Algebra Examples

$3 x - 2 y = 18$ $4 x + 5 y = 4$

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 $y$ opposite.

$(5) \cdot (3 x - 2 y) = (5) (18)$

$(2) \cdot (4 x + 5 y) = (2) (4)$

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 $(5) \cdot (3 x - 2 y)$ .

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Step 1.2.1.1.1

Apply the distributive property.

$5 (3 x) + 5 (- 2 y) = (5) (18)$

$(2) \cdot (4 x + 5 y) = (2) (4)$

Step 1.2.1.1.2

Multiply.

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Step 1.2.1.1.2.1

Multiply $3$ by $5$ .

$15 x + 5 (- 2 y) = (5) (18)$

$(2) \cdot (4 x + 5 y) = (2) (4)$

Step 1.2.1.1.2.2

Multiply $- 2$ by $5$ .

$15 x - 10 y = (5) (18)$

$(2) \cdot (4 x + 5 y) = (2) (4)$

$15 x - 10 y = (5) (18)$

$(2) \cdot (4 x + 5 y) = (2) (4)$

$15 x - 10 y = (5) (18)$

$(2) \cdot (4 x + 5 y) = (2) (4)$

$15 x - 10 y = (5) (18)$

$(2) \cdot (4 x + 5 y) = (2) (4)$

Step 1.2.2

Simplify the right side.

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Step 1.2.2.1

Multiply $5$ by $18$ .

$15 x - 10 y = 90$

$(2) \cdot (4 x + 5 y) = (2) (4)$

$15 x - 10 y = 90$

$(2) \cdot (4 x + 5 y) = (2) (4)$

Step 1.2.3

Simplify the left side.

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Step 1.2.3.1

Simplify $(2) \cdot (4 x + 5 y)$ .

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Step 1.2.3.1.1

Apply the distributive property.

$15 x - 10 y = 90$

$2 (4 x) + 2 (5 y) = (2) (4)$

Step 1.2.3.1.2

Multiply.

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Step 1.2.3.1.2.1

Multiply $4$ by $2$ .

$15 x - 10 y = 90$

$8 x + 2 (5 y) = (2) (4)$

Step 1.2.3.1.2.2

Multiply $5$ by $2$ .

$15 x - 10 y = 90$

$8 x + 10 y = (2) (4)$

$15 x - 10 y = 90$

$8 x + 10 y = (2) (4)$

$15 x - 10 y = 90$

$8 x + 10 y = (2) (4)$

$15 x - 10 y = 90$

$8 x + 10 y = (2) (4)$

Step 1.2.4

Simplify the right side.

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Step 1.2.4.1

Multiply $2$ by $4$ .

$15 x - 10 y = 90$

$8 x + 10 y = 8$

$15 x - 10 y = 90$

$8 x + 10 y = 8$

$15 x - 10 y = 90$

$8 x + 10 y = 8$

Step 1.3

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

	$1$	$5$	$x$	$-$	$1$	$0$	$y$	$=$	$9$	$0$
$+$		$8$	$x$	$+$	$1$	$0$	$y$	$=$		$8$
	$2$	$3$	$x$					$=$	$9$	$8$

Step 1.4

Divide each term in $23 x = 98$ by $23$ and simplify.

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Step 1.4.1

Divide each term in $23 x = 98$ by $23$ .

$\frac{23 x}{23} = \frac{98}{23}$

Step 1.4.2

Simplify the left side.

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Step 1.4.2.1

Cancel the common factor of $23$ .

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Step 1.4.2.1.1

Cancel the common factor.

$\frac{23 x}{23} = \frac{98}{23}$

Step 1.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{98}{23}$

Step 1.5

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

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Step 1.5.1

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

$15 (\frac{98}{23}) - 10 y = 90$

Step 1.5.2

Multiply $15 (\frac{98}{23})$ .

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Step 1.5.2.1

Combine $15$ and $\frac{98}{23}$ .

$\frac{15 \cdot 98}{23} - 10 y = 90$

Step 1.5.2.2

Multiply $15$ by $98$ .

