Solve the System of Equations 1/4x+1/3y=1/12 1/4(-x+y)=-5/4

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$ $\frac{1}{4} (- x + y) = - \frac{5}{4}$

Solve for $x$ in $- \frac{x}{4} + \frac{y}{4} = - \frac{5}{4}$ .

Subtract $\frac{y}{4}$ from both sides of the equation.

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

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Multiply both sides of the equation by $- 4$ .

$- 4 (- \frac{x}{4}) = - 4 (- \frac{5}{4} - \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- 4 (- \frac{x}{4})$ .

Cancel the common factor of $4$ .

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

$- 4 \frac{- x}{4} = - 4 (- \frac{5}{4} - \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Factor $4$ out of $- 4$ .

$4 (- 1) (\frac{- x}{4}) = - 4 (- \frac{5}{4} - \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Cancel the common factor.

$4 \cdot (- 1 \frac{- x}{4}) = - 4 (- \frac{5}{4} - \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Rewrite the expression.

$x = - 4 (- \frac{5}{4} - \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

$x = - 4 (- \frac{5}{4} - \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 x = - 4 (- \frac{5}{4} - \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Multiply $x$ by $1$ .

$x = - 4 (- \frac{5}{4} - \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

$x = - 4 (- \frac{5}{4} - \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

$x = - 4 (- \frac{5}{4} - \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

$x = - 4 (- \frac{5}{4} - \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Simplify the right side.

Simplify $- 4 (- \frac{5}{4} - \frac{y}{4})$ .

Apply the distributive property.

$x = - 4 (- \frac{5}{4}) - 4 (- \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{5}{4}$ into the numerator.

$x = - 4 (\frac{- 5}{4}) - 4 (- \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Factor $4$ out of $- 4$ .

$x = 4 (- 1) (\frac{- 5}{4}) - 4 (- \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Cancel the common factor.

$x = 4 \cdot (- 1 (\frac{- 5}{4})) - 4 (- \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Rewrite the expression.

$x = 5 - 4 (- \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

$x = 5 - 4 (- \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Multiply $- 1$ by $- 5$ .

$x = 5 - 4 (- \frac{y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Cancel the common factor of $4$ .

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

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

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Factor $4$ out of $- 4$ .

$x = 5 + 4 (- 1) (\frac{- y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Cancel the common factor.

$x = 5 + 4 \cdot (- 1 \frac{- y}{4})$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Rewrite the expression.

$x = 5 + y$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

$x = 5 + y$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Multiply.

Multiply $- 1$ by $- 1$ .

$x = 5 + 1 y$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Multiply $y$ by $1$ .

$x = 5 + y$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

$x = 5 + y$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

$x = 5 + y$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

$x = 5 + y$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

$x = 5 + y$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Reorder $5$ and $y$ .

$x = y + 5$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

$x = y + 5$

$\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$

Replace all occurrences of $x$ with $y + 5$ in each equation.

Replace all occurrences of $x$ in $\frac{1}{4} x + \frac{1}{3} y = \frac{1}{12}$ with $y + 5$ .

$\frac{1}{4} \cdot (y + 5) + \frac{1}{3} y = \frac{1}{12}$

$x = y + 5$

Simplify the left side.

Simplify $\frac{1}{4} \cdot (y + 5) + \frac{1}{3} y$ .

Simplify each term.

Apply the distributive property.

$\frac{1}{4} y + \frac{1}{4} \cdot 5 + \frac{1}{3} y = \frac{1}{12}$

$x = y + 5$

Combine $\frac{1}{4}$ and $y$ .

$\frac{y}{4} + \frac{1}{4} \cdot 5 + \frac{1}{3} y = \frac{1}{12}$

$x = y + 5$

Combine $\frac{1}{4}$ and $5$ .

$\frac{y}{4} + \frac{5}{4} + \frac{1}{3} y = \frac{1}{12}$

$x = y + 5$

Combine $\frac{1}{3}$ and $y$ .

