Solve by Substitution 2x-y=10 3x-2y=8

$2 x - y = 10$ $3 x - 2 y = 8$

Step 1

Solve for $x$ in $2 x - y = 10$ .

Add $y$ to both sides of the equation.

$2 x = 10 + y$

$3 x - 2 y = 8$

Divide each term in $2 x = 10 + y$ by $2$ and simplify.

Divide each term in $2 x = 10 + y$ by $2$ .

$\frac{2 x}{2} = \frac{10}{2} + \frac{y}{2}$

$3 x - 2 y = 8$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{10}{2} + \frac{y}{2}$

$3 x - 2 y = 8$

Divide $x$ by $1$ .

$x = \frac{10}{2} + \frac{y}{2}$

$3 x - 2 y = 8$

$x = \frac{10}{2} + \frac{y}{2}$

$3 x - 2 y = 8$

$x = \frac{10}{2} + \frac{y}{2}$

$3 x - 2 y = 8$

Simplify the right side.

Divide $10$ by $2$ .

$x = 5 + \frac{y}{2}$

$3 x - 2 y = 8$

$x = 5 + \frac{y}{2}$

$3 x - 2 y = 8$

$x = 5 + \frac{y}{2}$

$3 x - 2 y = 8$

$x = 5 + \frac{y}{2}$

$3 x - 2 y = 8$

Step 2

Replace all occurrences of $x$ with $5 + \frac{y}{2}$ in each equation.

Replace all occurrences of $x$ in $3 x - 2 y = 8$ with $5 + \frac{y}{2}$ .

$3 (5 + \frac{y}{2}) - 2 y = 8$

$x = 5 + \frac{y}{2}$

Simplify the left side.

Simplify $3 (5 + \frac{y}{2}) - 2 y$ .

Simplify each term.

Apply the distributive property.

$3 \cdot 5 + 3 (\frac{y}{2}) - 2 y = 8$

$x = 5 + \frac{y}{2}$

Multiply $3$ by $5$ .

$15 + 3 (\frac{y}{2}) - 2 y = 8$

$x = 5 + \frac{y}{2}$

Combine $3$ and $\frac{y}{2}$ .

$15 + \frac{3 y}{2} - 2 y = 8$

$x = 5 + \frac{y}{2}$

$15 + \frac{3 y}{2} - 2 y = 8$

$x = 5 + \frac{y}{2}$

To write $- 2 y$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$15 + \frac{3 y}{2} - 2 y \cdot \frac{2}{2} = 8$

$x = 5 + \frac{y}{2}$

Simplify terms.

Combine $- 2 y$ and $\frac{2}{2}$ .

$15 + \frac{3 y}{2} + \frac{- 2 y \cdot 2}{2} = 8$

$x = 5 + \frac{y}{2}$

Combine the numerators over the common denominator.

$15 + \frac{3 y - 2 y \cdot 2}{2} = 8$

$x = 5 + \frac{y}{2}$

$15 + \frac{3 y - 2 y \cdot 2}{2} = 8$

$x = 5 + \frac{y}{2}$

Simplify each term.

Simplify the numerator.

Factor $y$ out of $3 y - 2 y \cdot 2$ .

Factor $y$ out of $3 y$ .

$15 + \frac{y \cdot 3 - 2 y \cdot 2}{2} = 8$

$x = 5 + \frac{y}{2}$

Factor $y$ out of $- 2 y \cdot 2$ .

$15 + \frac{y \cdot 3 + y (- 2 \cdot 2)}{2} = 8$

$x = 5 + \frac{y}{2}$

Factor $y$ out of $y \cdot 3 + y (- 2 \cdot 2)$ .

$15 + \frac{y (3 - 2 \cdot 2)}{2} = 8$

$x = 5 + \frac{y}{2}$

$15 + \frac{y (3 - 2 \cdot 2)}{2} = 8$

$x = 5 + \frac{y}{2}$

Multiply $- 2$ by $2$ .

