Solve by Substitution 3x+4y=5 , 6x+8y=10

$3 x + 4 y = 5$ , $6 x + 8 y = 10$

Step 1

Solve for $x$ in $3 x + 4 y = 5$ .

Subtract $4 y$ from both sides of the equation.

$3 x = 5 - 4 y$

$6 x + 8 y = 10$

Divide each term in $3 x = 5 - 4 y$ by $3$ and simplify.

Divide each term in $3 x = 5 - 4 y$ by $3$ .

$\frac{3 x}{3} = \frac{5}{3} + \frac{- 4 y}{3}$

$6 x + 8 y = 10$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{5}{3} + \frac{- 4 y}{3}$

$6 x + 8 y = 10$

Divide $x$ by $1$ .

$x = \frac{5}{3} + \frac{- 4 y}{3}$

$6 x + 8 y = 10$

$x = \frac{5}{3} + \frac{- 4 y}{3}$

$6 x + 8 y = 10$

$x = \frac{5}{3} + \frac{- 4 y}{3}$

$6 x + 8 y = 10$

Simplify the right side.

Move the negative in front of the fraction.

$x = \frac{5}{3} - \frac{4 y}{3}$

$6 x + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

$6 x + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

$6 x + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

$6 x + 8 y = 10$

Step 2

Replace all occurrences of $x$ with $\frac{5}{3} - \frac{4 y}{3}$ in each equation.

Replace all occurrences of $x$ in $6 x + 8 y = 10$ with $\frac{5}{3} - \frac{4 y}{3}$ .

$6 (\frac{5}{3} - \frac{4 y}{3}) + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

Simplify the left side.

Simplify $6 (\frac{5}{3} - \frac{4 y}{3}) + 8 y$ .

Simplify each term.

Apply the distributive property.

$6 (\frac{5}{3}) + 6 (- \frac{4 y}{3}) + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

Cancel the common factor of $3$ .

Factor $3$ out of $6$ .

$3 (2) (\frac{5}{3}) + 6 (- \frac{4 y}{3}) + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

Cancel the common factor.

$3 \cdot (2 (\frac{5}{3})) + 6 (- \frac{4 y}{3}) + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

Rewrite the expression.

$2 \cdot 5 + 6 (- \frac{4 y}{3}) + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

$2 \cdot 5 + 6 (- \frac{4 y}{3}) + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

Multiply $2$ by $5$ .

$10 + 6 (- \frac{4 y}{3}) + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{4 y}{3}$ into the numerator.

$10 + 6 (\frac{- 4 y}{3}) + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

Factor $3$ out of $6$ .

$10 + 3 (2) (\frac{- 4 y}{3}) + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

Cancel the common factor.

$10 + 3 \cdot (2 (\frac{- 4 y}{3})) + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

Rewrite the expression.

$10 + 2 (- 4 y) + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

$10 + 2 (- 4 y) + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

Multiply $- 4$ by $2$ .

$10 - 8 y + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

$10 - 8 y + 8 y = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

Combine the opposite terms in $10 - 8 y + 8 y$ .

Add $- 8 y$ and $8 y$ .

$10 + 0 = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

Add $10$ and $0$ .

$10 = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

$10 = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

$10 = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

$10 = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

$10 = 10$

$x = \frac{5}{3} - \frac{4 y}{3}$

Step 3

Remove any equations from the system that are always true.

$x = \frac{5}{3} - \frac{4 y}{3}$

Step 4

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