Precalculus Examples

$- 3 x + 2 y = 10$ , $4 x - 7 y = 4$

Step 1

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

$(4) \cdot (- 3 x + 2 y) = (4) (10)$

$(3) \cdot (4 x - 7 y) = (3) (4)$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

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

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Step 2.1.1.1

Apply the distributive property.

$4 (- 3 x) + 4 (2 y) = (4) (10)$

$(3) \cdot (4 x - 7 y) = (3) (4)$

Step 2.1.1.2

Multiply.

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Step 2.1.1.2.1

Multiply $- 3$ by $4$ .

$- 12 x + 4 (2 y) = (4) (10)$

$(3) \cdot (4 x - 7 y) = (3) (4)$

Step 2.1.1.2.2

Multiply $2$ by $4$ .

$- 12 x + 8 y = (4) (10)$

$(3) \cdot (4 x - 7 y) = (3) (4)$

$- 12 x + 8 y = (4) (10)$

$(3) \cdot (4 x - 7 y) = (3) (4)$

$- 12 x + 8 y = (4) (10)$

$(3) \cdot (4 x - 7 y) = (3) (4)$

$- 12 x + 8 y = (4) (10)$

$(3) \cdot (4 x - 7 y) = (3) (4)$

Step 2.2

Simplify the right side.

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Step 2.2.1

Multiply $4$ by $10$ .

$- 12 x + 8 y = 40$

$(3) \cdot (4 x - 7 y) = (3) (4)$

$- 12 x + 8 y = 40$

$(3) \cdot (4 x - 7 y) = (3) (4)$

Step 2.3

Simplify the left side.

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Step 2.3.1

Simplify $(3) \cdot (4 x - 7 y)$ .

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Step 2.3.1.1

Apply the distributive property.

$- 12 x + 8 y = 40$

$3 (4 x) + 3 (- 7 y) = (3) (4)$

Step 2.3.1.2

Multiply.

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Step 2.3.1.2.1

Multiply $4$ by $3$ .

$- 12 x + 8 y = 40$

$12 x + 3 (- 7 y) = (3) (4)$

Step 2.3.1.2.2

Multiply $- 7$ by $3$ .

$- 12 x + 8 y = 40$

$12 x - 21 y = (3) (4)$

$- 12 x + 8 y = 40$

$12 x - 21 y = (3) (4)$

$- 12 x + 8 y = 40$

$12 x - 21 y = (3) (4)$

$- 12 x + 8 y = 40$

$12 x - 21 y = (3) (4)$

Step 2.4

Simplify the right side.

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Step 2.4.1

Multiply $3$ by $4$ .

$- 12 x + 8 y = 40$

$12 x - 21 y = 12$

$- 12 x + 8 y = 40$

$12 x - 21 y = 12$

$- 12 x + 8 y = 40$

$12 x - 21 y = 12$

Step 3

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

	$-$	$1$	$2$	$x$	$+$		$8$	$y$	$=$	$4$	$0$
$+$		$1$	$2$	$x$	$-$	$2$	$1$	$y$	$=$	$1$	$2$
					$-$	$1$	$3$	$y$	$=$	$5$	$2$

Step 4

Divide each term in $- 13 y = 52$ by $- 13$ and simplify.

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Step 4.1

Divide each term in $- 13 y = 52$ by $- 13$ .

$\frac{- 13 y}{- 13} = \frac{52}{- 13}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $- 13$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{- 13 y}{- 13} = \frac{52}{- 13}$

Step 4.2.1.2

Divide $y$ by $1$ .

$y = \frac{52}{- 13}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $52$ by $- 13$ .

$y = - 4$

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 $x$ .

$- 12 x + 8 (- 4) = 40$

Step 5.2

Multiply $8$ by $- 4$ .

$- 12 x - 32 = 40$

Step 5.3

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

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Step 5.3.1

Add $32$ to both sides of the equation.

$- 12 x = 40 + 32$

Step 5.3.2

Add $40$ and $32$ .

$- 12 x = 72$

Step 5.4

Divide each term in $- 12 x = 72$ by $- 12$ and simplify.

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Step 5.4.1

Divide each term in $- 12 x = 72$ by $- 12$ .

$\frac{- 12 x}{- 12} = \frac{72}{- 12}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $- 12$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{- 12 x}{- 12} = \frac{72}{- 12}$

Step 5.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{72}{- 12}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Divide $72$ by $- 12$ .

$x = - 6$

Step 6

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

$(- 6, - 4)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(- 6, - 4)$

Equation Form:

$x = - 6, y = - 4$

Step 8

	$-$	$1$	$2$	$x$	$+$		$8$	$y$	$=$	$4$	$0$
$+$		$1$	$2$	$x$	$-$	$2$	$1$	$y$	$=$	$1$	$2$
					$-$	$1$	$3$	$y$	$=$	$5$	$2$

	$-$	$1$	$2$	$x$	$+$		$8$	$y$	$=$	$4$	$0$
$+$		$1$	$2$	$x$	$-$	$2$	$1$	$y$	$=$	$1$	$2$
					$-$	$1$	$3$	$y$	$=$	$5$	$2$

	$-$	$1$	$2$	$x$	$+$		$8$	$y$	$=$	$4$	$0$
$+$		$1$	$2$	$x$	$-$	$2$	$1$	$y$	$=$	$1$	$2$
					$-$	$1$	$3$	$y$	$=$	$5$	$2$