$15 + \frac{y (3 - 4)}{2} = 8$

$x = 5 + \frac{y}{2}$

Subtract $4$ from $3$ .

$15 + \frac{y \cdot - 1}{2} = 8$

$x = 5 + \frac{y}{2}$

$15 + \frac{y \cdot - 1}{2} = 8$

$x = 5 + \frac{y}{2}$

Move $- 1$ to the left of $y$ .

$15 + \frac{- 1 \cdot y}{2} = 8$

$x = 5 + \frac{y}{2}$

Move the negative in front of the fraction.

$15 - \frac{y}{2} = 8$

$x = 5 + \frac{y}{2}$

$15 - \frac{y}{2} = 8$

$x = 5 + \frac{y}{2}$

$15 - \frac{y}{2} = 8$

$x = 5 + \frac{y}{2}$

$15 - \frac{y}{2} = 8$

$x = 5 + \frac{y}{2}$

$15 - \frac{y}{2} = 8$

$x = 5 + \frac{y}{2}$

Step 3

Solve for $y$ in $15 - \frac{y}{2} = 8$ .

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

Subtract $15$ from both sides of the equation.

$- \frac{y}{2} = 8 - 15$

$x = 5 + \frac{y}{2}$

Subtract $15$ from $8$ .

$- \frac{y}{2} = - 7$

$x = 5 + \frac{y}{2}$

$- \frac{y}{2} = - 7$

$x = 5 + \frac{y}{2}$

Multiply both sides of the equation by $- 2$ .

$- 2 (- \frac{y}{2}) = - 2 \cdot - 7$

$x = 5 + \frac{y}{2}$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- 2 (- \frac{y}{2})$ .

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{y}{2}$ into the numerator.

$- 2 \frac{- y}{2} = - 2 \cdot - 7$

$x = 5 + \frac{y}{2}$

Factor $2$ out of $- 2$ .

$2 (- 1) (\frac{- y}{2}) = - 2 \cdot - 7$

$x = 5 + \frac{y}{2}$

Cancel the common factor.

$2 \cdot (- 1 \frac{- y}{2}) = - 2 \cdot - 7$

$x = 5 + \frac{y}{2}$

Rewrite the expression.

$y = - 2 \cdot - 7$

$x = 5 + \frac{y}{2}$

$y = - 2 \cdot - 7$

$x = 5 + \frac{y}{2}$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 y = - 2 \cdot - 7$

$x = 5 + \frac{y}{2}$

Multiply $y$ by $1$ .

$y = - 2 \cdot - 7$

$x = 5 + \frac{y}{2}$

$y = - 2 \cdot - 7$

$x = 5 + \frac{y}{2}$

$y = - 2 \cdot - 7$

$x = 5 + \frac{y}{2}$

$y = - 2 \cdot - 7$

$x = 5 + \frac{y}{2}$

Simplify the right side.

Multiply $- 2$ by $- 7$ .

$y = 14$

$x = 5 + \frac{y}{2}$

$y = 14$

$x = 5 + \frac{y}{2}$

$y = 14$

$x = 5 + \frac{y}{2}$

$y = 14$

$x = 5 + \frac{y}{2}$

Step 4

Replace all occurrences of $y$ with $14$ in each equation.

Replace all occurrences of $y$ in $x = 5 + \frac{y}{2}$ with $14$ .

$x = 5 + \frac{14}{2}$

$y = 14$

Simplify the right side.

Simplify $5 + \frac{14}{2}$ .

Divide $14$ by $2$ .

$x = 5 + 7$

$y = 14$

Add $5$ and $7$ .

$x = 12$

$y = 14$

$x = 12$

$y = 14$

$x = 12$

$y = 14$

$x = 12$

$y = 14$

Step 5

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

$(12, 14)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(12, 14)$

Equation Form:

$x = 12, y = 14$

Step 7

Solve by Substitution 2x-y=10 3x-2y=8

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