Solve by Substitution 1/2x+1/3y=5 , 1/4x+y=10

$\frac{1}{2} x + \frac{1}{3} y = 5$ , $\frac{1}{4} x + y = 10$

Step 1

Solve for $y$ in $\frac{1}{4} x + y = 10$ .

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

$\frac{x}{4} + y = 10$

$\frac{1}{2} x + \frac{1}{3} y = 5$

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

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

$\frac{1}{2} x + \frac{1}{3} y = 5$

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

$\frac{1}{2} x + \frac{1}{3} y = 5$

Step 2

Replace all occurrences of $y$ with $10 - \frac{x}{4}$ in each equation.

Replace all occurrences of $y$ in $\frac{1}{2} x + \frac{1}{3} y = 5$ with $10 - \frac{x}{4}$ .

$\frac{1}{2} x + \frac{1}{3} \cdot (10 - \frac{x}{4}) = 5$

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

Simplify the left side.

Simplify $\frac{1}{2} x + \frac{1}{3} \cdot (10 - \frac{x}{4})$ .

Simplify each term.

Combine $\frac{1}{2}$ and $x$ .

$\frac{x}{2} + \frac{1}{3} \cdot (10 - \frac{x}{4}) = 5$

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

Apply the distributive property.

$\frac{x}{2} + \frac{1}{3} \cdot 10 + \frac{1}{3} \cdot (- \frac{x}{4}) = 5$

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

Combine $\frac{1}{3}$ and $10$ .

$\frac{x}{2} + \frac{10}{3} + \frac{1}{3} \cdot (- \frac{x}{4}) = 5$

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

Multiply $\frac{1}{3} (- \frac{x}{4})$ .

Multiply $\frac{1}{3}$ by $\frac{x}{4}$ .

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

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

Multiply $3$ by $4$ .

$\frac{x}{2} + \frac{10}{3} - \frac{x}{12} = 5$

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

$\frac{x}{2} + \frac{10}{3} - \frac{x}{12} = 5$

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

$\frac{x}{2} + \frac{10}{3} - \frac{x}{12} = 5$

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

To write $\frac{x}{2}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$\frac{x}{2} \cdot \frac{6}{6} - \frac{x}{12} + \frac{10}{3} = 5$

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

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

Multiply $\frac{x}{2}$ by $\frac{6}{6}$ .

$\frac{x \cdot 6}{2 \cdot 6} - \frac{x}{12} + \frac{10}{3} = 5$

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

Multiply $2$ by $6$ .

$\frac{x \cdot 6}{12} - \frac{x}{12} + \frac{10}{3} = 5$

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

$\frac{x \cdot 6}{12} - \frac{x}{12} + \frac{10}{3} = 5$

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

Combine the numerators over the common denominator.

$\frac{x \cdot 6 - x}{12} + \frac{10}{3} = 5$

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

Subtract $x$ from $x \cdot 6$ .

Reorder $x$ and $6$ .

$\frac{6 \cdot x - x}{12} + \frac{10}{3} = 5$

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

Subtract $x$ from $6 \cdot x$ .

$\frac{5 \cdot x}{12} + \frac{10}{3} = 5$

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

$\frac{5 x}{12} + \frac{10}{3} = 5$

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

$\frac{5 x}{12} + \frac{10}{3} = 5$

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

$\frac{5 x}{12} + \frac{10}{3} = 5$

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

$\frac{5 x}{12} + \frac{10}{3} = 5$

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

Step 3

Solve for $x$ in $\frac{5 x}{12} + \frac{10}{3} = 5$ .

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

Subtract $\frac{10}{3}$ from both sides of the equation.

$\frac{5 x}{12} = 5 - \frac{10}{3}$

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

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

$\frac{5 x}{12} = 5 \cdot \frac{3}{3} - \frac{10}{3}$

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

Combine $5$ and $\frac{3}{3}$ .

