Algebra Examples

$- (y - 4) = x + 9$ , $x - \frac{8}{3} y = 0$

Step 1

Solve the system of equations.

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Step 1.1

Multiply each term in $x - \frac{8}{3} y = 0$ by $3$ to eliminate the fractions.

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Step 1.1.1

Multiply each term in $x - \frac{8}{3} y = 0$ by $3$ .

$x \cdot 3 - \frac{8}{3} y \cdot 3 = 0 \cdot 3$

$- (y - 4) = x + 9$

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Simplify each term.

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Step 1.1.2.1.1

Move $3$ to the left of $x$ .

$3 \cdot x - \frac{8}{3} y \cdot 3 = 0 \cdot 3$

$- (y - 4) = x + 9$

Step 1.1.2.1.2

Combine $y$ and $\frac{8}{3}$ .

$3 x - \frac{y \cdot 8}{3} \cdot 3 = 0 \cdot 3$

$- (y - 4) = x + 9$

Step 1.1.2.1.3

Cancel the common factor of $3$ .

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Step 1.1.2.1.3.1

Move the leading negative in $- \frac{y \cdot 8}{3}$ into the numerator.

$3 x + \frac{- y \cdot 8}{3} \cdot 3 = 0 \cdot 3$

$- (y - 4) = x + 9$

Step 1.1.2.1.3.2

Cancel the common factor.

$3 x + \frac{- y \cdot 8}{3} \cdot 3 = 0 \cdot 3$

$- (y - 4) = x + 9$

Step 1.1.2.1.3.3

Rewrite the expression.

$3 x - y \cdot 8 = 0 \cdot 3$

$- (y - 4) = x + 9$

$3 x - y \cdot 8 = 0 \cdot 3$

$- (y - 4) = x + 9$

Step 1.1.2.1.4

Multiply $8$ by $- 1$ .

$3 x - 8 y = 0 \cdot 3$

$- (y - 4) = x + 9$

$3 x - 8 y = 0 \cdot 3$

$- (y - 4) = x + 9$

$3 x - 8 y = 0 \cdot 3$

$- (y - 4) = x + 9$

Step 1.1.3

Simplify the right side.

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Step 1.1.3.1

Multiply $0$ by $3$ .

$3 x - 8 y = 0$

$- (y - 4) = x + 9$

$3 x - 8 y = 0$

$- (y - 4) = x + 9$

$3 x - 8 y = 0$

$- (y - 4) = x + 9$

Step 1.2

Simplify the left side.

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Step 1.2.1

Simplify $- (y - 4)$ .

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Step 1.2.1.1

Apply the distributive property.

$- y + 4 = x + 9$

$3 x - 8 y = 0$

Step 1.2.1.2

Multiply $- 1$ by $- 4$ .

$- y + 4 = x + 9$

$3 x - 8 y = 0$

$- y + 4 = x + 9$

$3 x - 8 y = 0$

$- y + 4 = x + 9$

$3 x - 8 y = 0$

Step 1.3

Move all terms containing variables to the left.

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Step 1.3.1

Subtract $4$ from both sides of the equation.

$- y = x + 9 - 4, 3 x - 8 y = 0$

Step 1.3.2

Subtract $x$ from both sides of the equation.

$- y - x = 9 - 4, 3 x - 8 y = 0$

Step 1.4

Reorder the polynomial.

$- x - y = 9 - 4$

$3 x - 8 y = 0$

Step 1.5

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

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

$3 x - 8 y = 0$

Step 1.6

Simplify.

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Step 1.6.1

Simplify the left side.

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Step 1.6.1.1

Simplify $(3) \cdot (- x - y)$ .

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Step 1.6.1.1.1

Apply the distributive property.

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

$3 x - 8 y = 0$

Step 1.6.1.1.2

Multiply.

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Step 1.6.1.1.2.1

Multiply $- 1$ by $3$ .

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

$3 x - 8 y = 0$

Step 1.6.1.1.2.2

Multiply $- 1$ by $3$ .

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

$3 x - 8 y = 0$

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

$3 x - 8 y = 0$

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

$3 x - 8 y = 0$

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

$3 x - 8 y = 0$

Step 1.6.2

Simplify the right side.

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Step 1.6.2.1

Simplify $(3) (9 - 4)$ .

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Step 1.6.2.1.1

Subtract $4$ from $9$ .

$- 3 x - 3 y = 3 \cdot 5$

$3 x - 8 y = 0$

Step 1.6.2.1.2

Multiply $3$ by $5$ .

$- 3 x - 3 y = 15$

$3 x - 8 y = 0$

$- 3 x - 3 y = 15$

$3 x - 8 y = 0$

$- 3 x - 3 y = 15$

$3 x - 8 y = 0$

$- 3 x - 3 y = 15$

$3 x - 8 y = 0$

Step 1.7

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

	$-$	$3$	$x$		$-$	$3$	$y$	$=$	$1$	$5$
$+$		$3$	$x$		$-$	$8$	$y$	$=$		$0$
				$-$	$1$	$1$	$y$	$=$	$1$	$5$

Step 1.8

Divide each term in $- 11 y = 15$ by $- 11$ and simplify.

