Algebra Examples

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

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 $2 y$ from both sides of the equation.

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

Step 1.1.2

Subtract $2 x$ from both sides of the equation.

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

Step 1.2

Reorder the polynomial.

$- 2 x + y = 3$

$4 x - 2 y = - 6$

Step 1.3

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

$4 x - 2 y = - 6$

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

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

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Step 1.4.1.1.1

Apply the distributive property.

$4 x - 2 y = - 6$

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

Step 1.4.1.1.2

Multiply $- 2$ by $2$ .

$4 x - 2 y = - 6$

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

$4 x - 2 y = - 6$

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

$4 x - 2 y = - 6$

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

Step 1.4.2

Simplify the right side.

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Step 1.4.2.1

Multiply $2$ by $3$ .

$4 x - 2 y = - 6$

$- 4 x + 2 y = 6$

$4 x - 2 y = - 6$

$- 4 x + 2 y = 6$

$4 x - 2 y = - 6$

$- 4 x + 2 y = 6$

Step 1.5

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

		$4$	$x$	$-$	$2$	$y$	$=$	$-$	$6$
$+$	$-$	$4$	$x$	$+$	$2$	$y$	$=$		$6$
						$0$	$=$		$0$

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

Subtract $4 x$ from both sides of the equation.

$- 2 y = - 6 - 4 x$

Step 1.7.2

Divide each term in $- 2 y = - 6 - 4 x$ by $- 2$ and simplify.

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Step 1.7.2.1

Divide each term in $- 2 y = - 6 - 4 x$ by $- 2$ .

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

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.

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

Step 1.7.2.2.1.2

Divide $y$ by $1$ .

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

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 $- 6$ by $- 2$ .

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

Step 1.7.2.3.1.2

Cancel the common factor of $- 4$ and $- 2$ .

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Step 1.7.2.3.1.2.1

Factor $- 2$ out of $- 4 x$ .

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

Step 1.7.2.3.1.2.2

Cancel the common factors.

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Step 1.7.2.3.1.2.2.1

Factor $- 2$ out of $- 2$ .

$y = 3 + \frac{- 2 (2 x)}{- 2 (1)}$

Step 1.7.2.3.1.2.2.2

Cancel the common factor.

$y = 3 + \frac{- 2 (2 x)}{- 2 \cdot 1}$

Step 1.7.2.3.1.2.2.3

Rewrite the expression.

$y = 3 + \frac{2 x}{1}$

Step 1.7.2.3.1.2.2.4

Divide $2 x$ by $1$ .

$y = 3 + 2 x$

Step 1.8

The solution is the set of ordered pairs that make $y = 3 + 2 x$ true.

$(x, 3 + 2 x)$

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

		$4$	$x$	$-$	$2$	$y$	$=$	$-$	$6$
$+$	$-$	$4$	$x$	$+$	$2$	$y$	$=$		$6$
						$0$	$=$		$0$

		$4$	$x$	$-$	$2$	$y$	$=$	$-$	$6$
$+$	$-$	$4$	$x$	$+$	$2$	$y$	$=$		$6$
						$0$	$=$		$0$

		$4$	$x$	$-$	$2$	$y$	$=$	$-$	$6$
$+$	$-$	$4$	$x$	$+$	$2$	$y$	$=$		$6$
						$0$	$=$		$0$