Algebra Examples

$y = - x + 1$ $y = 4 x - 14$

Step 1

Move all terms containing variables to the left.

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Step 1.1

Add $x$ to both sides of the equation.

$y + x = 1, y = 4 x - 14$

Step 1.2

Subtract $4 x$ from both sides of the equation.

$y + x = 1, y - 4 x = - 14$

Step 2

Reorder the polynomial.

$x + y = 1$

$y - 4 x = - 14$

Step 3

Reorder the polynomial.

$- 4 x + y = - 14$

$x + y = 1$

Step 4

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

$x + y = 1$

$(- 1) \cdot (- 4 x + y) = (- 1) (- 14)$

Step 5

Simplify.

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Step 5.1

Simplify the left side.

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Step 5.1.1

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

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Step 5.1.1.1

Apply the distributive property.

$x + y = 1$

$- 1 (- 4 x) - 1 y = (- 1) (- 14)$

Step 5.1.1.2

Simplify the expression.

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Step 5.1.1.2.1

Multiply $- 4$ by $- 1$ .

$x + y = 1$

$4 x - 1 y = (- 1) (- 14)$

Step 5.1.1.2.2

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

$x + y = 1$

$4 x - y = (- 1) (- 14)$

$x + y = 1$

$4 x - y = (- 1) (- 14)$

$x + y = 1$

$4 x - y = (- 1) (- 14)$

$x + y = 1$

$4 x - y = (- 1) (- 14)$

Step 5.2

Simplify the right side.

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Step 5.2.1

Multiply $- 1$ by $- 14$ .

$x + y = 1$

$4 x - y = 14$

$x + y = 1$

$4 x - y = 14$

$x + y = 1$

$4 x - y = 14$

Step 6

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

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

Step 7

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

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Step 7.1

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

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

Step 7.2

Simplify the left side.

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Step 7.2.1

Cancel the common factor of $5$ .

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Step 7.2.1.1

Cancel the common factor.

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

Step 7.2.1.2

Divide $x$ by $1$ .

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

Step 7.3

Simplify the right side.

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Step 7.3.1

Divide $15$ by $5$ .

$x = 3$

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 $y$ .

$(3) + y = 1$

Step 8.2

Remove parentheses.

$3 + y = 1$

Step 8.3

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

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Step 8.3.1

Subtract $3$ from both sides of the equation.

$y = 1 - 3$

Step 8.3.2

Subtract $3$ from $1$ .

$y = - 2$

Step 9

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

$(3, - 2)$

Step 10

The result can be shown in multiple forms.

Point Form:

$(3, - 2)$

Equation Form:

$x = 3, y = - 2$

Step 11