$\frac{1470}{23} - 10 y = 90$

Step 1.5.3

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

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Step 1.5.3.1

Subtract $\frac{1470}{23}$ from both sides of the equation.

$- 10 y = 90 - \frac{1470}{23}$

Step 1.5.3.2

To write $90$ as a fraction with a common denominator, multiply by $\frac{23}{23}$ .

$- 10 y = 90 \cdot \frac{23}{23} - \frac{1470}{23}$

Step 1.5.3.3

Combine $90$ and $\frac{23}{23}$ .

$- 10 y = \frac{90 \cdot 23}{23} - \frac{1470}{23}$

Step 1.5.3.4

Combine the numerators over the common denominator.

$- 10 y = \frac{90 \cdot 23 - 1470}{23}$

Step 1.5.3.5

Simplify the numerator.

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Step 1.5.3.5.1

Multiply $90$ by $23$ .

$- 10 y = \frac{2070 - 1470}{23}$

Step 1.5.3.5.2

Subtract $1470$ from $2070$ .

$- 10 y = \frac{600}{23}$

Step 1.5.4

Divide each term in $- 10 y = \frac{600}{23}$ by $- 10$ and simplify.

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Step 1.5.4.1

Divide each term in $- 10 y = \frac{600}{23}$ by $- 10$ .

$\frac{- 10 y}{- 10} = \frac{\frac{600}{23}}{- 10}$

Step 1.5.4.2

Simplify the left side.

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Step 1.5.4.2.1

Cancel the common factor of $- 10$ .

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Step 1.5.4.2.1.1

Cancel the common factor.

$\frac{- 10 y}{- 10} = \frac{\frac{600}{23}}{- 10}$

Step 1.5.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{\frac{600}{23}}{- 10}$

Step 1.5.4.3

Simplify the right side.

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Step 1.5.4.3.1

Multiply the numerator by the reciprocal of the denominator.

$y = \frac{600}{23} \cdot \frac{1}{- 10}$

Step 1.5.4.3.2

Cancel the common factor of $10$ .

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Step 1.5.4.3.2.1

Factor $10$ out of $600$ .

$y = \frac{10 (60)}{23} \cdot \frac{1}{- 10}$

Step 1.5.4.3.2.2

Factor $10$ out of $- 10$ .

$y = \frac{10 \cdot 60}{23} \cdot \frac{1}{10 \cdot - 1}$

Step 1.5.4.3.2.3

Cancel the common factor.

$y = \frac{10 \cdot 60}{23} \cdot \frac{1}{10 \cdot - 1}$

Step 1.5.4.3.2.4

Rewrite the expression.

$y = \frac{60}{23} \cdot \frac{1}{- 1}$

Step 1.5.4.3.3

Multiply $\frac{60}{23}$ by $\frac{1}{- 1}$ .

$y = \frac{60}{23 \cdot - 1}$

Step 1.5.4.3.4

Simplify the expression.

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Step 1.5.4.3.4.1

Multiply $23$ by $- 1$ .

$y = \frac{60}{- 23}$

Step 1.5.4.3.4.2

Move the negative in front of the fraction.

$y = - \frac{60}{23}$

Step 1.6

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

$(\frac{98}{23}, - \frac{60}{23})$

Step 2

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

Independent

Step 3

	$1$	$5$	$x$	$-$	$1$	$0$	$y$	$=$	$9$	$0$
$+$		$8$	$x$	$+$	$1$	$0$	$y$	$=$		$8$
	$2$	$3$	$x$					$=$	$9$	$8$

	$1$	$5$	$x$	$-$	$1$	$0$	$y$	$=$	$9$	$0$
$+$		$8$	$x$	$+$	$1$	$0$	$y$	$=$		$8$
	$2$	$3$	$x$					$=$	$9$	$8$

	$1$	$5$	$x$	$-$	$1$	$0$	$y$	$=$	$9$	$0$
$+$		$8$	$x$	$+$	$1$	$0$	$y$	$=$		$8$
	$2$	$3$	$x$					$=$	$9$	$8$