Solve by Substitution x+y=3 , 2x-8y=19

$x + y = 3$ , $2 x - 8 y = 19$

Step 1

Subtract $y$ from both sides of the equation.

$x = 3 - y$

$2 x - 8 y = 19$

Step 2

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

Replace all occurrences of $x$ in $2 x - 8 y = 19$ with $3 - y$ .

$2 (3 - y) - 8 y = 19$

$x = 3 - y$

Simplify the left side.

Simplify $2 (3 - y) - 8 y$ .

Simplify each term.

Apply the distributive property.

$2 \cdot 3 + 2 (- y) - 8 y = 19$

$x = 3 - y$

Multiply $2$ by $3$ .

$6 + 2 (- y) - 8 y = 19$

$x = 3 - y$

Multiply $- 1$ by $2$ .

$6 - 2 y - 8 y = 19$

$x = 3 - y$

$6 - 2 y - 8 y = 19$

$x = 3 - y$

Subtract $8 y$ from $- 2 y$ .

$6 - 10 y = 19$

$x = 3 - y$

$6 - 10 y = 19$

$x = 3 - y$

$6 - 10 y = 19$

$x = 3 - y$

$6 - 10 y = 19$

$x = 3 - y$

Step 3

Solve for $y$ in $6 - 10 y = 19$ .

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

Subtract $6$ from both sides of the equation.

$- 10 y = 19 - 6$

$x = 3 - y$

Subtract $6$ from $19$ .

$- 10 y = 13$

$x = 3 - y$

$- 10 y = 13$

$x = 3 - y$

Divide each term in $- 10 y = 13$ by $- 10$ and simplify.

Divide each term in $- 10 y = 13$ by $- 10$ .

$\frac{- 10 y}{- 10} = \frac{13}{- 10}$

$x = 3 - y$

Simplify the left side.

Cancel the common factor of $- 10$ .

Cancel the common factor.

$\frac{- 10 y}{- 10} = \frac{13}{- 10}$

$x = 3 - y$

Divide $y$ by $1$ .

$y = \frac{13}{- 10}$

$x = 3 - y$

$y = \frac{13}{- 10}$

$x = 3 - y$

$y = \frac{13}{- 10}$

$x = 3 - y$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{13}{10}$

$x = 3 - y$

$y = - \frac{13}{10}$

$x = 3 - y$

$y = - \frac{13}{10}$

$x = 3 - y$

$y = - \frac{13}{10}$

$x = 3 - y$

Step 4

Replace all occurrences of $y$ with $- \frac{13}{10}$ in each equation.

Replace all occurrences of $y$ in $x = 3 - y$ with $- \frac{13}{10}$ .

$x = 3 - (- \frac{13}{10})$

$y = - \frac{13}{10}$

Simplify the right side.

Simplify $3 - (- \frac{13}{10})$ .

Multiply $- (- \frac{13}{10})$ .

Multiply $- 1$ by $- 1$ .

$x = 3 + 1 (\frac{13}{10})$

$y = - \frac{13}{10}$

Multiply $\frac{13}{10}$ by $1$ .

$x = 3 + \frac{13}{10}$

$y = - \frac{13}{10}$

$x = 3 + \frac{13}{10}$

$y = - \frac{13}{10}$

To write $3$ as a fraction with a common denominator, multiply by $\frac{10}{10}$ .

$x = 3 \cdot \frac{10}{10} + \frac{13}{10}$

$y = - \frac{13}{10}$

Combine $3$ and $\frac{10}{10}$ .

$x = \frac{3 \cdot 10}{10} + \frac{13}{10}$

$y = - \frac{13}{10}$

Combine the numerators over the common denominator.

$x = \frac{3 \cdot 10 + 13}{10}$

$y = - \frac{13}{10}$

Simplify the numerator.

Multiply $3$ by $10$ .

$x = \frac{30 + 13}{10}$

$y = - \frac{13}{10}$

Add $30$ and $13$ .

$x = \frac{43}{10}$

$y = - \frac{13}{10}$

$x = \frac{43}{10}$

$y = - \frac{13}{10}$

$x = \frac{43}{10}$

$y = - \frac{13}{10}$

$x = \frac{43}{10}$

$y = - \frac{13}{10}$

$x = \frac{43}{10}$

$y = - \frac{13}{10}$

Step 5

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

$(\frac{43}{10}, - \frac{13}{10})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{43}{10}, - \frac{13}{10})$

Equation Form:

$x = \frac{43}{10}, y = - \frac{13}{10}$

Step 7

Solve by Substitution x+y=3 , 2x-8y=19

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