Algebra Examples

$x = 4 y$ $x + 2 y = 12$

Step 1

Replace all occurrences of $x$ with $4 y$ in each equation.

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Step 1.1

Replace all occurrences of $x$ in $x + 2 y = 12$ with $4 y$ .

$(4 y) + 2 y = 12$

$x = 4 y$

Step 1.2

Simplify the left side.

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Step 1.2.1

Add $4 y$ and $2 y$ .

$6 y = 12$

$x = 4 y$

$6 y = 12$

$x = 4 y$

$6 y = 12$

$x = 4 y$

Step 2

Divide each term in $6 y = 12$ by $6$ and simplify.

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Step 2.1

Divide each term in $6 y = 12$ by $6$ .

$\frac{6 y}{6} = \frac{12}{6}$

$x = 4 y$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $6$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{6 y}{6} = \frac{12}{6}$

$x = 4 y$

Step 2.2.1.2

Divide $y$ by $1$ .

$y = \frac{12}{6}$

$x = 4 y$

$y = \frac{12}{6}$

$x = 4 y$

$y = \frac{12}{6}$

$x = 4 y$

Step 2.3

Simplify the right side.

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Step 2.3.1

Divide $12$ by $6$ .

$y = 2$

$x = 4 y$

$y = 2$

$x = 4 y$

$y = 2$

$x = 4 y$

Step 3

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

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Step 3.1

Replace all occurrences of $y$ in $x = 4 y$ with $2$ .

$x = 4 (2)$

$y = 2$

Step 3.2

Simplify the right side.

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Step 3.2.1

Multiply $4$ by $2$ .

$x = 8$

$y = 2$

$x = 8$

$y = 2$

$x = 8$

$y = 2$

Step 4

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

$(8, 2)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(8, 2)$

Equation Form:

$x = 8, y = 2$

Step 6