Algebra Examples

$- x + y = 8$ , $2 x - 2 y = - 16$

Step 1

Solve the system of equations.

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Step 1.1

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

$(2) \cdot (- x + y) = (2) (8)$

$2 x - 2 y = - 16$

Step 1.2

Simplify.

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Step 1.2.1

Simplify the left side.

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Step 1.2.1.1

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

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Step 1.2.1.1.1

Apply the distributive property.

$2 (- x) + 2 y = (2) (8)$

$2 x - 2 y = - 16$

Step 1.2.1.1.2

Multiply $- 1$ by $2$ .

$- 2 x + 2 y = (2) (8)$

$2 x - 2 y = - 16$

$- 2 x + 2 y = (2) (8)$

$2 x - 2 y = - 16$

$- 2 x + 2 y = (2) (8)$

$2 x - 2 y = - 16$

Step 1.2.2

Simplify the right side.

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Step 1.2.2.1

Multiply $2$ by $8$ .

$- 2 x + 2 y = 16$

$2 x - 2 y = - 16$

$- 2 x + 2 y = 16$

$2 x - 2 y = - 16$

$- 2 x + 2 y = 16$

$2 x - 2 y = - 16$

Step 1.3

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

	$-$	$2$	$x$	$+$	$2$	$y$	$=$		$1$	$6$
$+$		$2$	$x$	$-$	$2$	$y$	$=$	$-$	$1$	$6$
						$0$	$=$			$0$

Step 1.4

Since $0 = 0$ , the equations intersect at an infinite number of points.

Infinite number of solutions

Step 1.5

Solve one of the equations for $y$ .

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Step 1.5.1

Add $2 x$ to both sides of the equation.

$2 y = 16 + 2 x$

Step 1.5.2

Divide each term in $2 y = 16 + 2 x$ by $2$ and simplify.

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Step 1.5.2.1

Divide each term in $2 y = 16 + 2 x$ by $2$ .

$\frac{2 y}{2} = \frac{16}{2} + \frac{2 x}{2}$

Step 1.5.2.2

Simplify the left side.

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Step 1.5.2.2.1

Cancel the common factor of $2$ .

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Step 1.5.2.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{16}{2} + \frac{2 x}{2}$

Step 1.5.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{16}{2} + \frac{2 x}{2}$

Step 1.5.2.3

Simplify the right side.

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Step 1.5.2.3.1

Simplify each term.

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Step 1.5.2.3.1.1

Divide $16$ by $2$ .

$y = 8 + \frac{2 x}{2}$

Step 1.5.2.3.1.2

Cancel the common factor of $2$ .

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Step 1.5.2.3.1.2.1

Cancel the common factor.

$y = 8 + \frac{2 x}{2}$

Step 1.5.2.3.1.2.2

Divide $x$ by $1$ .

$y = 8 + x$

Step 1.6

The solution is the set of ordered pairs that make $y = 8 + x$ true.

$(x, 8 + x)$

Step 2

Since the system is always true, the equations are equal and the graphs are the same line. Thus, the system is dependent.

Dependent

Step 3

Enter YOUR Problem

	$-$	$2$	$x$	$+$	$2$	$y$	$=$		$1$	$6$
$+$		$2$	$x$	$-$	$2$	$y$	$=$	$-$	$1$	$6$
						$0$	$=$			$0$

	$-$	$2$	$x$	$+$	$2$	$y$	$=$		$1$	$6$
$+$		$2$	$x$	$-$	$2$	$y$	$=$	$-$	$1$	$6$
						$0$	$=$			$0$

	$-$	$2$	$x$	$+$	$2$	$y$	$=$		$1$	$6$
$+$		$2$	$x$	$-$	$2$	$y$	$=$	$-$	$1$	$6$
						$0$	$=$			$0$