Examples

x + y = 2

$x + y = 2$ ,

x - 2 y = 4

$x - 2 y = 4$

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

$x$ opposite.

x + y = 2

$x + y = 2$

(- 1) \cdot (x - 2 y) = (- 1) (4)

$(- 1) \cdot (x - 2 y) = (- 1) (4)$

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

(- 1) \cdot (x - 2 y)

$(- 1) \cdot (x - 2 y)$ .

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Step 1.2.1.1.1

Apply the distributive property.

x + y = 2

$x + y = 2$

- 1 x - 1 (- 2 y) = (- 1) (4)

$- 1 x - 1 (- 2 y) = (- 1) (4)$

Step 1.2.1.1.2

Simplify the expression.

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Step 1.2.1.1.2.1

Rewrite

- 1 x

$- 1 x$ as

- x

$- x$ .

x + y = 2

$x + y = 2$

- x - 1 (- 2 y) = (- 1) (4)

$- x - 1 (- 2 y) = (- 1) (4)$

Step 1.2.1.1.2.2

Multiply

- 2

$- 2$ by

- 1

$- 1$ .

x + y = 2

$x + y = 2$

- x + 2 y = (- 1) (4)

$- x + 2 y = (- 1) (4)$

x + y = 2

$x + y = 2$

- x + 2 y = (- 1) (4)

$- x + 2 y = (- 1) (4)$

x + y = 2

$x + y = 2$

- x + 2 y = (- 1) (4)

$- x + 2 y = (- 1) (4)$

x + y = 2

$x + y = 2$

- x + 2 y = (- 1) (4)

$- x + 2 y = (- 1) (4)$

Step 1.2.2

Simplify the right side.

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Step 1.2.2.1

Multiply

- 1

$- 1$ by

4

$4$ .

x + y = 2

$x + y = 2$

- x + 2 y = - 4

$- x + 2 y = - 4$

x + y = 2

$x + y = 2$

- x + 2 y = - 4

$- x + 2 y = - 4$

x + y = 2

$x + y = 2$

- x + 2 y = - 4

$- x + 2 y = - 4$

Step 1.3

Add the two equations together to eliminate

x

$x$ from the system.

		$x$ $x$	$+$ $+$		$y$ $y$	$=$ $=$		$2$ $2$
$+$ $+$	$-$ $-$	$x$ $x$	$+$ $+$	$2$ $2$	$y$ $y$	$=$ $=$	$-$ $-$	$4$ $4$
				$3$ $3$	$y$ $y$	$=$ $=$	$-$ $-$	$2$ $2$

Step 1.4

Divide each term in

3 y = - 2

$3 y = - 2$ by

3

$3$ and simplify.

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Step 1.4.1

Divide each term in

3 y = - 2

$3 y = - 2$ by

3

$3$ .

\frac{3 y}{3} = \frac{- 2}{3}

$\frac{3 y}{3} = \frac{- 2}{3}$

Step 1.4.2

Simplify the left side.

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Step 1.4.2.1

Cancel the common factor of

3

$3$ .

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Step 1.4.2.1.1

Cancel the common factor.

$\frac{3 y}{3} = \frac{- 2}{3}$

Step 1.4.2.1.2

Divide

$y$ by

$1$ .

$y = \frac{- 2}{3}$

Step 1.4.3

Simplify the right side.

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Step 1.4.3.1

Move the negative in front of the fraction.

$y = - \frac{2}{3}$

Step 1.5

Substitute the value found for

$y$ into one of the original equations, then solve for

$x$ .

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Step 1.5.1

Substitute the value found for

$y$ into one of the original equations to solve for

$x$ .

$x - \frac{2}{3} = 2$

Step 1.5.2

Move all terms not containing

$x$ to the right side of the equation.

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Step 1.5.2.1

Add

$\frac{2}{3}$ to both sides of the equation.

$x = 2 + \frac{2}{3}$

Step 1.5.2.2

To write

$2$ as a fraction with a common denominator, multiply by

$\frac{3}{3}$ .

$x = 2 \cdot \frac{3}{3} + \frac{2}{3}$

Step 1.5.2.3

Combine

$2$ and

$\frac{3}{3}$ .

$x = \frac{2 \cdot 3}{3} + \frac{2}{3}$

Step 1.5.2.4

Combine the numerators over the common denominator.

$x = \frac{2 \cdot 3 + 2}{3}$

Step 1.5.2.5

Simplify the numerator.

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Step 1.5.2.5.1

Multiply

$2$ by

$3$ .

$x = \frac{6 + 2}{3}$

Step 1.5.2.5.2

Add

$6$ and

$2$ .

$x = \frac{8}{3}$

Step 1.6

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

$(\frac{8}{3}, - \frac{2}{3})$

Step 2

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

Independent

Step 3

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