Algebra Examples

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

Step 1

Solve the system of equations.

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Step 1.1

Move all terms containing variables to the left.

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Step 1.1.1

Subtract $3 x$ from both sides of the equation.

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

Step 1.1.2

Add $2 x$ to both sides of the equation.

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

Step 1.2

Reorder the polynomial.

$- 3 x + y = - 4$

$y + 2 x = 4$

Step 1.3

Reorder the polynomial.

$2 x + y = 4$

$- 3 x + y = - 4$

Step 1.4

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

$- 3 x + y = - 4$

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

Step 1.5

Simplify.

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Step 1.5.1

Simplify the left side.

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Step 1.5.1.1

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

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Step 1.5.1.1.1

Apply the distributive property.

$- 3 x + y = - 4$

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

Step 1.5.1.1.2

Simplify the expression.

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Step 1.5.1.1.2.1

Multiply $2$ by $- 1$ .

$- 3 x + y = - 4$

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

Step 1.5.1.1.2.2

Rewrite $- 1 y$ as $- y$ .

$- 3 x + y = - 4$

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

$- 3 x + y = - 4$

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

$- 3 x + y = - 4$

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

$- 3 x + y = - 4$

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

Step 1.5.2

Simplify the right side.

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Step 1.5.2.1

Multiply $- 1$ by $4$ .

$- 3 x + y = - 4$

$- 2 x - y = - 4$

$- 3 x + y = - 4$

$- 2 x - y = - 4$

$- 3 x + y = - 4$

$- 2 x - y = - 4$

Step 1.6

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

	$-$	$3$	$x$	$+$	$y$	$=$	$-$	$4$
$+$	$-$	$2$	$x$	$-$	$y$	$=$	$-$	$4$
	$-$	$5$	$x$			$=$	$-$	$8$

Step 1.7

Divide each term in $- 5 x = - 8$ by $- 5$ and simplify.

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Step 1.7.1

Divide each term in $- 5 x = - 8$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{- 8}{- 5}$

Step 1.7.2

Simplify the left side.

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Step 1.7.2.1

Cancel the common factor of $- 5$ .

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Step 1.7.2.1.1

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{- 8}{- 5}$

Step 1.7.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 8}{- 5}$

Step 1.7.3

Simplify the right side.

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Step 1.7.3.1

Dividing two negative values results in a positive value.

$x = \frac{8}{5}$

Step 1.8

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

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Step 1.8.1

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

$- 3 (\frac{8}{5}) + y = - 4$

Step 1.8.2

Simplify each term.

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Step 1.8.2.1

Multiply $- 3 (\frac{8}{5})$ .

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Step 1.8.2.1.1

Combine $- 3$ and $\frac{8}{5}$ .

$\frac{- 3 \cdot 8}{5} + y = - 4$

Step 1.8.2.1.2

Multiply $- 3$ by $8$ .

$\frac{- 24}{5} + y = - 4$

Step 1.8.2.2

Move the negative in front of the fraction.

$- \frac{24}{5} + y = - 4$

Step 1.8.3

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

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Step 1.8.3.1

Add $\frac{24}{5}$ to both sides of the equation.

$y = - 4 + \frac{24}{5}$

Step 1.8.3.2

To write $- 4$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$y = - 4 \cdot \frac{5}{5} + \frac{24}{5}$

Step 1.8.3.3

Combine $- 4$ and $\frac{5}{5}$ .

$y = \frac{- 4 \cdot 5}{5} + \frac{24}{5}$

Step 1.8.3.4

Combine the numerators over the common denominator.

$y = \frac{- 4 \cdot 5 + 24}{5}$

Step 1.8.3.5

Simplify the numerator.

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Step 1.8.3.5.1

Multiply $- 4$ by $5$ .

$y = \frac{- 20 + 24}{5}$

Step 1.8.3.5.2

Add $- 20$ and $24$ .

$y = \frac{4}{5}$

Step 1.9

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

$(\frac{8}{5}, \frac{4}{5})$

Step 2

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

Independent

Step 3

	$-$	$3$	$x$	$+$	$y$	$=$	$-$	$4$
$+$	$-$	$2$	$x$	$-$	$y$	$=$	$-$	$4$
	$-$	$5$	$x$			$=$	$-$	$8$

	$-$	$3$	$x$	$+$	$y$	$=$	$-$	$4$
$+$	$-$	$2$	$x$	$-$	$y$	$=$	$-$	$4$
	$-$	$5$	$x$			$=$	$-$	$8$

	$-$	$3$	$x$	$+$	$y$	$=$	$-$	$4$
$+$	$-$	$2$	$x$	$-$	$y$	$=$	$-$	$4$
	$-$	$5$	$x$			$=$	$-$	$8$