$\frac{y}{4} + \frac{5}{4} + \frac{y}{3} = \frac{1}{12}$

$x = y + 5$

$\frac{y}{4} + \frac{5}{4} + \frac{y}{3} = \frac{1}{12}$

$x = y + 5$

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

$\frac{y}{4} \cdot \frac{3}{3} + \frac{y}{3} + \frac{5}{4} = \frac{1}{12}$

$x = y + 5$

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

$\frac{y}{4} \cdot \frac{3}{3} + \frac{y}{3} \cdot \frac{4}{4} + \frac{5}{4} = \frac{1}{12}$

$x = y + 5$

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{y}{4}$ by $\frac{3}{3}$ .

$\frac{y \cdot 3}{4 \cdot 3} + \frac{y}{3} \cdot \frac{4}{4} + \frac{5}{4} = \frac{1}{12}$

$x = y + 5$

Multiply $4$ by $3$ .

$\frac{y \cdot 3}{12} + \frac{y}{3} \cdot \frac{4}{4} + \frac{5}{4} = \frac{1}{12}$

$x = y + 5$

Multiply $\frac{y}{3}$ by $\frac{4}{4}$ .

$\frac{y \cdot 3}{12} + \frac{y \cdot 4}{3 \cdot 4} + \frac{5}{4} = \frac{1}{12}$

$x = y + 5$

Multiply $3$ by $4$ .

$\frac{y \cdot 3}{12} + \frac{y \cdot 4}{12} + \frac{5}{4} = \frac{1}{12}$

$x = y + 5$

$\frac{y \cdot 3}{12} + \frac{y \cdot 4}{12} + \frac{5}{4} = \frac{1}{12}$

$x = y + 5$

Combine the numerators over the common denominator.

$\frac{y \cdot 3 + y \cdot 4}{12} + \frac{5}{4} = \frac{1}{12}$

$x = y + 5$

Reorder $y$ and $4$ .

$\frac{3 \cdot y + 4 \cdot y}{12} + \frac{5}{4} = \frac{1}{12}$

$x = y + 5$

Add $3 \cdot y$ and $4 \cdot y$ .

$\frac{7 y}{12} + \frac{5}{4} = \frac{1}{12}$

$x = y + 5$

$\frac{7 y}{12} + \frac{5}{4} = \frac{1}{12}$

$x = y + 5$

$\frac{7 y}{12} + \frac{5}{4} = \frac{1}{12}$

$x = y + 5$

$\frac{7 y}{12} + \frac{5}{4} = \frac{1}{12}$

$x = y + 5$

Solve for $y$ in $\frac{7 y}{12} + \frac{5}{4} = \frac{1}{12}$ .

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

Subtract $\frac{5}{4}$ from both sides of the equation.

$\frac{7 y}{12} = \frac{1}{12} - \frac{5}{4}$

$x = y + 5$

To write $- \frac{5}{4}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{7 y}{12} = \frac{1}{12} - \frac{5}{4} \cdot \frac{3}{3}$

$x = y + 5$

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{5}{4}$ by $\frac{3}{3}$ .

$\frac{7 y}{12} = \frac{1}{12} - \frac{5 \cdot 3}{4 \cdot 3}$

$x = y + 5$

Multiply $4$ by $3$ .

$\frac{7 y}{12} = \frac{1}{12} - \frac{5 \cdot 3}{12}$

$x = y + 5$

$\frac{7 y}{12} = \frac{1}{12} - \frac{5 \cdot 3}{12}$

$x = y + 5$

Combine the numerators over the common denominator.

$\frac{7 y}{12} = \frac{1 - 5 \cdot 3}{12}$

$x = y + 5$

Simplify the numerator.

Multiply $- 5$ by $3$ .

$\frac{7 y}{12} = \frac{1 - 15}{12}$

$x = y + 5$

Subtract $15$ from $1$ .

$\frac{7 y}{12} = \frac{- 14}{12}$

$x = y + 5$

$\frac{7 y}{12} = \frac{- 14}{12}$

$x = y + 5$

Cancel the common factor of $- 14$ and $12$ .

Factor $2$ out of $- 14$ .

$\frac{7 y}{12} = \frac{2 (- 7)}{12}$

$x = y + 5$

Cancel the common factors.

Factor $2$ out of $12$ .

$\frac{7 y}{12} = \frac{2 \cdot - 7}{2 \cdot 6}$

$x = y + 5$

Cancel the common factor.

