Precalculus Examples

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

Solve the system of equations.

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

$x + y = 2$

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

Simplify.

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Simplify the left side.

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Apply the distributive property.

$x + y = 2$

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

Simplify the expression.

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

$x + y = 2$

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

Multiply $- 2$ by $- 1$ .

$x + y = 2$

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

$x + y = 2$

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

$x + y = 2$

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

Multiply $- 1$ by $4$ .

$x + y = 2$

$- x + 2 y = - 4$

$x + y = 2$

$- x + 2 y = - 4$

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

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

Divide each term by $3$ and simplify.

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

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

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

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

Divide $y$ by $1$ .

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

Move the negative in front of the fraction.

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

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

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

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

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

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

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

Simplify the right side of the equation.

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To write $\frac{2}{1}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Write each expression with a common denominator of $3$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

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

Multiply $3$ by $1$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

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Multiply $2$ by $3$ .

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

Add $6$ and $2$ .

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

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

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

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

Independent

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