Solve by Substitution 2x+y=5 , x+y=4

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

Step 1

Subtract $2 x$ from both sides of the equation.

$y = 5 - 2 x$

$x + y = 4$

Step 2

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

Replace all occurrences of $y$ in $x + y = 4$ with $5 - 2 x$ .

$x + 5 - 2 x = 4$

$y = 5 - 2 x$

Simplify the left side.

Simplify $x + 5 - 2 x$ .

Remove parentheses.

$x + 5 - 2 x = 4$

$y = 5 - 2 x$

Subtract $2 x$ from $x$ .

$- x + 5 = 4$

$y = 5 - 2 x$

$- x + 5 = 4$

$y = 5 - 2 x$

$- x + 5 = 4$

$y = 5 - 2 x$

$- x + 5 = 4$

$y = 5 - 2 x$

Step 3

Solve for $x$ in $- x + 5 = 4$ .

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

Subtract $5$ from both sides of the equation.

$- x = 4 - 5$

$y = 5 - 2 x$

Subtract $5$ from $4$ .

$- x = - 1$

$y = 5 - 2 x$

$- x = - 1$

$y = 5 - 2 x$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

$y = 5 - 2 x$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

$y = 5 - 2 x$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

$y = 5 - 2 x$

$x = \frac{- 1}{- 1}$

$y = 5 - 2 x$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

$y = 5 - 2 x$

$x = 1$

$y = 5 - 2 x$

$x = 1$

$y = 5 - 2 x$

$x = 1$

$y = 5 - 2 x$

Step 4

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

Replace all occurrences of $x$ in $y = 5 - 2 x$ with $1$ .

$y = 5 - 2 \cdot 1$

$x = 1$

Simplify the right side.

Simplify $5 - 2 \cdot 1$ .

Multiply $- 2$ by $1$ .

$y = 5 - 2$

$x = 1$

Subtract $2$ from $5$ .

$y = 3$

$x = 1$

$y = 3$

$x = 1$

$y = 3$

$x = 1$

$y = 3$

$x = 1$

Step 5

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

$(1, 3)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(1, 3)$

Equation Form:

$x = 1, y = 3$

Step 7

Solve by Substitution 2x+y=5 , x+y=4

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