Solve by Substitution p=6-2q , p=4+q

$p = 6 - 2 q$ , $p = 4 + q$

Step 1

Eliminate the equal sides of each equation and combine.

$6 - 2 q = 4 + q$

Step 2

Solve $6 - 2 q = 4 + q$ for $q$ .

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

Subtract $q$ from both sides of the equation.

$6 - 2 q - q = 4$

Subtract $q$ from $- 2 q$ .

$6 - 3 q = 4$

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

Subtract $6$ from both sides of the equation.

$- 3 q = 4 - 6$

Subtract $6$ from $4$ .

$- 3 q = - 2$

Divide each term in $- 3 q = - 2$ by $- 3$ and simplify.

Divide each term in $- 3 q = - 2$ by $- 3$ .

$\frac{- 3 q}{- 3} = \frac{- 2}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 q}{- 3} = \frac{- 2}{- 3}$

Divide $q$ by $1$ .

$q = \frac{- 2}{- 3}$

Simplify the right side.

Dividing two negative values results in a positive value.

$q = \frac{2}{3}$

Step 3

Evaluate $p$ when $q = \frac{2}{3}$ .

Substitute $\frac{2}{3}$ for $q$ .

$p = 4 + \frac{2}{3}$

Substitute $\frac{2}{3}$ for $q$ in $p = 4 + \frac{2}{3}$ and solve for $p$ .

Remove parentheses.

$p = 4 + \frac{2}{3}$

Simplify $4 + \frac{2}{3}$ .

To write $4$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$p = 4 \cdot \frac{3}{3} + \frac{2}{3}$

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

$p = \frac{4 \cdot 3}{3} + \frac{2}{3}$

Combine the numerators over the common denominator.

$p = \frac{4 \cdot 3 + 2}{3}$

Simplify the numerator.

Multiply $4$ by $3$ .

$p = \frac{12 + 2}{3}$

Add $12$ and $2$ .

$p = \frac{14}{3}$

Step 4

The solution of the system of equations is all the values that make the system true.

$p = \frac{14}{3}, q = \frac{2}{3}$

Solve by Substitution p=6-2q , p=4+q

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