Solve by Substitution x+y=3 , x-y=1

$x + y = 3$ , $x - y = 1$

Step 1

Subtract $y$ from both sides of the equation.

$x = 3 - y$

$x - y = 1$

Step 2

Replace all occurrences of $x$ with $3 - y$ in each equation.

Replace all occurrences of $x$ in $x - y = 1$ with $3 - y$ .

$(3 - y) - y = 1$

$x = 3 - y$

Simplify the left side.

Subtract $y$ from $- y$ .

$3 - 2 y = 1$

$x = 3 - y$

$3 - 2 y = 1$

$x = 3 - y$

$3 - 2 y = 1$

$x = 3 - y$

Step 3

Solve for $y$ in $3 - 2 y = 1$ .

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

Subtract $3$ from both sides of the equation.

$- 2 y = 1 - 3$

$x = 3 - y$

Subtract $3$ from $1$ .

$- 2 y = - 2$

$x = 3 - y$

$- 2 y = - 2$

$x = 3 - y$

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

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

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

$x = 3 - y$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

$x = 3 - y$

Divide $y$ by $1$ .

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

$x = 3 - y$

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

$x = 3 - y$

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

$x = 3 - y$

Simplify the right side.

Divide $- 2$ by $- 2$ .

$y = 1$

$x = 3 - y$

$y = 1$

$x = 3 - y$

$y = 1$

$x = 3 - y$

$y = 1$

$x = 3 - y$

Step 4

Replace all occurrences of $y$ with $1$ in each equation.

Replace all occurrences of $y$ in $x = 3 - y$ with $1$ .

$x = 3 - (1)$

$y = 1$

Simplify the right side.

Simplify $3 - (1)$ .

Multiply $- 1$ by $1$ .

$x = 3 - 1$

$y = 1$

Subtract $1$ from $3$ .

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

Step 5

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

$(2, 1)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(2, 1)$

Equation Form:

$x = 2, y = 1$

Step 7

Solve by Substitution x+y=3 , x-y=1

Related Questions

Support LCMGCF.com by disabling your adblocker.