Solve the System of @WORD x+y=2 , x-y=5

$x + y = 2$ , $x - y = 5$

Step 1

Subtract $y$ from both sides of the equation.

$x = 2 - y$

$x - y = 5$

Step 2

Replace all occurrences of $x$ with $2 - y$ in each equation.

Replace all occurrences of $x$ in $x - y = 5$ with $2 - y$ .

$(2 - y) - y = 5$

$x = 2 - y$

Simplify the left side.

Subtract $y$ from $- y$ .

$2 - 2 y = 5$

$x = 2 - y$

$2 - 2 y = 5$

$x = 2 - y$

$2 - 2 y = 5$

$x = 2 - y$

Step 3

Solve for $y$ in $2 - 2 y = 5$ .

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

Subtract $2$ from both sides of the equation.

$- 2 y = 5 - 2$

$x = 2 - y$

Subtract $2$ from $5$ .

$- 2 y = 3$

$x = 2 - y$

$- 2 y = 3$

$x = 2 - y$

Divide each term in $- 2 y = 3$ by $- 2$ and simplify.

Divide each term in $- 2 y = 3$ by $- 2$ .

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

$x = 2 - y$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

$x = 2 - y$

Divide $y$ by $1$ .

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

$x = 2 - y$

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

$x = 2 - y$

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

$x = 2 - y$

Simplify the right side.

Move the negative in front of the fraction.

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

$x = 2 - y$

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

$x = 2 - y$

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

$x = 2 - y$

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

$x = 2 - y$

Step 4

Replace all occurrences of $y$ with $- \frac{3}{2}$ in each equation.

Replace all occurrences of $y$ in $x = 2 - y$ with $- \frac{3}{2}$ .

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

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

Simplify the right side.

Simplify $2 - (- \frac{3}{2})$ .

Multiply $- (- \frac{3}{2})$ .

Multiply $- 1$ by $- 1$ .

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

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

Multiply $\frac{3}{2}$ by $1$ .

$x = 2 + \frac{3}{2}$

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

$x = 2 + \frac{3}{2}$

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

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = 2 \cdot \frac{2}{2} + \frac{3}{2}$

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

Combine $2$ and $\frac{2}{2}$ .

$x = \frac{2 \cdot 2}{2} + \frac{3}{2}$

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

Combine the numerators over the common denominator.

$x = \frac{2 \cdot 2 + 3}{2}$

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

Simplify the numerator.

Multiply $2$ by $2$ .

$x = \frac{4 + 3}{2}$

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

Add $4$ and $3$ .

$x = \frac{7}{2}$

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

$x = \frac{7}{2}$

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

$x = \frac{7}{2}$

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

$x = \frac{7}{2}$

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

$x = \frac{7}{2}$

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

Step 5

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(\frac{7}{2}, - \frac{3}{2})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{7}{2}, - \frac{3}{2})$

Equation Form:

$x = \frac{7}{2}, y = - \frac{3}{2}$

Step 7

Solve the System of @WORD x+y=2 , x-y=5

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