$\frac{7 y}{12} = \frac{2 \cdot - 7}{2 \cdot 6}$

$x = y + 5$

Rewrite the expression.

$\frac{7 y}{12} = \frac{- 7}{6}$

$x = y + 5$

$\frac{7 y}{12} = \frac{- 7}{6}$

$x = y + 5$

$\frac{7 y}{12} = \frac{- 7}{6}$

$x = y + 5$

Move the negative in front of the fraction.

$\frac{7 y}{12} = - \frac{7}{6}$

$x = y + 5$

$\frac{7 y}{12} = - \frac{7}{6}$

$x = y + 5$

Multiply both sides of the equation by $\frac{12}{7}$ .

$\frac{12}{7} \cdot \frac{7 y}{12} = \frac{12}{7} \cdot (- \frac{7}{6})$

$x = y + 5$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{12}{7} \cdot \frac{7 y}{12}$ .

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12}{7} \cdot \frac{7 y}{12} = \frac{12}{7} \cdot (- \frac{7}{6})$

$x = y + 5$

Rewrite the expression.

$\frac{1}{7} \cdot (7 y) = \frac{12}{7} \cdot (- \frac{7}{6})$

$x = y + 5$

$\frac{1}{7} \cdot (7 y) = \frac{12}{7} \cdot (- \frac{7}{6})$

$x = y + 5$

Cancel the common factor of $7$ .

Factor $7$ out of $7 y$ .

$\frac{1}{7} \cdot (7 (y)) = \frac{12}{7} \cdot (- \frac{7}{6})$

$x = y + 5$

Cancel the common factor.

$\frac{1}{7} \cdot (7 y) = \frac{12}{7} \cdot (- \frac{7}{6})$

$x = y + 5$

Rewrite the expression.

$y = \frac{12}{7} \cdot (- \frac{7}{6})$

$x = y + 5$

$y = \frac{12}{7} \cdot (- \frac{7}{6})$

$x = y + 5$

$y = \frac{12}{7} \cdot (- \frac{7}{6})$

$x = y + 5$

$y = \frac{12}{7} \cdot (- \frac{7}{6})$

$x = y + 5$

Simplify the right side.

Simplify $\frac{12}{7} (- \frac{7}{6})$ .

Cancel the common factor of $6$ .

Move the leading negative in $- \frac{7}{6}$ into the numerator.

$y = \frac{12}{7} \cdot \frac{- 7}{6}$

$x = y + 5$

Factor $6$ out of $12$ .

$y = \frac{6 (2)}{7} \cdot \frac{- 7}{6}$

$x = y + 5$

Cancel the common factor.

$y = \frac{6 \cdot 2}{7} \cdot \frac{- 7}{6}$

$x = y + 5$

Rewrite the expression.

$y = \frac{2}{7} \cdot - 7$

$x = y + 5$

$y = \frac{2}{7} \cdot - 7$

$x = y + 5$

Cancel the common factor of $7$ .

Factor $7$ out of $- 7$ .

$y = \frac{2}{7} \cdot (7 (- 1))$

$x = y + 5$

Cancel the common factor.

$y = \frac{2}{7} \cdot (7 \cdot - 1)$

$x = y + 5$

Rewrite the expression.

$y = 2 \cdot - 1$

$x = y + 5$

$y = 2 \cdot - 1$

$x = y + 5$

Multiply $2$ by $- 1$ .

$y = - 2$

$x = y + 5$

$y = - 2$

$x = y + 5$

$y = - 2$

$x = y + 5$

$y = - 2$

$x = y + 5$

$y = - 2$

$x = y + 5$

Replace all occurrences of $y$ with $- 2$ in each equation.

Replace all occurrences of $y$ in $x = y + 5$ with $- 2$ .

$x = (- 2) + 5$

$y = - 2$

Simplify the right side.

Add $- 2$ and $5$ .

$x = 3$

$y = - 2$

$x = 3$

$y = - 2$

$x = 3$

$y = - 2$

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

$(3, - 2)$

The result can be shown in multiple forms.

Point Form:

$(3, - 2)$

Equation Form:

$x = 3, y = - 2$

Solve the System of Equations 1/4x+1/3y=1/12 1/4(-x+y)=-5/4

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