Solve by Addition/Elimination 3x-4y=1 , -6x+8y=-2

$3 x - 4 y = 1$ , $- 6 x + 8 y = - 2$

Step 1

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

$(2) \cdot (3 x - 4 y) = (2) (1)$

$- 6 x + 8 y = - 2$

Step 2

Simplify.

Simplify the left side.

Simplify $(2) \cdot (3 x - 4 y)$ .

Apply the distributive property.

$2 (3 x) + 2 (- 4 y) = (2) (1)$

$- 6 x + 8 y = - 2$

Multiply.

Multiply $3$ by $2$ .

$6 x + 2 (- 4 y) = (2) (1)$

$- 6 x + 8 y = - 2$

Multiply $- 4$ by $2$ .

$6 x - 8 y = (2) (1)$

$- 6 x + 8 y = - 2$

$6 x - 8 y = (2) (1)$

$- 6 x + 8 y = - 2$

$6 x - 8 y = (2) (1)$

$- 6 x + 8 y = - 2$

$6 x - 8 y = (2) (1)$

$- 6 x + 8 y = - 2$

Simplify the right side.

Multiply $2$ by $1$ .

$6 x - 8 y = 2$

$- 6 x + 8 y = - 2$

$6 x - 8 y = 2$

$- 6 x + 8 y = - 2$

$6 x - 8 y = 2$

$- 6 x + 8 y = - 2$

Step 3

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

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

Step 4

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

Infinite number of solutions

Step 5

Solve one of the equations for $y$ .

Subtract $6 x$ from both sides of the equation.

$- 8 y = 2 - 6 x$

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

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

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

Simplify the left side.

Cancel the common factor of $- 8$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Cancel the common factor of $2$ and $- 8$ .

Factor $2$ out of $2$ .

$y = \frac{2 (1)}{- 8} + \frac{- 6 x}{- 8}$

Cancel the common factors.

Factor $2$ out of $- 8$ .

$y = \frac{2 \cdot 1}{2 \cdot - 4} + \frac{- 6 x}{- 8}$

Cancel the common factor.

$y = \frac{2 \cdot 1}{2 \cdot - 4} + \frac{- 6 x}{- 8}$

Rewrite the expression.

$y = \frac{1}{- 4} + \frac{- 6 x}{- 8}$

Move the negative in front of the fraction.

$y = - \frac{1}{4} + \frac{- 6 x}{- 8}$

Cancel the common factor of $- 6$ and $- 8$ .

Factor $- 2$ out of $- 6 x$ .

$y = - \frac{1}{4} + \frac{- 2 (3 x)}{- 8}$

Cancel the common factors.

Factor $- 2$ out of $- 8$ .

$y = - \frac{1}{4} + \frac{- 2 (3 x)}{- 2 (4)}$

Cancel the common factor.

$y = - \frac{1}{4} + \frac{- 2 (3 x)}{- 2 \cdot 4}$

Rewrite the expression.

$y = - \frac{1}{4} + \frac{3 x}{4}$

Step 6

The solution is the set of ordered pairs that make $y = - \frac{1}{4} + \frac{3 x}{4}$ true.

$(x, - \frac{1}{4} + \frac{3 x}{4})$

Step 7

The result can be shown in multiple forms.

Point Form:

$(x, - \frac{1}{4} + \frac{3 x}{4})$

Equation Form:

$x = x, y = - \frac{1}{4} + \frac{3 x}{4}$

Step 8

Solve by Addition/Elimination 3x-4y=1 , -6x+8y=-2

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