Precalculus Examples

$3 x + y = 4$ , $6 x - 7 y = 2$

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

$(- 2) \cdot (3 x + y) = (- 2) (4)$

$6 x - 7 y = 2$

Simplify.

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Simplify $(- 2) \cdot (3 x + y)$ .

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Apply the distributive property.

$- 2 (3 x) - 2 y = (- 2) (4)$

$6 x - 7 y = 2$

Multiply $3$ by $- 2$ .

$- 6 x - 2 y = (- 2) (4)$

$6 x - 7 y = 2$

$- 6 x - 2 y = (- 2) (4)$

$6 x - 7 y = 2$

Multiply $- 2$ by $4$ .

$- 6 x - 2 y = - 8$

$6 x - 7 y = 2$

$- 6 x - 2 y = - 8$

$6 x - 7 y = 2$

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

	$-$	$6$	$x$	$-$	$2$	$y$	$=$	$-$	$8$
$+$		$6$	$x$	$-$	$7$	$y$	$=$		$2$
				$-$	$9$	$y$	$=$	$-$	$6$

Divide each term by $- 9$ and simplify.

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Divide each term in $- 9 y = - 6$ by $- 9$ .

$\frac{- 9 y}{- 9} = \frac{- 6}{- 9}$

Cancel the common factor of $- 9$ .

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Cancel the common factor.

$\frac{- 9 y}{- 9} = \frac{- 6}{- 9}$

Divide $y$ by $1$ .

$y = \frac{- 6}{- 9}$

Cancel the common factor of $- 6$ and $- 9$ .

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Factor $- 3$ out of $- 6$ .

$y = \frac{- 3 (2)}{- 9}$

Cancel the common factors.

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Factor $- 3$ out of $- 9$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

$- 6 x - 2 (\frac{2}{3}) = - 8$

Simplify each term.

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

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

$- 6 x + \frac{- 2 \cdot 2}{3} = - 8$

Multiply $- 2$ by $2$ .

$- 6 x + \frac{- 4}{3} = - 8$

Move the negative in front of the fraction.

$- 6 x - \frac{4}{3} = - 8$

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

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Add $\frac{4}{3}$ to both sides of the equation.

$- 6 x = - 8 + \frac{4}{3}$

To write $- 8$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- 6 x = - 8 \cdot \frac{3}{3} + \frac{4}{3}$

Combine $- 8$ and $\frac{3}{3}$ .

$- 6 x = \frac{- 8 \cdot 3}{3} + \frac{4}{3}$

Combine the numerators over the common denominator.

$- 6 x = \frac{- 8 \cdot 3 + 4}{3}$

Simplify the numerator.

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Multiply $- 8$ by $3$ .

$- 6 x = \frac{- 24 + 4}{3}$

Add $- 24$ and $4$ .

$- 6 x = \frac{- 20}{3}$

Move the negative in front of the fraction.

$- 6 x = - \frac{20}{3}$

Divide each term by $- 6$ and simplify.

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Divide each term in $- 6 x = - \frac{20}{3}$ by $- 6$ .

$\frac{- 6 x}{- 6} = - \frac{20}{3} \cdot (- \frac{1}{6})$

Cancel the common factor of $- 6$ .

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Cancel the common factor.

$\frac{- 6 x}{- 6} = - \frac{20}{3} \cdot (- \frac{1}{6})$

Divide $x$ by $1$ .

$x = - \frac{20}{3} \cdot (- \frac{1}{6})$

Simplify $- \frac{20}{3} \cdot (- \frac{1}{6})$ .

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Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{20}{3}$ into the numerator.

$x = \frac{- 20}{3} \cdot (- \frac{1}{6})$

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

$x = \frac{- 20}{3} \cdot \frac{- 1}{6}$

Factor $2$ out of $- 20$ .

$x = \frac{2 (- 10)}{3} \cdot \frac{- 1}{6}$

Factor $2$ out of $6$ .

$x = \frac{2 \cdot - 10}{3} \cdot \frac{- 1}{2 \cdot 3}$

Cancel the common factor.

$x = \frac{2 \cdot - 10}{3} \cdot \frac{- 1}{2 \cdot 3}$

Rewrite the expression.

$x = \frac{- 10}{3} \cdot \frac{- 1}{3}$

Multiply $\frac{- 10}{3}$ and $\frac{- 1}{3}$ .

$x = \frac{- 10 \cdot - 1}{3 \cdot 3}$

Multiply.

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Multiply $- 10$ by $- 1$ .

$x = \frac{10}{3 \cdot 3}$

Multiply $3$ by $3$ .

$x = \frac{10}{9}$

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

$(\frac{10}{9}, \frac{2}{3})$

The result can be shown in multiple forms.

Point Form:

$(\frac{10}{9}, \frac{2}{3})$

Equation Form:

$x = \frac{10}{9}, y = \frac{2}{3}$

Enter YOUR Problem

	$-$	$6$	$x$	$-$	$2$	$y$	$=$	$-$	$8$
$+$		$6$	$x$	$-$	$7$	$y$	$=$		$2$
				$-$	$9$	$y$	$=$	$-$	$6$

	$-$	$6$	$x$	$-$	$2$	$y$	$=$	$-$	$8$
$+$		$6$	$x$	$-$	$7$	$y$	$=$		$2$
				$-$	$9$	$y$	$=$	$-$	$6$

	$-$	$6$	$x$	$-$	$2$	$y$	$=$	$-$	$8$
$+$		$6$	$x$	$-$	$7$	$y$	$=$		$2$
				$-$	$9$	$y$	$=$	$-$	$6$