$\frac{5 x}{12} = \frac{5 \cdot 3}{3} - \frac{10}{3}$

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

Combine the numerators over the common denominator.

$\frac{5 x}{12} = \frac{5 \cdot 3 - 10}{3}$

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

Simplify the numerator.

Multiply $5$ by $3$ .

$\frac{5 x}{12} = \frac{15 - 10}{3}$

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

Subtract $10$ from $15$ .

$\frac{5 x}{12} = \frac{5}{3}$

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

$\frac{5 x}{12} = \frac{5}{3}$

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

$\frac{5 x}{12} = \frac{5}{3}$

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

Multiply both sides of the equation by $\frac{12}{5}$ .

$\frac{12}{5} \cdot \frac{5 x}{12} = \frac{12}{5} \cdot \frac{5}{3}$

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

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{12}{5} \cdot \frac{5 x}{12}$ .

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12}{5} \cdot \frac{5 x}{12} = \frac{12}{5} \cdot \frac{5}{3}$

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

Rewrite the expression.

$\frac{1}{5} \cdot (5 x) = \frac{12}{5} \cdot \frac{5}{3}$

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

$\frac{1}{5} \cdot (5 x) = \frac{12}{5} \cdot \frac{5}{3}$

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

Cancel the common factor of $5$ .

Factor $5$ out of $5 x$ .

$\frac{1}{5} \cdot (5 (x)) = \frac{12}{5} \cdot \frac{5}{3}$

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

Cancel the common factor.

$\frac{1}{5} \cdot (5 x) = \frac{12}{5} \cdot \frac{5}{3}$

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

Rewrite the expression.

$x = \frac{12}{5} \cdot \frac{5}{3}$

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

$x = \frac{12}{5} \cdot \frac{5}{3}$

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

$x = \frac{12}{5} \cdot \frac{5}{3}$

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

$x = \frac{12}{5} \cdot \frac{5}{3}$

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

Simplify the right side.

Simplify $\frac{12}{5} \cdot \frac{5}{3}$ .

Cancel the common factor of $3$ .

Factor $3$ out of $12$ .

$x = \frac{3 (4)}{5} \cdot \frac{5}{3}$

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

Cancel the common factor.

$x = \frac{3 \cdot 4}{5} \cdot \frac{5}{3}$

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

Rewrite the expression.

$x = \frac{4}{5} \cdot 5$

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

$x = \frac{4}{5} \cdot 5$

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

Cancel the common factor of $5$ .

Cancel the common factor.

$x = \frac{4}{5} \cdot 5$

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

Rewrite the expression.

$x = 4$

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

$x = 4$

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

$x = 4$

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

$x = 4$

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

$x = 4$

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

$x = 4$

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

Step 4

Replace all occurrences of $x$ with $4$ in each equation.

Replace all occurrences of $x$ in $y = 10 - \frac{x}{4}$ with $4$ .

$y = 10 - \frac{4}{4}$

$x = 4$

Simplify the right side.

Simplify $10 - \frac{4}{4}$ .

Simplify each term.

Cancel the common factor of $4$ .

Cancel the common factor.

$y = 10 - \frac{4}{4}$

$x = 4$

Rewrite the expression.

$y = 10 - 1 \cdot 1$

$x = 4$

$y = 10 - 1 \cdot 1$

$x = 4$

Multiply $- 1$ by $1$ .

$y = 10 - 1$

$x = 4$

$y = 10 - 1$

$x = 4$

Subtract $1$ from $10$ .

$y = 9$

$x = 4$

$y = 9$

$x = 4$

$y = 9$

$x = 4$

$y = 9$

$x = 4$

Step 5

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

$(4, 9)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 9)$

Equation Form:

$x = 4, y = 9$

Step 7

Solve by Substitution 1/2x+1/3y=5 , 1/4x+y=10

Related Questions

Support LCMGCF.com by disabling your adblocker.