Algebra Examples

$y = - 3 x + 4$ , $6 x + 2 y = 8$

Step 1

Solve the system of equations.

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Step 1.1

Add $3 x$ to both sides of the equation.

$y + 3 x = 4, 6 x + 2 y = 8$

Step 1.2

Reorder the polynomial.

$3 x + y = 4$

$6 x + 2 y = 8$

Step 1.3

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

$(- 2) \cdot (3 x + y) = (- 2) (4)$

$6 x + 2 y = 8$

Step 1.4

Simplify.

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Step 1.4.1

Simplify the left side.

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Step 1.4.1.1

Simplify $(- 2) \cdot (3 x + y)$ .

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Step 1.4.1.1.1

Apply the distributive property.

$- 2 (3 x) - 2 y = (- 2) (4)$

$6 x + 2 y = 8$

Step 1.4.1.1.2

Multiply $3$ by $- 2$ .

$- 6 x - 2 y = (- 2) (4)$

$6 x + 2 y = 8$

$- 6 x - 2 y = (- 2) (4)$

$6 x + 2 y = 8$

$- 6 x - 2 y = (- 2) (4)$

$6 x + 2 y = 8$

Step 1.4.2

Simplify the right side.

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Step 1.4.2.1

Multiply $- 2$ by $4$ .

$- 6 x - 2 y = - 8$

$6 x + 2 y = 8$

$- 6 x - 2 y = - 8$

$6 x + 2 y = 8$

$- 6 x - 2 y = - 8$

$6 x + 2 y = 8$

Step 1.5

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

	$-$	$6$	$x$	$-$	$2$	$y$	$=$	$-$	$8$
$+$		$6$	$x$	$+$	$2$	$y$	$=$		$8$
						$0$	$=$		$0$

Step 1.6

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

Infinite number of solutions

Step 1.7

Solve one of the equations for $y$ .

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Step 1.7.1

Add $6 x$ to both sides of the equation.

$- 2 y = - 8 + 6 x$

Step 1.7.2

Divide each term in $- 2 y = - 8 + 6 x$ by $- 2$ and simplify.

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Step 1.7.2.1

Divide each term in $- 2 y = - 8 + 6 x$ by $- 2$ .

$\frac{- 2 y}{- 2} = \frac{- 8}{- 2} + \frac{6 x}{- 2}$

Step 1.7.2.2

Simplify the left side.

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Step 1.7.2.2.1

Cancel the common factor of $- 2$ .

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Step 1.7.2.2.1.1

Cancel the common factor.

$\frac{- 2 y}{- 2} = \frac{- 8}{- 2} + \frac{6 x}{- 2}$

Step 1.7.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 8}{- 2} + \frac{6 x}{- 2}$

Step 1.7.2.3

Simplify the right side.

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Step 1.7.2.3.1

Simplify each term.

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Step 1.7.2.3.1.1

Divide $- 8$ by $- 2$ .

$y = 4 + \frac{6 x}{- 2}$

Step 1.7.2.3.1.2

Cancel the common factor of $6$ and $- 2$ .

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Step 1.7.2.3.1.2.1

Factor $2$ out of $6 x$ .

$y = 4 + \frac{2 (3 x)}{- 2}$

Step 1.7.2.3.1.2.2

Move the negative one from the denominator of $\frac{3 x}{- 1}$ .

$y = 4 - 1 \cdot (3 x)$

Step 1.7.2.3.1.3

Rewrite $- 1 \cdot (3 x)$ as $- (3 x)$ .

$y = 4 - (3 x)$

Step 1.7.2.3.1.4

Multiply $3$ by $- 1$ .

$y = 4 - 3 x$

Step 1.8

The solution is the set of ordered pairs that make $y = 4 - 3 x$ true.

$(x, 4 - 3 x)$

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

	$-$	$6$	$x$	$-$	$2$	$y$	$=$	$-$	$8$
$+$		$6$	$x$	$+$	$2$	$y$	$=$		$8$
						$0$	$=$		$0$

	$-$	$6$	$x$	$-$	$2$	$y$	$=$	$-$	$8$
$+$		$6$	$x$	$+$	$2$	$y$	$=$		$8$
						$0$	$=$		$0$

	$-$	$6$	$x$	$-$	$2$	$y$	$=$	$-$	$8$
$+$		$6$	$x$	$+$	$2$	$y$	$=$		$8$
						$0$	$=$		$0$