Algebra Examples

$y - x = 2$ $y = - 14 x + 7$

Step 1

Simplify the left side.

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Step 1.1

Reorder $y$ and $- x$ .

$- x + y = 2$

$y = - 14 x + 7$

$- x + y = 2$

$y = - 14 x + 7$

Step 2

Add $14 x$ to both sides of the equation.

$- x + y = 2, y + 14 x = 7$

Step 3

Reorder the polynomial.

$14 x + y = 7$

$- x + y = 2$

Step 4

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

$- x + y = 2$

$(- 1) \cdot (14 x + y) = (- 1) (7)$

Step 5

Simplify.

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Step 5.1

Simplify the left side.

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Step 5.1.1

Simplify $(- 1) \cdot (14 x + y)$ .

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Step 5.1.1.1

Apply the distributive property.

$- x + y = 2$

$- 1 (14 x) - 1 y = (- 1) (7)$

Step 5.1.1.2

Simplify the expression.

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Step 5.1.1.2.1

Multiply $14$ by $- 1$ .

$- x + y = 2$

$- 14 x - 1 y = (- 1) (7)$

Step 5.1.1.2.2

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

$- x + y = 2$

$- 14 x - y = (- 1) (7)$

$- x + y = 2$

$- 14 x - y = (- 1) (7)$

$- x + y = 2$

$- 14 x - y = (- 1) (7)$

$- x + y = 2$

$- 14 x - y = (- 1) (7)$

Step 5.2

Simplify the right side.

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Step 5.2.1

Multiply $- 1$ by $7$ .

$- x + y = 2$

$- 14 x - y = - 7$

$- x + y = 2$

$- 14 x - y = - 7$

$- x + y = 2$

$- 14 x - y = - 7$

Step 6

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

			$-$	$x$	$+$	$y$	$=$		$2$
$+$	$-$	$1$	$4$	$x$	$-$	$y$	$=$	$-$	$7$
	$-$	$1$	$5$	$x$			$=$	$-$	$5$

Step 7

Divide each term in $- 15 x = - 5$ by $- 15$ and simplify.

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Step 7.1

Divide each term in $- 15 x = - 5$ by $- 15$ .

$\frac{- 15 x}{- 15} = \frac{- 5}{- 15}$

Step 7.2

Simplify the left side.

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Step 7.2.1

Cancel the common factor of $- 15$ .

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Step 7.2.1.1

Cancel the common factor.

$\frac{- 15 x}{- 15} = \frac{- 5}{- 15}$

Step 7.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 5}{- 15}$

Step 7.3

Simplify the right side.

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Step 7.3.1

Cancel the common factor of $- 5$ and $- 15$ .

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Step 7.3.1.1

Factor $- 5$ out of $- 5$ .

$x = \frac{- 5 (1)}{- 15}$

Step 7.3.1.2

Cancel the common factors.

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Step 7.3.1.2.1

Factor $- 5$ out of $- 15$ .

$x = \frac{- 5 \cdot 1}{- 5 \cdot 3}$

Step 7.3.1.2.2

Cancel the common factor.

$x = \frac{- 5 \cdot 1}{- 5 \cdot 3}$

Step 7.3.1.2.3

Rewrite the expression.

$x = \frac{1}{3}$

Step 8

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

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Step 8.1

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

$- (\frac{1}{3}) + y = 2$

Step 8.2

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

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Step 8.2.1

Add $\frac{1}{3}$ to both sides of the equation.

$y = 2 + \frac{1}{3}$

Step 8.2.2

To write $2$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = 2 \cdot \frac{3}{3} + \frac{1}{3}$

Step 8.2.3

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

$y = \frac{2 \cdot 3}{3} + \frac{1}{3}$

Step 8.2.4

Combine the numerators over the common denominator.

$y = \frac{2 \cdot 3 + 1}{3}$

Step 8.2.5

Simplify the numerator.

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Step 8.2.5.1

Multiply $2$ by $3$ .

$y = \frac{6 + 1}{3}$

Step 8.2.5.2

Add $6$ and $1$ .

$y = \frac{7}{3}$

Step 9

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

$(\frac{1}{3}, \frac{7}{3})$

Step 10

The result can be shown in multiple forms.

Point Form:

$(\frac{1}{3}, \frac{7}{3})$

Equation Form:

$x = \frac{1}{3}, y = \frac{7}{3}$

Step 11

			$-$	$x$	$+$	$y$	$=$		$2$
$+$	$-$	$1$	$4$	$x$	$-$	$y$	$=$	$-$	$7$
	$-$	$1$	$5$	$x$			$=$	$-$	$5$

			$-$	$x$	$+$	$y$	$=$		$2$
$+$	$-$	$1$	$4$	$x$	$-$	$y$	$=$	$-$	$7$
	$-$	$1$	$5$	$x$			$=$	$-$	$5$

			$-$	$x$	$+$	$y$	$=$		$2$
$+$	$-$	$1$	$4$	$x$	$-$	$y$	$=$	$-$	$7$
	$-$	$1$	$5$	$x$			$=$	$-$	$5$