Solve by Substitution 2x+y=1 , x-y=3

$2 x + y = 1$ , $x - y = 3$

Step 1

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

$y = 1 - 2 x$

$x - y = 3$

Step 2

Replace all occurrences of $y$ with $1 - 2 x$ in each equation.

Replace all occurrences of $y$ in $x - y = 3$ with $1 - 2 x$ .

$x - (1 - 2 x) = 3$

$y = 1 - 2 x$

Simplify the left side.

Simplify $x - (1 - 2 x)$ .

Simplify each term.

Apply the distributive property.

$x - 1 \cdot 1 - (- 2 x) = 3$

$y = 1 - 2 x$

Multiply $- 1$ by $1$ .

$x - 1 - (- 2 x) = 3$

$y = 1 - 2 x$

Multiply $- 2$ by $- 1$ .

$x - 1 + 2 x = 3$

$y = 1 - 2 x$

$x - 1 + 2 x = 3$

$y = 1 - 2 x$

Add $x$ and $2 x$ .

$3 x - 1 = 3$

$y = 1 - 2 x$

$3 x - 1 = 3$

$y = 1 - 2 x$

$3 x - 1 = 3$

$y = 1 - 2 x$

$3 x - 1 = 3$

$y = 1 - 2 x$

Step 3

Solve for $x$ in $3 x - 1 = 3$ .

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

Add $1$ to both sides of the equation.

$3 x = 3 + 1$

$y = 1 - 2 x$

Add $3$ and $1$ .

$3 x = 4$

$y = 1 - 2 x$

$3 x = 4$

$y = 1 - 2 x$

Divide each term in $3 x = 4$ by $3$ and simplify.

Divide each term in $3 x = 4$ by $3$ .

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

$y = 1 - 2 x$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

$y = 1 - 2 x$

Divide $x$ by $1$ .

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

$y = 1 - 2 x$

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

$y = 1 - 2 x$

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

$y = 1 - 2 x$

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

$y = 1 - 2 x$

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

$y = 1 - 2 x$

Step 4

Replace all occurrences of $x$ with $\frac{4}{3}$ in each equation.

Replace all occurrences of $x$ in $y = 1 - 2 x$ with $\frac{4}{3}$ .

$y = 1 - 2 (\frac{4}{3})$

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

Simplify the right side.

Simplify $1 - 2 (\frac{4}{3})$ .

Simplify each term.

Multiply $- 2 (\frac{4}{3})$ .

Combine $- 2$ and $\frac{4}{3}$ .

$y = 1 + \frac{- 2 \cdot 4}{3}$

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

Multiply $- 2$ by $4$ .

$y = 1 + \frac{- 8}{3}$

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

$y = 1 + \frac{- 8}{3}$

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

Move the negative in front of the fraction.

$y = 1 - \frac{8}{3}$

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

$y = 1 - \frac{8}{3}$

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

Simplify the expression.

Write $1$ as a fraction with a common denominator.

$y = \frac{3}{3} - \frac{8}{3}$

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

Combine the numerators over the common denominator.

$y = \frac{3 - 8}{3}$

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

Subtract $8$ from $3$ .

$y = \frac{- 5}{3}$

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

Move the negative in front of the fraction.

$y = - \frac{5}{3}$

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

$y = - \frac{5}{3}$

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

$y = - \frac{5}{3}$

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

$y = - \frac{5}{3}$

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

$y = - \frac{5}{3}$

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

Step 5

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

$(\frac{4}{3}, - \frac{5}{3})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{4}{3}, - \frac{5}{3})$

Equation Form:

$x = \frac{4}{3}, y = - \frac{5}{3}$

Step 7

Solve by Substitution 2x+y=1 , x-y=3

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