Algebra Examples

$3 x - 7 y = 18$ , $- 5 x + 4 y = - 30$

Step 1

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

$(5) \cdot (3 x - 7 y) = (5) (18)$

$(3) \cdot (- 5 x + 4 y) = (3) (- 30)$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Simplify $(5) \cdot (3 x - 7 y)$ .

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Step 2.1.1.1

Apply the distributive property.

$5 (3 x) + 5 (- 7 y) = (5) (18)$

$(3) \cdot (- 5 x + 4 y) = (3) (- 30)$

Step 2.1.1.2

Multiply.

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Step 2.1.1.2.1

Multiply $3$ by $5$ .

$15 x + 5 (- 7 y) = (5) (18)$

$(3) \cdot (- 5 x + 4 y) = (3) (- 30)$

Step 2.1.1.2.2

Multiply $- 7$ by $5$ .

$15 x - 35 y = (5) (18)$

$(3) \cdot (- 5 x + 4 y) = (3) (- 30)$

$15 x - 35 y = (5) (18)$

$(3) \cdot (- 5 x + 4 y) = (3) (- 30)$

$15 x - 35 y = (5) (18)$

$(3) \cdot (- 5 x + 4 y) = (3) (- 30)$

$15 x - 35 y = (5) (18)$

$(3) \cdot (- 5 x + 4 y) = (3) (- 30)$

Step 2.2

Simplify the right side.

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Step 2.2.1

Multiply $5$ by $18$ .

$15 x - 35 y = 90$

$(3) \cdot (- 5 x + 4 y) = (3) (- 30)$

$15 x - 35 y = 90$

$(3) \cdot (- 5 x + 4 y) = (3) (- 30)$

Step 2.3

Simplify the left side.

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Step 2.3.1

Simplify $(3) \cdot (- 5 x + 4 y)$ .

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Step 2.3.1.1

Apply the distributive property.

$15 x - 35 y = 90$

$3 (- 5 x) + 3 (4 y) = (3) (- 30)$

Step 2.3.1.2

Multiply.

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Step 2.3.1.2.1

Multiply $- 5$ by $3$ .

$15 x - 35 y = 90$

$- 15 x + 3 (4 y) = (3) (- 30)$

Step 2.3.1.2.2

Multiply $4$ by $3$ .

$15 x - 35 y = 90$

$- 15 x + 12 y = (3) (- 30)$

$15 x - 35 y = 90$

$- 15 x + 12 y = (3) (- 30)$

$15 x - 35 y = 90$

$- 15 x + 12 y = (3) (- 30)$

$15 x - 35 y = 90$

$- 15 x + 12 y = (3) (- 30)$

Step 2.4

Simplify the right side.

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Step 2.4.1

Multiply $3$ by $- 30$ .

$15 x - 35 y = 90$

$- 15 x + 12 y = - 90$

$15 x - 35 y = 90$

$- 15 x + 12 y = - 90$

$15 x - 35 y = 90$

$- 15 x + 12 y = - 90$

Step 3

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

		$1$	$5$	$x$	$-$	$3$	$5$	$y$	$=$		$9$	$0$
$+$	$-$	$1$	$5$	$x$	$+$	$1$	$2$	$y$	$=$	$-$	$9$	$0$
					$-$	$2$	$3$	$y$	$=$			$0$

Step 4

Divide each term in $- 23 y = 0$ by $- 23$ and simplify.

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Step 4.1

Divide each term in $- 23 y = 0$ by $- 23$ .

$\frac{- 23 y}{- 23} = \frac{0}{- 23}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $- 23$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{- 23 y}{- 23} = \frac{0}{- 23}$

Step 4.2.1.2

Divide $y$ by $1$ .

$y = \frac{0}{- 23}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $- 23$ .

$y = 0$

Step 5

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

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Step 5.1

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

$15 x - 35 \cdot 0 = 90$

Step 5.2

Simplify $15 x - 35 \cdot 0$ .

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Step 5.2.1

Multiply $- 35$ by $0$ .

$15 x + 0 = 90$

Step 5.2.2

Add $15 x$ and $0$ .

$15 x = 90$

Step 5.3

Divide each term in $15 x = 90$ by $15$ and simplify.

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Step 5.3.1

Divide each term in $15 x = 90$ by $15$ .

$\frac{15 x}{15} = \frac{90}{15}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $15$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{15 x}{15} = \frac{90}{15}$

Step 5.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{90}{15}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Divide $90$ by $15$ .

$x = 6$

Step 6

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

$(6, 0)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(6, 0)$

Equation Form:

$x = 6, y = 0$

Step 8

		$1$	$5$	$x$	$-$	$3$	$5$	$y$	$=$		$9$	$0$
$+$	$-$	$1$	$5$	$x$	$+$	$1$	$2$	$y$	$=$	$-$	$9$	$0$
					$-$	$2$	$3$	$y$	$=$			$0$

		$1$	$5$	$x$	$-$	$3$	$5$	$y$	$=$		$9$	$0$
$+$	$-$	$1$	$5$	$x$	$+$	$1$	$2$	$y$	$=$	$-$	$9$	$0$
					$-$	$2$	$3$	$y$	$=$			$0$

		$1$	$5$	$x$	$-$	$3$	$5$	$y$	$=$		$9$	$0$
$+$	$-$	$1$	$5$	$x$	$+$	$1$	$2$	$y$	$=$	$-$	$9$	$0$
					$-$	$2$	$3$	$y$	$=$			$0$