frac{1}{3})$

Step 5

The result can be shown in multiple forms.

Point Form:

$(- \frac{1}{6}, \frac{1}{3})$

Equation Form:

$x = - \frac{1}{6}, y = \frac{1}{3}$

Step 6