Determine if Dependent, Independent, or Inconsistent x+y=2 , x-y=10

$x + y = 2$ , $x - y = 10$

Step 1

Solve the system of equations.

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

$x + y = 2$

$(- 1) \cdot (x - y) = (- 1) (10)$

Simplify.

Simplify the left side.

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

Apply the distributive property.

$x + y = 2$

$- 1 x - 1 (- y) = (- 1) (10)$

Rewrite $- 1 x$ as $- x$ .

$x + y = 2$

$- x - 1 (- y) = (- 1) (10)$

Multiply $- 1 (- y)$ .

Multiply $- 1$ by $- 1$ .

$x + y = 2$

$- x + 1 y = (- 1) (10)$

Multiply $y$ by $1$ .

$x + y = 2$

$- x + y = (- 1) (10)$

$x + y = 2$

$- x + y = (- 1) (10)$

$x + y = 2$

$- x + y = (- 1) (10)$

$x + y = 2$

$- x + y = (- 1) (10)$

Simplify the right side.

Multiply $- 1$ by $10$ .

$x + y = 2$

$- x + y = - 10$

$x + y = 2$

$- x + y = - 10$

$x + y = 2$

$- x + y = - 10$

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

		$x$	$+$	$y$	$=$			$2$
$+$	$-$	$x$	$+$	$y$	$=$	$-$	$1$	$0$
			$2$	$y$	$=$		$-$	$8$

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

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

$\frac{2 y}{2} = \frac{- 8}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{- 8}{2}$

Divide $y$ by $1$ .

$y = \frac{- 8}{2}$

Simplify the right side.

Divide $- 8$ by $2$ .

$y = - 4$

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

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

$x - 4 = 2$

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

Add $4$ to both sides of the equation.

$x = 2 + 4$

Add $2$ and $4$ .

$x = 6$

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

$(6, - 4)$

Step 2

Since the system has a point of intersection, the system is independent.

Independent

Step 3

Determine if Dependent, Independent, or Inconsistent x+y=2 , x-y=10

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