Solve by Substitution 3y=-1/2x+2 , y=-x+9

$3 y = - \frac{1}{2} x + 2$ , $y = - x + 9$

Step 1

Replace all occurrences of $y$ with $- x + 9$ in each equation.

Replace all occurrences of $y$ in $3 y = - \frac{1}{2} x + 2$ with $- x + 9$ .

$3 (- x + 9) = - \frac{1}{2} x + 2$

$y = - x + 9$

Simplify $3 (- x + 9) = - \frac{1}{2} x + 2$ .

Simplify the left side.

Simplify $3 (- x + 9)$ .

Apply the distributive property.

$3 (- x) + 3 \cdot 9 = - \frac{1}{2} x + 2$

$y = - x + 9$

Multiply.

Multiply $- 1$ by $3$ .

$- 3 x + 3 \cdot 9 = - \frac{1}{2} x + 2$

$y = - x + 9$

Multiply $3$ by $9$ .

$- 3 x + 27 = - \frac{1}{2} x + 2$

$y = - x + 9$

$- 3 x + 27 = - \frac{1}{2} x + 2$

$y = - x + 9$

$- 3 x + 27 = - \frac{1}{2} x + 2$

$y = - x + 9$

$- 3 x + 27 = - \frac{1}{2} x + 2$

$y = - x + 9$

Simplify the right side.

Combine $x$ and $\frac{1}{2}$ .

$- 3 x + 27 = - \frac{x}{2} + 2$

$y = - x + 9$

$- 3 x + 27 = - \frac{x}{2} + 2$

$y = - x + 9$

$- 3 x + 27 = - \frac{x}{2} + 2$

$y = - x + 9$

$- 3 x + 27 = - \frac{x}{2} + 2$

$y = - x + 9$

Step 2

Solve for $x$ in $- 3 x + 27 = - \frac{x}{2} + 2$ .

Move all terms containing $x$ to the left side of the equation.

Add $\frac{x}{2}$ to both sides of the equation.

$- 3 x + 27 + \frac{x}{2} = 2$

$y = - x + 9$

To write $- 3 x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- 3 x \cdot \frac{2}{2} + \frac{x}{2} + 27 = 2$

$y = - x + 9$

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

$\frac{- 3 x \cdot 2}{2} + \frac{x}{2} + 27 = 2$

$y = - x + 9$

Combine the numerators over the common denominator.

$\frac{- 3 x \cdot 2 + x}{2} + 27 = 2$

$y = - x + 9$

Simplify each term.

Simplify the numerator.

Factor $x$ out of $- 3 x \cdot 2 + x$ .

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

$\frac{x (- 3 \cdot 2) + x}{2} + 27 = 2$

$y = - x + 9$

Raise $x$ to the power of $1$ .

$\frac{x (- 3 \cdot 2) + x}{2} + 27 = 2$

$y = - x + 9$

Factor $x$ out of $x^{1}$ .

$\frac{x (- 3 \cdot 2) + x \cdot 1}{2} + 27 = 2$

$y = - x + 9$

Factor $x$ out of $x (- 3 \cdot 2) + x \cdot 1$ .

$\frac{x (- 3 \cdot 2 + 1)}{2} + 27 = 2$

$y = - x + 9$

$\frac{x (- 3 \cdot 2 + 1)}{2} + 27 = 2$

$y = - x + 9$

Multiply $- 3$ by $2$ .

$\frac{x (- 6 + 1)}{2} + 27 = 2$

$y = - x + 9$

Add $- 6$ and $1$ .

$\frac{x \cdot - 5}{2} + 27 = 2$

$y = - x + 9$

$\frac{x \cdot - 5}{2} + 27 = 2$

$y = - x + 9$

Move $- 5$ to the left of $x$ .

$\frac{- 5 \cdot x}{2} + 27 = 2$

$y = - x + 9$

Move the negative in front of the fraction.

$- \frac{5 x}{2} + 27 = 2$

$y = - x + 9$

$- \frac{5 x}{2} + 27 = 2$

$y = - x + 9$

$- \frac{5 x}{2} + 27 = 2$

$y = - x + 9$

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

Subtract $27$ from both sides of the equation.

$- \frac{5 x}{2} = 2 - 27$

$y = - x + 9$

Subtract $27$ from $2$ .

$- \frac{5 x}{2} = - 25$

$y = - x + 9$

$- \frac{5 x}{2} = - 25$

$y = - x + 9$

Multiply both sides of the equation by $- \frac{2}{5}$ .

$- \frac{2}{5} \cdot (- \frac{5 x}{2}) = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- \frac{2}{5} (- \frac{5 x}{2})$ .

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{2}{5}$ into the numerator.

$\frac{- 2}{5} \cdot (- \frac{5 x}{2}) = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

Move the leading negative in $- \frac{5 x}{2}$ into the numerator.

$\frac{- 2}{5} \cdot \frac{- 5 x}{2} = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

Factor $2$ out of $- 2$ .

$\frac{2 (- 1)}{5} \cdot \frac{- 5 x}{2} = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

Cancel the common factor.

$\frac{2 \cdot - 1}{5} \cdot \frac{- 5 x}{2} = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

Rewrite the expression.

$\frac{- 1}{5} \cdot (- 5 x) = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

$\frac{- 1}{5} \cdot (- 5 x) = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

Cancel the common factor of $5$ .

Factor $5$ out of $- 5 x$ .

$\frac{- 1}{5} \cdot (5 (- x)) = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

Cancel the common factor.

$\frac{- 1}{5} \cdot (5 (- x)) = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

Rewrite the expression.

$x = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

$x = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 x = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

Multiply $x$ by $1$ .

$x = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

$x = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

$x = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

$x = - \frac{2}{5} \cdot - 25$

$y = - x + 9$

Simplify the right side.

Simplify $- \frac{2}{5} \cdot - 25$ .

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{2}{5}$ into the numerator.

$x = \frac{- 2}{5} \cdot - 25$

$y = - x + 9$

Factor $5$ out of $- 25$ .

$x = \frac{- 2}{5} \cdot (5 (- 5))$

$y = - x + 9$

Cancel the common factor.

$x = \frac{- 2}{5} \cdot (5 \cdot - 5)$

$y = - x + 9$

Rewrite the expression.

$x = - 2 \cdot - 5$

$y = - x + 9$

$x = - 2 \cdot - 5$

$y = - x + 9$

Multiply $- 2$ by $- 5$ .

$x = 10$

$y = - x + 9$

$x = 10$

$y = - x + 9$

$x = 10$

$y = - x + 9$

$x = 10$

$y = - x + 9$

$x = 10$

$y = - x + 9$

Step 3

Replace all occurrences of $x$ with $10$ in each equation.

Replace all occurrences of $x$ in $y = - x + 9$ with $10$ .

$y = - (10) + 9$

$x = 10$

Simplify the right side.

Simplify $- (10) + 9$ .

Multiply $- 1$ by $10$ .

$y = - 10 + 9$

$x = 10$

Add $- 10$ and $9$ .

$y = - 1$

$x = 10$

$y = - 1$

$x = 10$

$y = - 1$

$x = 10$

$y = - 1$

$x = 10$

Step 4

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

$(10, - 1)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(10, - 1)$

Equation Form:

$x = 10, y = - 1$

Step 6

Solve by Substitution 3y=-1/2x+2 , y=-x+9

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