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

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

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

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

$4 x + 6 y = 2$

Simplify.

Simplify the left side.

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

Apply the distributive property.

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

$4 x + 6 y = 2$

Multiply.

Multiply $2$ by $- 2$ .

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

$4 x + 6 y = 2$

Multiply $3$ by $- 2$ .

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

$4 x + 6 y = 2$

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

$4 x + 6 y = 2$

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

$4 x + 6 y = 2$

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

$4 x + 6 y = 2$

Simplify the right side.

Multiply $- 2$ by $1$ .

$- 4 x - 6 y = - 2$

$4 x + 6 y = 2$

$- 4 x - 6 y = - 2$

$4 x + 6 y = 2$

$- 4 x - 6 y = - 2$

$4 x + 6 y = 2$

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

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

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

Infinite number of solutions

Solve one of the equations for $y$ .

Add $4 x$ to both sides of the equation.

$- 6 y = - 2 + 4 x$

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

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

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

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

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

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

$y = \frac{- 2 (1)}{- 6} + \frac{4 x}{- 6}$

Cancel the common factors.

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

$y = \frac{- 2 \cdot 1}{- 2 \cdot 3} + \frac{4 x}{- 6}$

Cancel the common factor.

$y = \frac{- 2 \cdot 1}{- 2 \cdot 3} + \frac{4 x}{- 6}$

Rewrite the expression.

$y = \frac{1}{3} + \frac{4 x}{- 6}$

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

Factor $2$ out of $4 x$ .

$y = \frac{1}{3} + \frac{2 (2 x)}{- 6}$

Cancel the common factors.

Factor $2$ out of $- 6$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

The solution is the set of ordered pairs that make $y = \frac{1}{3} - \frac{2 x}{3}$ true.

$(x, \frac{1}{3} - \frac{2 x}{3})$

The result can be shown in multiple forms.

Point Form:

$(x, \frac{1}{3} - \frac{2 x}{3})$

Equation Form:

$x = x, y = \frac{1}{3} - \frac{2 x}{3}$

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

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