Solve by Linear Combination 3x+y=4 2x+y=5

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

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

$3 x + y = 4$

$(- 1) \cdot (2 x + y) = (- 1) (5)$

Simplify.

Simplify the left side.

Simplify $(- 1) \cdot (2 x + y)$ .

Apply the distributive property.

$3 x + y = 4$

$- 1 (2 x) - 1 y = (- 1) (5)$

Simplify the expression.

Multiply $2$ by $- 1$ .

$3 x + y = 4$

$- 2 x - 1 y = (- 1) (5)$

Rewrite $- 1 y$ as $- y$ .

$3 x + y = 4$

$- 2 x - y = (- 1) (5)$

$3 x + y = 4$

$- 2 x - y = (- 1) (5)$

$3 x + y = 4$

$- 2 x - y = (- 1) (5)$

$3 x + y = 4$

$- 2 x - y = (- 1) (5)$

Simplify the right side.

Multiply $- 1$ by $5$ .

$3 x + y = 4$

$- 2 x - y = - 5$

$3 x + y = 4$

$- 2 x - y = - 5$

$3 x + y = 4$

$- 2 x - y = - 5$

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

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

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 $y$ .

$3 (- 1) + y = 4$

Multiply $3$ by $- 1$ .

$- 3 + y = 4$

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

Add $3$ to both sides of the equation.

$y = 4 + 3$

Add $4$ and $3$ .

$y = 7$

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

$(- 1, 7)$

The result can be shown in multiple forms.

Point Form:

$(- 1, 7)$

Equation Form:

$x = - 1, y = 7$

Solve by Linear Combination 3x+y=4 2x+y=5

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