Solve by Substitution x/2-y/3=5/6 , x/5-y/4=29/10

$\frac{x}{2} - \frac{y}{3} = \frac{5}{6}$ , $\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

Step 1

Solve for $x$ in $\frac{x}{2} - \frac{y}{3} = \frac{5}{6}$ .

Add $\frac{y}{3}$ to both sides of the equation.

$\frac{x}{2} = \frac{5}{6} + \frac{y}{3}$

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

Multiply both sides of the equation by $2$ .

$2 (\frac{x}{2}) = 2 (\frac{5}{6} + \frac{y}{3})$

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$2 (\frac{x}{2}) = 2 (\frac{5}{6} + \frac{y}{3})$

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

Rewrite the expression.

$x = 2 (\frac{5}{6} + \frac{y}{3})$

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

$x = 2 (\frac{5}{6} + \frac{y}{3})$

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

$x = 2 (\frac{5}{6} + \frac{y}{3})$

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

Simplify the right side.

Simplify $2 (\frac{5}{6} + \frac{y}{3})$ .

Apply the distributive property.

$x = 2 (\frac{5}{6}) + 2 (\frac{y}{3})$

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

Cancel the common factor of $2$ .

Factor $2$ out of $6$ .

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

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

Cancel the common factor.

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

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

Rewrite the expression.

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

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

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

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

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

$x = \frac{5}{3} + \frac{2 y}{3}$

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

$x = \frac{5}{3} + \frac{2 y}{3}$

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

$x = \frac{5}{3} + \frac{2 y}{3}$

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

$x = \frac{5}{3} + \frac{2 y}{3}$

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

Reorder $\frac{5}{3}$ and $\frac{2 y}{3}$ .

$x = \frac{2 y}{3} + \frac{5}{3}$

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

$\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$

Step 2

Replace all occurrences of $x$ with $\frac{2 y}{3} + \frac{5}{3}$ in each equation.

Replace all occurrences of $x$ in $\frac{x}{5} - \frac{y}{4} = \frac{29}{10}$ with $\frac{2 y}{3} + \frac{5}{3}$ .

$\frac{\frac{2 y}{3} + \frac{5}{3}}{5} - \frac{y}{4} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Simplify the left side.

Simplify $\frac{\frac{2 y}{3} + \frac{5}{3}}{5} - \frac{y}{4}$ .

Simplify each term.

Combine the numerators over the common denominator.

$\frac{\frac{2 y + 5}{3}}{5} - \frac{y}{4} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Multiply the numerator by the reciprocal of the denominator.

$\frac{2 y + 5}{3} \cdot \frac{1}{5} - \frac{y}{4} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Multiply $\frac{2 y + 5}{3} \cdot \frac{1}{5}$ .

Multiply $\frac{2 y + 5}{3}$ by $\frac{1}{5}$ .

$\frac{2 y + 5}{3 \cdot 5} - \frac{y}{4} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Multiply $3$ by $5$ .

$\frac{2 y + 5}{15} - \frac{y}{4} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

$\frac{2 y + 5}{15} - \frac{y}{4} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

$\frac{2 y + 5}{15} - \frac{y}{4} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

To write $\frac{2 y + 5}{15}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{2 y + 5}{15} \cdot \frac{4}{4} - \frac{y}{4} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

To write $- \frac{y}{4}$ as a fraction with a common denominator, multiply by $\frac{15}{15}$ .

$\frac{2 y + 5}{15} \cdot \frac{4}{4} - \frac{y}{4} \cdot \frac{15}{15} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Write each expression with a common denominator of $60$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2 y + 5}{15}$ by $\frac{4}{4}$ .

$\frac{(2 y + 5) \cdot 4}{15 \cdot 4} - \frac{y}{4} \cdot \frac{15}{15} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Multiply $15$ by $4$ .

