Solve by Substitution x+y=3 , 2x-y=0

$x + y = 3$ , $2 x - y = 0$

Step 1

Subtract $y$ from both sides of the equation.

$x = 3 - y$

$2 x - y = 0$

Step 2

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

Replace all occurrences of $x$ in $2 x - y = 0$ with $3 - y$ .

$2 (3 - y) - y = 0$

$x = 3 - y$

Simplify the left side.

Simplify $2 (3 - y) - y$ .

Simplify each term.

Apply the distributive property.

$2 \cdot 3 + 2 (- y) - y = 0$

$x = 3 - y$

Multiply $2$ by $3$ .

$6 + 2 (- y) - y = 0$

$x = 3 - y$

Multiply $- 1$ by $2$ .

$6 - 2 y - y = 0$

$x = 3 - y$

$6 - 2 y - y = 0$

$x = 3 - y$

Subtract $y$ from $- 2 y$ .

$6 - 3 y = 0$

$x = 3 - y$

$6 - 3 y = 0$

$x = 3 - y$

$6 - 3 y = 0$

$x = 3 - y$

$6 - 3 y = 0$

$x = 3 - y$

Step 3

Solve for $y$ in $6 - 3 y = 0$ .

Subtract $6$ from both sides of the equation.

$- 3 y = - 6$

$x = 3 - y$

Divide each term in $- 3 y = - 6$ by $- 3$ and simplify.

Divide each term in $- 3 y = - 6$ by $- 3$ .

$\frac{- 3 y}{- 3} = \frac{- 6}{- 3}$

$x = 3 - y$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 y}{- 3} = \frac{- 6}{- 3}$

$x = 3 - y$

Divide $y$ by $1$ .

$y = \frac{- 6}{- 3}$

$x = 3 - y$

$y = \frac{- 6}{- 3}$

$x = 3 - y$

$y = \frac{- 6}{- 3}$

$x = 3 - y$

Simplify the right side.

Divide $- 6$ by $- 3$ .

$y = 2$

$x = 3 - y$

$y = 2$

$x = 3 - y$

$y = 2$

$x = 3 - y$

$y = 2$

$x = 3 - y$

Step 4

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

Replace all occurrences of $y$ in $x = 3 - y$ with $2$ .

$x = 3 - (2)$

$y = 2$

Simplify the right side.

Simplify $3 - (2)$ .

Multiply $- 1$ by $2$ .

$x = 3 - 2$

$y = 2$

Subtract $2$ from $3$ .

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

Step 5

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

$(1, 2)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(1, 2)$

Equation Form:

$x = 1, y = 2$

Step 7

Solve by Substitution x+y=3 , 2x-y=0

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