Algebra Examples

$y = 2 x + 3$ $x + y = - 9$

Step 1

Solve the system of equations.

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Step 1.1

Subtract $2 x$ from both sides of the equation.

$y - 2 x = 3, x + y = - 9$

Step 1.2

Reorder the polynomial.

$- 2 x + y = 3$

$x + y = - 9$

Step 1.3

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

$- 2 x + y = 3$

$(- 1) \cdot (x + y) = (- 1) (- 9)$

Step 1.4

Simplify.

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Step 1.4.1

Simplify the left side.

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Step 1.4.1.1

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

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Step 1.4.1.1.1

Apply the distributive property.

$- 2 x + y = 3$

$- 1 x - 1 y = (- 1) (- 9)$

Step 1.4.1.1.2

Rewrite negatives.

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Step 1.4.1.1.2.1

Rewrite $- 1 x$ as $- x$ .

$- 2 x + y = 3$

$- x - 1 y = (- 1) (- 9)$

Step 1.4.1.1.2.2

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

$- 2 x + y = 3$

$- x - y = (- 1) (- 9)$

$- 2 x + y = 3$

$- x - y = (- 1) (- 9)$

$- 2 x + y = 3$

$- x - y = (- 1) (- 9)$

$- 2 x + y = 3$

$- x - y = (- 1) (- 9)$

Step 1.4.2

Simplify the right side.

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Step 1.4.2.1

Multiply $- 1$ by $- 9$ .

$- 2 x + y = 3$

$- x - y = 9$

$- 2 x + y = 3$

$- x - y = 9$

$- 2 x + y = 3$

$- x - y = 9$

Step 1.5

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

	$-$	$2$	$x$	$+$	$y$	$=$	$3$
$+$		$-$	$x$	$-$	$y$	$=$	$9$
	$-$	$3$	$x$			$=$	$1$	$2$

Step 1.6

Divide each term in $- 3 x = 12$ by $- 3$ and simplify.

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Step 1.6.1

Divide each term in $- 3 x = 12$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{12}{- 3}$

Step 1.6.2

Simplify the left side.

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Step 1.6.2.1

Cancel the common factor of $- 3$ .

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Step 1.6.2.1.1

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{12}{- 3}$

Step 1.6.2.1.2

Divide $x$ by $1$ .

$x = \frac{12}{- 3}$

Step 1.6.3

Simplify the right side.

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Step 1.6.3.1

Divide $12$ by $- 3$ .

$x = - 4$

Step 1.7

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

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Step 1.7.1

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

$- 2 \cdot - 4 + y = 3$

Step 1.7.2

Multiply $- 2$ by $- 4$ .

$8 + y = 3$

Step 1.7.3

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

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Step 1.7.3.1

Subtract $8$ from both sides of the equation.

$y = 3 - 8$

Step 1.7.3.2

Subtract $8$ from $3$ .

$y = - 5$

Step 1.8

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

$(- 4, - 5)$

Step 2

Since the system has a point of intersection, the system is independent.

Independent

Step 3

	$-$	$2$	$x$	$+$	$y$	$=$	$3$
$+$		$-$	$x$	$-$	$y$	$=$	$9$
	$-$	$3$	$x$			$=$	$1$	$2$

	$-$	$2$	$x$	$+$	$y$	$=$	$3$
$+$		$-$	$x$	$-$	$y$	$=$	$9$
	$-$	$3$	$x$			$=$	$1$	$2$

	$-$	$2$	$x$	$+$	$y$	$=$	$3$
$+$		$-$	$x$	$-$	$y$	$=$	$9$
	$-$	$3$	$x$			$=$	$1$	$2$