Solve by Substitution 4x+3y=15 , 2x-5y=1

$4 x + 3 y = 15$ , $2 x - 5 y = 1$

Step 1

Solve for $x$ in $4 x + 3 y = 15$ .

Subtract $3 y$ from both sides of the equation.

$4 x = 15 - 3 y$

$2 x - 5 y = 1$

Divide each term in $4 x = 15 - 3 y$ by $4$ and simplify.

Divide each term in $4 x = 15 - 3 y$ by $4$ .

$\frac{4 x}{4} = \frac{15}{4} + \frac{- 3 y}{4}$

$2 x - 5 y = 1$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{15}{4} + \frac{- 3 y}{4}$

$2 x - 5 y = 1$

Divide $x$ by $1$ .

$x = \frac{15}{4} + \frac{- 3 y}{4}$

$2 x - 5 y = 1$

$x = \frac{15}{4} + \frac{- 3 y}{4}$

$2 x - 5 y = 1$

$x = \frac{15}{4} + \frac{- 3 y}{4}$

$2 x - 5 y = 1$

Simplify the right side.

Move the negative in front of the fraction.

$x = \frac{15}{4} - \frac{3 y}{4}$

$2 x - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

$2 x - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

$2 x - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

$2 x - 5 y = 1$

Step 2

Replace all occurrences of $x$ with $\frac{15}{4} - \frac{3 y}{4}$ in each equation.

Replace all occurrences of $x$ in $2 x - 5 y = 1$ with $\frac{15}{4} - \frac{3 y}{4}$ .

$2 (\frac{15}{4} - \frac{3 y}{4}) - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Simplify the left side.

Simplify $2 (\frac{15}{4} - \frac{3 y}{4}) - 5 y$ .

Simplify each term.

Apply the distributive property.

$2 (\frac{15}{4}) + 2 (- \frac{3 y}{4}) - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

$2 (\frac{15}{2 (2)}) + 2 (- \frac{3 y}{4}) - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Cancel the common factor.

$2 (\frac{15}{2 \cdot 2}) + 2 (- \frac{3 y}{4}) - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Rewrite the expression.

$\frac{15}{2} + 2 (- \frac{3 y}{4}) - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

$\frac{15}{2} + 2 (- \frac{3 y}{4}) - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Cancel the common factor of $2$ .

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

$\frac{15}{2} + 2 (\frac{- 3 y}{4}) - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Factor $2$ out of $4$ .

$\frac{15}{2} + 2 (\frac{- 3 y}{2 (2)}) - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Cancel the common factor.

$\frac{15}{2} + 2 (\frac{- 3 y}{2 \cdot 2}) - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Rewrite the expression.

$\frac{15}{2} + \frac{- 3 y}{2} - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

$\frac{15}{2} + \frac{- 3 y}{2} - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Move the negative in front of the fraction.

$\frac{15}{2} - \frac{3 y}{2} - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

$\frac{15}{2} - \frac{3 y}{2} - 5 y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

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

$\frac{15}{2} - \frac{3 y}{2} - 5 y \cdot \frac{2}{2} = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Combine $- 5 y$ and $\frac{2}{2}$ .

$\frac{15}{2} - \frac{3 y}{2} + \frac{- 5 y \cdot 2}{2} = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Combine the numerators over the common denominator.

$\frac{15}{2} + \frac{- 3 y - 5 y \cdot 2}{2} = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Combine the numerators over the common denominator.

$\frac{15 - 3 y - 5 y \cdot 2}{2} = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Multiply $2$ by $- 5$ .

$\frac{15 - 3 y - 10 y}{2} = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Subtract $10 y$ from $- 3 y$ .

$\frac{15 - 13 y}{2} = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

$\frac{15 - 13 y}{2} = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

$\frac{15 - 13 y}{2} = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

$\frac{15 - 13 y}{2} = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Step 3

Solve for $y$ in $\frac{15 - 13 y}{2} = 1$ .

Multiply both sides by $2$ .