$\frac{(2 y + 5) \cdot 4}{60} - \frac{y}{4} \cdot \frac{15}{15} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Multiply $\frac{y}{4}$ by $\frac{15}{15}$ .

$\frac{(2 y + 5) \cdot 4}{60} - \frac{y \cdot 15}{4 \cdot 15} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Multiply $4$ by $15$ .

$\frac{(2 y + 5) \cdot 4}{60} - \frac{y \cdot 15}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

$\frac{(2 y + 5) \cdot 4}{60} - \frac{y \cdot 15}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Combine the numerators over the common denominator.

$\frac{(2 y + 5) \cdot 4 - y \cdot 15}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Simplify the numerator.

Apply the distributive property.

$\frac{2 y \cdot 4 + 5 \cdot 4 - y \cdot 15}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Multiply $4$ by $2$ .

$\frac{8 y + 5 \cdot 4 - y \cdot 15}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Multiply $5$ by $4$ .

$\frac{8 y + 20 - y \cdot 15}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Multiply $15$ by $- 1$ .

$\frac{8 y + 20 - 15 y}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Subtract $15 y$ from $8 y$ .

$\frac{- 7 y + 20}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

$\frac{- 7 y + 20}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Simplify with factoring out.

Factor $- 1$ out of $- 7 y$ .

$\frac{- (7 y) + 20}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Rewrite $20$ as $- 1 (- 20)$ .

$\frac{- (7 y) - 1 \cdot - 20}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Factor $- 1$ out of $- (7 y) - 1 (- 20)$ .

$\frac{- (7 y - 20)}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Simplify the expression.

Rewrite $- (7 y - 20)$ as $- 1 (7 y - 20)$ .

$\frac{- 1 (7 y - 20)}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Move the negative in front of the fraction.

$- \frac{7 y - 20}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

$- \frac{7 y - 20}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

$- \frac{7 y - 20}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

$- \frac{7 y - 20}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

$- \frac{7 y - 20}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

$- \frac{7 y - 20}{60} = \frac{29}{10}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Step 3

Solve for $y$ in $- \frac{7 y - 20}{60} = \frac{29}{10}$ .

Multiply both sides of the equation by $- 60$ .

$- 60 (- \frac{7 y - 20}{60}) = - 60 (\frac{29}{10})$

$x = \frac{2 y}{3} + \frac{5}{3}$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- 60 (- \frac{7 y - 20}{60})$ .

Cancel the common factor of $60$ .

Move the leading negative in $- \frac{7 y - 20}{60}$ into the numerator.

$- 60 \frac{- (7 y - 20)}{60} = - 60 (\frac{29}{10})$

$x = \frac{2 y}{3} + \frac{5}{3}$

Factor $60$ out of $- 60$ .

$60 (- 1) (\frac{- (7 y - 20)}{60}) = - 60 (\frac{29}{10})$

$x = \frac{2 y}{3} + \frac{5}{3}$

Cancel the common factor.

$60 \cdot (- 1 \frac{- (7 y - 20)}{60}) = - 60 (\frac{29}{10})$

$x = \frac{2 y}{3} + \frac{5}{3}$

Rewrite the expression.

$7 y - 20 = - 60 (\frac{29}{10})$

$x = \frac{2 y}{3} + \frac{5}{3}$

$7 y - 20 = - 60 (\frac{29}{10})$

$x = \frac{2 y}{3} + \frac{5}{3}$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 (7 y - 20) = - 60 (\frac{29}{10})$

$x = \frac{2 y}{3} + \frac{5}{3}$

Multiply $7 y - 20$ by $1$ .

$7 y - 20 = - 60 (\frac{29}{10})$

$x = \frac{2 y}{3} + \frac{5}{3}$

$7 y - 20 = - 60 (\frac{29}{10})$

$x = \frac{2 y}{3} + \frac{5}{3}$

$7 y - 20 = - 60 (\frac{29}{10})$

$x = \frac{2 y}{3} + \frac{5}{3}$

$7 y - 20 = - 60 (\frac{29}{10})$

$x = \frac{2 y}{3} + \frac{5}{3}$

Simplify the right side.

