Algebra Examples

$x + 3 y = 5$ $x = - 6 y + 14$

Step 1

Add $6 y$ to both sides of the equation.

$x + 3 y = 5, x + 6 y = 14$

Step 2

Multiply each equation by the value that makes the coefficients of $x$ opposite.

$x + 3 y = 5$

$(- 1) \cdot (x + 6 y) = (- 1) (14)$

Step 3

Simplify.

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Step 3.1

Simplify the left side.

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Step 3.1.1

Simplify $(- 1) \cdot (x + 6 y)$ .

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Step 3.1.1.1

Apply the distributive property.

$x + 3 y = 5$

$- 1 x - 1 (6 y) = (- 1) (14)$

Step 3.1.1.2

Simplify the expression.

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Step 3.1.1.2.1

Rewrite $- 1 x$ as $- x$ .

$x + 3 y = 5$

$- x - 1 (6 y) = (- 1) (14)$

Step 3.1.1.2.2

Multiply $6$ by $- 1$ .

$x + 3 y = 5$

$- x - 6 y = (- 1) (14)$

$x + 3 y = 5$

$- x - 6 y = (- 1) (14)$

$x + 3 y = 5$

$- x - 6 y = (- 1) (14)$

$x + 3 y = 5$

$- x - 6 y = (- 1) (14)$

Step 3.2

Simplify the right side.

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Step 3.2.1

Multiply $- 1$ by $14$ .

$x + 3 y = 5$

$- x - 6 y = - 14$

$x + 3 y = 5$

$- x - 6 y = - 14$

$x + 3 y = 5$

$- x - 6 y = - 14$

Step 4

Add the two equations together to eliminate $x$ from the system.

		$x$	$+$	$3$	$y$	$=$			$5$
$+$	$-$	$x$	$-$	$6$	$y$	$=$	$-$	$1$	$4$
			$-$	$3$	$y$	$=$		$-$	$9$

Step 5

Divide each term in $- 3 y = - 9$ by $- 3$ and simplify.

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Step 5.1

Divide each term in $- 3 y = - 9$ by $- 3$ .

$\frac{- 3 y}{- 3} = \frac{- 9}{- 3}$

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of $- 3$ .

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Step 5.2.1.1

Cancel the common factor.

$\frac{- 3 y}{- 3} = \frac{- 9}{- 3}$

Step 5.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 9}{- 3}$

Step 5.3

Simplify the right side.

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Step 5.3.1

Divide $- 9$ by $- 3$ .

$y = 3$

Step 6

Substitute the value found for $y$ into one of the original equations, then solve for $x$ .

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Step 6.1

Substitute the value found for $y$ into one of the original equations to solve for $x$ .

$x + 3 (3) = 5$

Step 6.2

Multiply $3$ by $3$ .

$x + 9 = 5$

Step 6.3

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

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Step 6.3.1

Subtract $9$ from both sides of the equation.

$x = 5 - 9$

Step 6.3.2

Subtract $9$ from $5$ .

$x = - 4$

Step 7

The solution to the independent system of equations can be represented as a point.

$(- 4, 3)$

Step 8

The result can be shown in multiple forms.

Point Form:

$(- 4, 3)$

Equation Form:

$x = - 4, y = 3$

Step 9

		$x$	$+$	$3$	$y$	$=$			$5$
$+$	$-$	$x$	$-$	$6$	$y$	$=$	$-$	$1$	$4$
			$-$	$3$	$y$	$=$		$-$	$9$

		$x$	$+$	$3$	$y$	$=$			$5$
$+$	$-$	$x$	$-$	$6$	$y$	$=$	$-$	$1$	$4$
			$-$	$3$	$y$	$=$		$-$	$9$

		$x$	$+$	$3$	$y$	$=$			$5$
$+$	$-$	$x$	$-$	$6$	$y$	$=$	$-$	$1$	$4$
			$-$	$3$	$y$	$=$		$-$	$9$