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

$4 y = 2$ $2 x + y = \frac{x}{2} + 1$

Step 1

Multiply each term in $2 x + y = \frac{x}{2} + 1$ by $2$ to eliminate the fractions.

Multiply each term in $2 x + y = \frac{x}{2} + 1$ by $2$ .

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

$4 y = 2$

Simplify the left side.

Simplify each term.

Multiply $2$ by $2$ .

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

$4 y = 2$

Move $2$ to the left of $y$ .

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

$4 y = 2$

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

$4 y = 2$

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

$4 y = 2$

Simplify the right side.

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

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

$4 y = 2$

Rewrite the expression.

$4 x + 2 y = x + 1 \cdot 2$

$4 y = 2$

$4 x + 2 y = x + 1 \cdot 2$

$4 y = 2$

Multiply $2$ by $1$ .

$4 x + 2 y = x + 2$

$4 y = 2$

$4 x + 2 y = x + 2$

$4 y = 2$

$4 x + 2 y = x + 2$

$4 y = 2$

$4 x + 2 y = x + 2$

$4 y = 2$

Step 2

Move all terms containing variables to the left side of the equation.

Subtract $x$ from both sides of the equation.

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

Subtract $x$ from $4 x$ .

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

Step 3

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

$4 y = 2$

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

Step 4

Simplify.

Simplify the left side.

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

Apply the distributive property.

$4 y = 2$

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

Multiply.

Multiply $3$ by $- 2$ .

$4 y = 2$

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

Multiply $2$ by $- 2$ .

$4 y = 2$

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

$4 y = 2$

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

$4 y = 2$

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

$4 y = 2$

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

Simplify the right side.

Multiply $- 2$ by $2$ .

$4 y = 2$

$- 6 x - 4 y = - 4$

$4 y = 2$

$- 6 x - 4 y = - 4$

$4 y = 2$

$- 6 x - 4 y = - 4$

Step 5

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

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

Step 6

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

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

$\frac{- 6 x}{- 6} = \frac{- 2}{- 6}$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 x}{- 6} = \frac{- 2}{- 6}$

Divide $x$ by $1$ .

$x = \frac{- 2}{- 6}$

Simplify the right side.

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

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

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

$x = \frac{1}{3}$

Step 7

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 $y$ .

$- 6 (\frac{1}{3}) - 4 y = - 4$

Cancel the common factor of $3$ .

Factor $3$ out of $- 6$ .

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

Cancel the common factor.

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

Rewrite the expression.

$- 2 - 4 y = - 4$

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

Add $2$ to both sides of the equation.

$- 4 y = - 4 + 2$

Add $- 4$ and $2$ .

$- 4 y = - 2$

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

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

$\frac{- 4 y}{- 4} = \frac{- 2}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 y}{- 4} = \frac{- 2}{- 4}$

Divide $y$ by $1$ .

$y = \frac{- 2}{- 4}$

Simplify the right side.

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

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

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

Cancel the common factors.

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

$y = \frac{- 2 \cdot 1}{- 2 \cdot 2}$

Cancel the common factor.

$y = \frac{- 2 \cdot 1}{- 2 \cdot 2}$

Rewrite the expression.

$y = \frac{1}{2}$

Step 8

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

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

Step 9

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

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

Step 10

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

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