Algebra Examples

$- 2 x - 5 y = 9$ and $3 x + 11 y = 4$

Step 1

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

$(3) \cdot (- 2 x - 5 y) = (3) (9)$

$(2) \cdot (3 x + 11 y) = (2) (4)$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Simplify $(3) \cdot (- 2 x - 5 y)$ .

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Step 2.1.1.1

Apply the distributive property.

$3 (- 2 x) + 3 (- 5 y) = (3) (9)$

$(2) \cdot (3 x + 11 y) = (2) (4)$

Step 2.1.1.2

Multiply.

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Step 2.1.1.2.1

Multiply $- 2$ by $3$ .

$- 6 x + 3 (- 5 y) = (3) (9)$

$(2) \cdot (3 x + 11 y) = (2) (4)$

Step 2.1.1.2.2

Multiply $- 5$ by $3$ .

$- 6 x - 15 y = (3) (9)$

$(2) \cdot (3 x + 11 y) = (2) (4)$

$- 6 x - 15 y = (3) (9)$

$(2) \cdot (3 x + 11 y) = (2) (4)$

$- 6 x - 15 y = (3) (9)$

$(2) \cdot (3 x + 11 y) = (2) (4)$

$- 6 x - 15 y = (3) (9)$

$(2) \cdot (3 x + 11 y) = (2) (4)$

Step 2.2

Simplify the right side.

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Step 2.2.1

Multiply $3$ by $9$ .

$- 6 x - 15 y = 27$

$(2) \cdot (3 x + 11 y) = (2) (4)$

$- 6 x - 15 y = 27$

$(2) \cdot (3 x + 11 y) = (2) (4)$

Step 2.3

Simplify the left side.

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Step 2.3.1

Simplify $(2) \cdot (3 x + 11 y)$ .

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Step 2.3.1.1

Apply the distributive property.

$- 6 x - 15 y = 27$

$2 (3 x) + 2 (11 y) = (2) (4)$

Step 2.3.1.2

Multiply.

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Step 2.3.1.2.1

Multiply $3$ by $2$ .

$- 6 x - 15 y = 27$

$6 x + 2 (11 y) = (2) (4)$

Step 2.3.1.2.2

Multiply $11$ by $2$ .

$- 6 x - 15 y = 27$

$6 x + 22 y = (2) (4)$

$- 6 x - 15 y = 27$

$6 x + 22 y = (2) (4)$

$- 6 x - 15 y = 27$

$6 x + 22 y = (2) (4)$

$- 6 x - 15 y = 27$

$6 x + 22 y = (2) (4)$

Step 2.4

Simplify the right side.

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Step 2.4.1

Multiply $2$ by $4$ .

$- 6 x - 15 y = 27$

$6 x + 22 y = 8$

$- 6 x - 15 y = 27$

$6 x + 22 y = 8$

$- 6 x - 15 y = 27$

$6 x + 22 y = 8$

Step 3

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

	$-$	$6$	$x$	$-$	$1$	$5$	$y$	$=$	$2$	$7$
$+$		$6$	$x$	$+$	$2$	$2$	$y$	$=$		$8$
						$7$	$y$	$=$	$3$	$5$

Step 4

Divide each term in $7 y = 35$ by $7$ and simplify.

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Step 4.1

Divide each term in $7 y = 35$ by $7$ .

$\frac{7 y}{7} = \frac{35}{7}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $7$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{7 y}{7} = \frac{35}{7}$

Step 4.2.1.2

Divide $y$ by $1$ .

$y = \frac{35}{7}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $35$ by $7$ .

$y = 5$

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 $x$ .

$- 6 x - 15 \cdot 5 = 27$

Step 5.2

Multiply $- 15$ by $5$ .

$- 6 x - 75 = 27$

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 $75$ to both sides of the equation.

$- 6 x = 27 + 75$

Step 5.3.2

Add $27$ and $75$ .

$- 6 x = 102$

Step 5.4

Divide each term in $- 6 x = 102$ by $- 6$ and simplify.

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Step 5.4.1

Divide each term in $- 6 x = 102$ by $- 6$ .

$\frac{- 6 x}{- 6} = \frac{102}{- 6}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $- 6$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{- 6 x}{- 6} = \frac{102}{- 6}$

Step 5.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{102}{- 6}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Divide $102$ by $- 6$ .

$x = - 17$

Step 6

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

$(- 17, 5)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(- 17, 5)$

Equation Form:

$x = - 17, y = 5$

Step 8

	$-$	$6$	$x$	$-$	$1$	$5$	$y$	$=$	$2$	$7$
$+$		$6$	$x$	$+$	$2$	$2$	$y$	$=$		$8$
						$7$	$y$	$=$	$3$	$5$

	$-$	$6$	$x$	$-$	$1$	$5$	$y$	$=$	$2$	$7$
$+$		$6$	$x$	$+$	$2$	$2$	$y$	$=$		$8$
						$7$	$y$	$=$	$3$	$5$

	$-$	$6$	$x$	$-$	$1$	$5$	$y$	$=$	$2$	$7$
$+$		$6$	$x$	$+$	$2$	$2$	$y$	$=$		$8$
						$7$	$y$	$=$	$3$	$5$