Simplify $- 60 (\frac{29}{10})$ .

Cancel the common factor of $10$ .

Factor $10$ out of $- 60$ .

$7 y - 20 = 10 (- 6) (\frac{29}{10})$

$x = \frac{2 y}{3} + \frac{5}{3}$

Cancel the common factor.

$7 y - 20 = 10 \cdot (- 6 (\frac{29}{10}))$

$x = \frac{2 y}{3} + \frac{5}{3}$

Rewrite the expression.

$7 y - 20 = - 6 \cdot 29$

$x = \frac{2 y}{3} + \frac{5}{3}$

$7 y - 20 = - 6 \cdot 29$

$x = \frac{2 y}{3} + \frac{5}{3}$

Multiply $- 6$ by $29$ .

$7 y - 20 = - 174$

$x = \frac{2 y}{3} + \frac{5}{3}$

$7 y - 20 = - 174$

$x = \frac{2 y}{3} + \frac{5}{3}$

$7 y - 20 = - 174$

$x = \frac{2 y}{3} + \frac{5}{3}$

$7 y - 20 = - 174$

$x = \frac{2 y}{3} + \frac{5}{3}$

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

Add $20$ to both sides of the equation.

$7 y = - 174 + 20$

$x = \frac{2 y}{3} + \frac{5}{3}$

Add $- 174$ and $20$ .

$7 y = - 154$

$x = \frac{2 y}{3} + \frac{5}{3}$

$7 y = - 154$

$x = \frac{2 y}{3} + \frac{5}{3}$

Divide each term in $7 y = - 154$ by $7$ and simplify.

Divide each term in $7 y = - 154$ by $7$ .

$\frac{7 y}{7} = \frac{- 154}{7}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 y}{7} = \frac{- 154}{7}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Divide $y$ by $1$ .

$y = \frac{- 154}{7}$

$x = \frac{2 y}{3} + \frac{5}{3}$

$y = \frac{- 154}{7}$

$x = \frac{2 y}{3} + \frac{5}{3}$

$y = \frac{- 154}{7}$

$x = \frac{2 y}{3} + \frac{5}{3}$

Simplify the right side.

Divide $- 154$ by $7$ .

$y = - 22$

$x = \frac{2 y}{3} + \frac{5}{3}$

$y = - 22$

$x = \frac{2 y}{3} + \frac{5}{3}$

$y = - 22$

$x = \frac{2 y}{3} + \frac{5}{3}$

$y = - 22$

$x = \frac{2 y}{3} + \frac{5}{3}$

Step 4

Replace all occurrences of $y$ with $- 22$ in each equation.

Replace all occurrences of $y$ in $x = \frac{2 y}{3} + \frac{5}{3}$ with $- 22$ .

$x = \frac{2 (- 22)}{3} + \frac{5}{3}$

$y = - 22$

Simplify the right side.

Simplify $\frac{2 (- 22)}{3} + \frac{5}{3}$ .

Combine the numerators over the common denominator.

$x = \frac{2 (- 22) + 5}{3}$

$y = - 22$

Simplify the expression.

Multiply $2$ by $- 22$ .

$x = \frac{- 44 + 5}{3}$

$y = - 22$

Add $- 44$ and $5$ .

$x = \frac{- 39}{3}$

$y = - 22$

Divide $- 39$ by $3$ .

$x = - 13$

$y = - 22$

$x = - 13$

$y = - 22$

$x = - 13$

$y = - 22$

$x = - 13$

$y = - 22$

$x = - 13$

$y = - 22$

Step 5

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

$(- 13, - 22)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 13, - 22)$

Equation Form:

$x = - 13, y = - 22$

Step 7

Solve by Substitution x/2-y/3=5/6 , x/5-y/4=29/10

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