Solve by Substitution 4p+2q=8 , q=2p+1

$4 p + 2 q = 8$ , $q = 2 p + 1$

Step 1

Replace all occurrences of $q$ with $2 p + 1$ in each equation.

Replace all occurrences of $q$ in $4 p + 2 q = 8$ with $2 p + 1$ .

$4 p + 2 (2 p + 1) = 8$

$q = 2 p + 1$

Simplify the left side.

Simplify $4 p + 2 (2 p + 1)$ .

Simplify each term.

Apply the distributive property.

$4 p + 2 (2 p) + 2 \cdot 1 = 8$

$q = 2 p + 1$

Multiply $2$ by $2$ .

$4 p + 4 p + 2 \cdot 1 = 8$

$q = 2 p + 1$

Multiply $2$ by $1$ .

$4 p + 4 p + 2 = 8$

$q = 2 p + 1$

$4 p + 4 p + 2 = 8$

$q = 2 p + 1$

Add $4 p$ and $4 p$ .

$8 p + 2 = 8$

$q = 2 p + 1$

$8 p + 2 = 8$

$q = 2 p + 1$

$8 p + 2 = 8$

$q = 2 p + 1$

$8 p + 2 = 8$

$q = 2 p + 1$

Step 2

Solve for $p$ in $8 p + 2 = 8$ .

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

Subtract $2$ from both sides of the equation.

$8 p = 8 - 2$

$q = 2 p + 1$

Subtract $2$ from $8$ .

$8 p = 6$

$q = 2 p + 1$

$8 p = 6$

$q = 2 p + 1$

Divide each term in $8 p = 6$ by $8$ and simplify.

Divide each term in $8 p = 6$ by $8$ .

$\frac{8 p}{8} = \frac{6}{8}$

$q = 2 p + 1$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 p}{8} = \frac{6}{8}$

$q = 2 p + 1$

Divide $p$ by $1$ .

$p = \frac{6}{8}$

$q = 2 p + 1$

$p = \frac{6}{8}$

$q = 2 p + 1$

$p = \frac{6}{8}$

$q = 2 p + 1$

Simplify the right side.

Cancel the common factor of $6$ and $8$ .

Factor $2$ out of $6$ .

$p = \frac{2 (3)}{8}$

$q = 2 p + 1$

Cancel the common factors.

Factor $2$ out of $8$ .

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

$q = 2 p + 1$

Cancel the common factor.

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

$q = 2 p + 1$

Rewrite the expression.

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

$q = 2 p + 1$

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

$q = 2 p + 1$

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

$q = 2 p + 1$

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

$q = 2 p + 1$

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

$q = 2 p + 1$

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

$q = 2 p + 1$

Step 3

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

Replace all occurrences of $p$ in $q = 2 p + 1$ with $\frac{3}{4}$ .

$q = 2 (\frac{3}{4}) + 1$

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

Simplify the right side.

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

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

$q = 2 (\frac{3}{2 (2)}) + 1$

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

Cancel the common factor.

$q = 2 (\frac{3}{2 \cdot 2}) + 1$

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

Rewrite the expression.

$q = \frac{3}{2} + 1$

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

$q = \frac{3}{2} + 1$

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

Simplify the expression.

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

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

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

Combine the numerators over the common denominator.

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

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

Add $3$ and $2$ .

$q = \frac{5}{2}$

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

$q = \frac{5}{2}$

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

$q = \frac{5}{2}$

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

$q = \frac{5}{2}$

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

$q = \frac{5}{2}$

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

Step 4

List all of the solutions.

$q = \frac{5}{2}, p = \frac{3}{4}$

Solve by Substitution 4p+2q=8 , q=2p+1

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