Precalculus Examples

$4 x - 9 y = 21$ , $12 x - 27 y = 63$

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.

$(- 3) \cdot (4 x - 9 y) = (- 3) (21)$

$12 x - 27 y = 63$

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 $(- 3) \cdot (4 x - 9 y)$ .

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Step 1.2.1.1.1

Apply the distributive property.

$- 3 (4 x) - 3 (- 9 y) = (- 3) (21)$

$12 x - 27 y = 63$

Step 1.2.1.1.2

Multiply.

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Step 1.2.1.1.2.1

Multiply $4$ by $- 3$ .

$- 12 x - 3 (- 9 y) = (- 3) (21)$

$12 x - 27 y = 63$

Step 1.2.1.1.2.2

Multiply $- 9$ by $- 3$ .

$- 12 x + 27 y = (- 3) (21)$

$12 x - 27 y = 63$

$- 12 x + 27 y = (- 3) (21)$

$12 x - 27 y = 63$

$- 12 x + 27 y = (- 3) (21)$

$12 x - 27 y = 63$

$- 12 x + 27 y = (- 3) (21)$

$12 x - 27 y = 63$

Step 1.2.2

Simplify the right side.

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Step 1.2.2.1

Multiply $- 3$ by $21$ .

$- 12 x + 27 y = - 63$

$12 x - 27 y = 63$

$- 12 x + 27 y = - 63$

$12 x - 27 y = 63$

$- 12 x + 27 y = - 63$

$12 x - 27 y = 63$

Step 1.3

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

	$-$	$1$	$2$	$x$	$+$	$2$	$7$	$y$	$=$	$-$	$6$	$3$
$+$		$1$	$2$	$x$	$-$	$2$	$7$	$y$	$=$		$6$	$3$
								$0$	$=$			$0$

Step 1.4

Since $0 = 0$ , the equations intersect at an infinite number of points.

Infinite number of solutions

Step 1.5

Solve one of the equations for $y$ .

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Step 1.5.1

Add $12 x$ to both sides of the equation.

$27 y = - 63 + 12 x$

Step 1.5.2

Divide each term in $27 y = - 63 + 12 x$ by $27$ and simplify.

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Step 1.5.2.1

Divide each term in $27 y = - 63 + 12 x$ by $27$ .

$\frac{27 y}{27} = \frac{- 63}{27} + \frac{12 x}{27}$

Step 1.5.2.2

Simplify the left side.

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Step 1.5.2.2.1

Cancel the common factor of $27$ .

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Step 1.5.2.2.1.1

Cancel the common factor.

$\frac{27 y}{27} = \frac{- 63}{27} + \frac{12 x}{27}$

Step 1.5.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 63}{27} + \frac{12 x}{27}$

Step 1.5.2.3

Simplify the right side.

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Step 1.5.2.3.1

Simplify each term.

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Step 1.5.2.3.1.1

Cancel the common factor of $- 63$ and $27$ .

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Step 1.5.2.3.1.1.1

Factor $9$ out of $- 63$ .

$y = \frac{9 (- 7)}{27} + \frac{12 x}{27}$

Step 1.5.2.3.1.1.2

Cancel the common factors.

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Step 1.5.2.3.1.1.2.1

Factor $9$ out of $27$ .

$y = \frac{9 \cdot - 7}{9 \cdot 3} + \frac{12 x}{27}$

Step 1.5.2.3.1.1.2.2

Cancel the common factor.

$y = \frac{9 \cdot - 7}{9 \cdot 3} + \frac{12 x}{27}$

Step 1.5.2.3.1.1.2.3

Rewrite the expression.

$y = \frac{- 7}{3} + \frac{12 x}{27}$

Step 1.5.2.3.1.2

Move the negative in front of the fraction.

$y = - \frac{7}{3} + \frac{12 x}{27}$

Step 1.5.2.3.1.3

Cancel the common factor of $12$ and $27$ .

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Step 1.5.2.3.1.3.1

Factor $3$ out of $12 x$ .

$y = - \frac{7}{3} + \frac{3 (4 x)}{27}$

Step 1.5.2.3.1.3.2

Cancel the common factors.

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Step 1.5.2.3.1.3.2.1

Factor $3$ out of $27$ .

$y = - \frac{7}{3} + \frac{3 (4 x)}{3 (9)}$

Step 1.5.2.3.1.3.2.2

Cancel the common factor.

$y = - \frac{7}{3} + \frac{3 (4 x)}{3 \cdot 9}$

Step 1.5.2.3.1.3.2.3

Rewrite the expression.

$y = - \frac{7}{3} + \frac{4 x}{9}$

Step 1.6

The solution is the set of ordered pairs that make $y = - \frac{7}{3} + \frac{4 x}{9}$ true.

$(x, - \frac{7}{3} + \frac{4 x}{9})$

Step 2

Since the system is always true, the equations are equal and the graphs are the same line. Thus, the system is dependent.

Dependent

Step 3

Enter YOUR Problem

	$-$	$1$	$2$	$x$	$+$	$2$	$7$	$y$	$=$	$-$	$6$	$3$
$+$		$1$	$2$	$x$	$-$	$2$	$7$	$y$	$=$		$6$	$3$
								$0$	$=$			$0$

	$-$	$1$	$2$	$x$	$+$	$2$	$7$	$y$	$=$	$-$	$6$	$3$
$+$		$1$	$2$	$x$	$-$	$2$	$7$	$y$	$=$		$6$	$3$
								$0$	$=$			$0$

	$-$	$1$	$2$	$x$	$+$	$2$	$7$	$y$	$=$	$-$	$6$	$3$
$+$		$1$	$2$	$x$	$-$	$2$	$7$	$y$	$=$		$6$	$3$
								$0$	$=$			$0$