Precalculus Examples

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

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 $x$ opposite.

$x + y = 2$

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

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

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Step 1.2.1.1.1

Apply the distributive property.

$x + y = 2$

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

Step 1.2.1.1.2

Simplify the expression.

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Step 1.2.1.1.2.1

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

$x + y = 2$

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

Step 1.2.1.1.2.2

Multiply $- 2$ by $- 1$ .

$x + y = 2$

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

$x + y = 2$

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

$x + y = 2$

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

$x + y = 2$

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

Step 1.2.2

Simplify the right side.

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Step 1.2.2.1

Multiply $- 1$ by $4$ .

$x + y = 2$

$- x + 2 y = - 4$

$x + y = 2$

$- x + 2 y = - 4$

$x + y = 2$

$- x + 2 y = - 4$

Step 1.3

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

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

Step 1.4

Divide each term in $3 y = - 2$ by $3$ and simplify.

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Step 1.4.1

Divide each term in $3 y = - 2$ by $3$ .

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

Step 1.4.2

Simplify the left side.

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Step 1.4.2.1

Cancel the common factor of $3$ .

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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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