Algebra Examples

$5 x + y = 2$ $4 x + y = 4$

Step 1

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

$5 x + y = 2$

$(- 1) \cdot (4 x + y) = (- 1) (4)$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Simplify $(- 1) \cdot (4 x + y)$ .

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Step 2.1.1.1

Apply the distributive property.

$5 x + y = 2$

$- 1 (4 x) - 1 y = (- 1) (4)$

Step 2.1.1.2

Simplify the expression.

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Step 2.1.1.2.1

Multiply $4$ by $- 1$ .

$5 x + y = 2$

$- 4 x - 1 y = (- 1) (4)$

Step 2.1.1.2.2

Rewrite $- 1 y$ as $- y$ .

$5 x + y = 2$

$- 4 x - y = (- 1) (4)$

$5 x + y = 2$

$- 4 x - y = (- 1) (4)$

$5 x + y = 2$

$- 4 x - y = (- 1) (4)$

$5 x + y = 2$

$- 4 x - y = (- 1) (4)$

Step 2.2

Simplify the right side.

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Step 2.2.1

Multiply $- 1$ by $4$ .

$5 x + y = 2$

$- 4 x - y = - 4$

$5 x + y = 2$

$- 4 x - y = - 4$

$5 x + y = 2$

$- 4 x - y = - 4$

Step 3

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

		$5$	$x$	$+$	$y$	$=$		$2$
$+$	$-$	$4$	$x$	$-$	$y$	$=$	$-$	$4$
			$x$			$=$	$-$	$2$

Step 4

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

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Step 4.1

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

$5 (- 2) + y = 2$

Step 4.2

Multiply $5$ by $- 2$ .

$- 10 + y = 2$

Step 4.3

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

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Step 4.3.1

Add $10$ to both sides of the equation.

$y = 2 + 10$

Step 4.3.2

Add $2$ and $10$ .

$y = 12$

Step 5

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

$(- 2, 12)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 2, 12)$

Equation Form:

$x = - 2, y = 12$

Step 7