Solve by Addition/Elimination 3x+4y=3 and 2x-4y=12

$3 x + 4 y = 3$ and $2 x - 4 y = 12$

Step 1

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

	$3$	$x$	$+$	$4$	$y$	$=$		$3$
$+$	$2$	$x$	$-$	$4$	$y$	$=$	$1$	$2$
	$5$	$x$				$=$	$1$	$5$

Step 2

Divide each term in $5 x = 15$ by $5$ and simplify.

Divide each term in $5 x = 15$ by $5$ .

$\frac{5 x}{5} = \frac{15}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{15}{5}$

Divide $x$ by $1$ .

$x = \frac{15}{5}$

Simplify the right side.

Divide $15$ by $5$ .

$x = 3$

Step 3

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

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

$3 (3) + 4 y = 3$

Multiply $3$ by $3$ .

$9 + 4 y = 3$

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

Subtract $9$ from both sides of the equation.

$4 y = 3 - 9$

Subtract $9$ from $3$ .

$4 y = - 6$

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

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

$\frac{4 y}{4} = \frac{- 6}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{- 6}{4}$

Divide $y$ by $1$ .

$y = \frac{- 6}{4}$

Simplify the right side.

Cancel the common factor of $- 6$ and $4$ .

Factor $2$ out of $- 6$ .

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

Cancel the common factors.

Factor $2$ out of $4$ .

$y = \frac{2 \cdot - 3}{2 \cdot 2}$

Cancel the common factor.

$y = \frac{2 \cdot - 3}{2 \cdot 2}$

Rewrite the expression.

$y = \frac{- 3}{2}$

Move the negative in front of the fraction.

$y = - \frac{3}{2}$

Step 4

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

$(3, - \frac{3}{2})$

Step 5

The result can be shown in multiple forms.

Point Form:

$(3, - \frac{3}{2})$

Equation Form:

$x = 3, y = - \frac{3}{2}$

Step 6

Solve by Addition/Elimination 3x+4y=3 and 2x-4y=12

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