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Step 1.8.1

Divide each term in $- 11 y = 15$ by $- 11$ .

$\frac{- 11 y}{- 11} = \frac{15}{- 11}$

Step 1.8.2

Simplify the left side.

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Step 1.8.2.1

Cancel the common factor of $- 11$ .

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Step 1.8.2.1.1

Cancel the common factor.

$\frac{- 11 y}{- 11} = \frac{15}{- 11}$

Step 1.8.2.1.2

Divide $y$ by $1$ .

$y = \frac{15}{- 11}$

Step 1.8.3

Simplify the right side.

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Step 1.8.3.1

Move the negative in front of the fraction.

$y = - \frac{15}{11}$

Step 1.9

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

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Step 1.9.1

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

$- 3 x - 3 (- \frac{15}{11}) = 15$

Step 1.9.2

Multiply $- 3 (- \frac{15}{11})$ .

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Step 1.9.2.1

Multiply $- 1$ by $- 3$ .

$- 3 x + 3 (\frac{15}{11}) = 15$

Step 1.9.2.2

Combine $3$ and $\frac{15}{11}$ .

$- 3 x + \frac{3 \cdot 15}{11} = 15$

Step 1.9.2.3

Multiply $3$ by $15$ .

$- 3 x + \frac{45}{11} = 15$

Step 1.9.3

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

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Step 1.9.3.1

Subtract $\frac{45}{11}$ from both sides of the equation.

$- 3 x = 15 - \frac{45}{11}$

Step 1.9.3.2

To write $15$ as a fraction with a common denominator, multiply by $\frac{11}{11}$ .

$- 3 x = 15 \cdot \frac{11}{11} - \frac{45}{11}$

Step 1.9.3.3

Combine $15$ and $\frac{11}{11}$ .

$- 3 x = \frac{15 \cdot 11}{11} - \frac{45}{11}$

Step 1.9.3.4

Combine the numerators over the common denominator.

$- 3 x = \frac{15 \cdot 11 - 45}{11}$

Step 1.9.3.5

Simplify the numerator.

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Step 1.9.3.5.1

Multiply $15$ by $11$ .

$- 3 x = \frac{165 - 45}{11}$

Step 1.9.3.5.2

Subtract $45$ from $165$ .

$- 3 x = \frac{120}{11}$

Step 1.9.4

Divide each term in $- 3 x = \frac{120}{11}$ by $- 3$ and simplify.

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Step 1.9.4.1

Divide each term in $- 3 x = \frac{120}{11}$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{\frac{120}{11}}{- 3}$

Step 1.9.4.2

Simplify the left side.

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Step 1.9.4.2.1

Cancel the common factor of $- 3$ .

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Step 1.9.4.2.1.1

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{\frac{120}{11}}{- 3}$

Step 1.9.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{120}{11}}{- 3}$

Step 1.9.4.3

Simplify the right side.

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Step 1.9.4.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{120}{11} \cdot \frac{1}{- 3}$

Step 1.9.4.3.2

Cancel the common factor of $3$ .

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Step 1.9.4.3.2.1

Factor $3$ out of $120$ .

$x = \frac{3 (40)}{11} \cdot \frac{1}{- 3}$

Step 1.9.4.3.2.2

Factor $3$ out of $- 3$ .

$x = \frac{3 \cdot 40}{11} \cdot \frac{1}{3 \cdot - 1}$

Step 1.9.4.3.2.3

Cancel the common factor.

$x = \frac{3 \cdot 40}{11} \cdot \frac{1}{3 \cdot - 1}$

Step 1.9.4.3.2.4

Rewrite the expression.

$x = \frac{40}{11} \cdot \frac{1}{- 1}$

Step 1.9.4.3.3

Multiply $\frac{40}{11}$ by $\frac{1}{- 1}$ .

$x = \frac{40}{11 \cdot - 1}$

Step 1.9.4.3.4

Simplify the expression.

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Step 1.9.4.3.4.1

Multiply $11$ by $- 1$ .

$x = \frac{40}{- 11}$

Step 1.9.4.3.4.2

Move the negative in front of the fraction.

$x = - \frac{40}{11}$

Step 1.10

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

$(- \frac{40}{11}, - \frac{15}{11})$

Step 2

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

Independent

Step 3

	$-$	$3$	$x$		$-$	$3$	$y$	$=$	$1$	$5$
$+$		$3$	$x$		$-$	$8$	$y$	$=$		$0$
				$-$	$1$	$1$	$y$	$=$	$1$	$5$

	$-$	$3$	$x$		$-$	$3$	$y$	$=$	$1$	$5$
$+$		$3$	$x$		$-$	$8$	$y$	$=$		$0$
				$-$	$1$	$1$	$y$	$=$	$1$	$5$

	$-$	$3$	$x$		$-$	$3$	$y$	$=$	$1$	$5$
$+$		$3$	$x$		$-$	$8$	$y$	$=$		$0$
				$-$	$1$	$1$	$y$	$=$	$1$	$5$