$\frac{15 - 13 y}{2} \cdot 2 = 1 \cdot 2$

$x = \frac{15}{4} - \frac{3 y}{4}$

Simplify.

Simplify the left side.

Simplify $\frac{15 - 13 y}{2} \cdot 2$ .

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{15 - 13 y}{2} \cdot 2 = 1 \cdot 2$

$x = \frac{15}{4} - \frac{3 y}{4}$

Rewrite the expression.

$15 - 13 y = 1 \cdot 2$

$x = \frac{15}{4} - \frac{3 y}{4}$

$15 - 13 y = 1 \cdot 2$

$x = \frac{15}{4} - \frac{3 y}{4}$

Reorder $15$ and $- 13 y$ .

$- 13 y + 15 = 1 \cdot 2$

$x = \frac{15}{4} - \frac{3 y}{4}$

$- 13 y + 15 = 1 \cdot 2$

$x = \frac{15}{4} - \frac{3 y}{4}$

$- 13 y + 15 = 1 \cdot 2$

$x = \frac{15}{4} - \frac{3 y}{4}$

Simplify the right side.

Multiply $2$ by $1$ .

$- 13 y + 15 = 2$

$x = \frac{15}{4} - \frac{3 y}{4}$

$- 13 y + 15 = 2$

$x = \frac{15}{4} - \frac{3 y}{4}$

$- 13 y + 15 = 2$

$x = \frac{15}{4} - \frac{3 y}{4}$

Solve for $y$ .

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

Subtract $15$ from both sides of the equation.

$- 13 y = 2 - 15$

$x = \frac{15}{4} - \frac{3 y}{4}$

Subtract $15$ from $2$ .

$- 13 y = - 13$

$x = \frac{15}{4} - \frac{3 y}{4}$

$- 13 y = - 13$

$x = \frac{15}{4} - \frac{3 y}{4}$

Divide each term in $- 13 y = - 13$ by $- 13$ and simplify.

Divide each term in $- 13 y = - 13$ by $- 13$ .

$\frac{- 13 y}{- 13} = \frac{- 13}{- 13}$

$x = \frac{15}{4} - \frac{3 y}{4}$

Simplify the left side.

Cancel the common factor of $- 13$ .

Cancel the common factor.

$\frac{- 13 y}{- 13} = \frac{- 13}{- 13}$

$x = \frac{15}{4} - \frac{3 y}{4}$

Divide $y$ by $1$ .

$y = \frac{- 13}{- 13}$

$x = \frac{15}{4} - \frac{3 y}{4}$

$y = \frac{- 13}{- 13}$

$x = \frac{15}{4} - \frac{3 y}{4}$

$y = \frac{- 13}{- 13}$

$x = \frac{15}{4} - \frac{3 y}{4}$

Simplify the right side.

Divide $- 13$ by $- 13$ .

$y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

$y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

$y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

$y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

$y = 1$

$x = \frac{15}{4} - \frac{3 y}{4}$

Step 4

Replace all occurrences of $y$ with $1$ in each equation.

Replace all occurrences of $y$ in $x = \frac{15}{4} - \frac{3 y}{4}$ with $1$ .

$x = \frac{15}{4} - \frac{3 (1)}{4}$

$y = 1$

Simplify the right side.

Simplify $\frac{15}{4} - \frac{3 (1)}{4}$ .

Combine the numerators over the common denominator.

$x = \frac{15 - 3 \cdot 1}{4}$

$y = 1$

Simplify the expression.

Multiply $- 3$ by $1$ .

$x = \frac{15 - 3}{4}$

$y = 1$

Subtract $3$ from $15$ .

$x = \frac{12}{4}$

$y = 1$

Divide $12$ by $4$ .

$x = 3$

$y = 1$

$x = 3$

$y = 1$

$x = 3$

$y = 1$

$x = 3$

$y = 1$

$x = 3$

$y = 1$

Step 5

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

$(3, 1)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(3, 1)$

Equation Form:

$x = 3, y = 1$

Step 7

Solve by Substitution 4x+3y=15 , 2x-5y=1

Related Questions

Support LCMGCF.com by disabling your adblocker.