Solve by Substitution 4y-x=5+2y 3x+7y=24

$4 y - x = 5 + 2 y$ $3 x + 7 y = 24$

Solve for $x$ in $4 y - x = 5 + 2 y$ .

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

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

$- x = 5 + 2 y - 4 y$

$3 x + 7 y = 24$

Subtract $4 y$ from $2 y$ .

$- x = 5 - 2 y$

$3 x + 7 y = 24$

$- x = 5 - 2 y$

$3 x + 7 y = 24$

Divide each term in $- x = 5 - 2 y$ by $- 1$ and simplify.

Divide each term in $- x = 5 - 2 y$ by $- 1$ .

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

$3 x + 7 y = 24$

Simplify the left side.

Dividing two negative values results in a positive value.

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

$3 x + 7 y = 24$

Divide $x$ by $1$ .

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

$3 x + 7 y = 24$

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

$3 x + 7 y = 24$

Simplify the right side.

Simplify each term.

Divide $5$ by $- 1$ .

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

$3 x + 7 y = 24$

Move the negative one from the denominator of $\frac{- 2 y}{- 1}$ .

$x = - 5 - 1 \cdot (- 2 y)$

$3 x + 7 y = 24$

Rewrite $- 1 \cdot (- 2 y)$ as $- (- 2 y)$ .

$x = - 5 - (- 2 y)$

$3 x + 7 y = 24$

Multiply $- 2$ by $- 1$ .

$x = - 5 + 2 y$

$3 x + 7 y = 24$

$x = - 5 + 2 y$

$3 x + 7 y = 24$

$x = - 5 + 2 y$

$3 x + 7 y = 24$

$x = - 5 + 2 y$

$3 x + 7 y = 24$

$x = - 5 + 2 y$

$3 x + 7 y = 24$

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

Replace all occurrences of $x$ in $3 x + 7 y = 24$ with $- 5 + 2 y$ .

$3 (- 5 + 2 y) + 7 y = 24$

$x = - 5 + 2 y$

Simplify the left side.

Simplify $3 (- 5 + 2 y) + 7 y$ .

Simplify each term.

Apply the distributive property.

$3 \cdot - 5 + 3 (2 y) + 7 y = 24$

$x = - 5 + 2 y$

Multiply $3$ by $- 5$ .

$- 15 + 3 (2 y) + 7 y = 24$

$x = - 5 + 2 y$

Multiply $2$ by $3$ .

$- 15 + 6 y + 7 y = 24$

$x = - 5 + 2 y$

$- 15 + 6 y + 7 y = 24$

$x = - 5 + 2 y$

Add $6 y$ and $7 y$ .

$- 15 + 13 y = 24$

$x = - 5 + 2 y$

$- 15 + 13 y = 24$

$x = - 5 + 2 y$

$- 15 + 13 y = 24$

$x = - 5 + 2 y$

$- 15 + 13 y = 24$

$x = - 5 + 2 y$

Solve for $y$ in $- 15 + 13 y = 24$ .

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

Add $15$ to both sides of the equation.

$13 y = 24 + 15$

$x = - 5 + 2 y$

Add $24$ and $15$ .

$13 y = 39$

$x = - 5 + 2 y$

$13 y = 39$

$x = - 5 + 2 y$

Divide each term in $13 y = 39$ by $13$ and simplify.

Divide each term in $13 y = 39$ by $13$ .

$\frac{13 y}{13} = \frac{39}{13}$

$x = - 5 + 2 y$

Simplify the left side.

Cancel the common factor of $13$ .

Cancel the common factor.

$\frac{13 y}{13} = \frac{39}{13}$

$x = - 5 + 2 y$

Divide $y$ by $1$ .

$y = \frac{39}{13}$

$x = - 5 + 2 y$

$y = \frac{39}{13}$

$x = - 5 + 2 y$

$y = \frac{39}{13}$

$x = - 5 + 2 y$

Simplify the right side.

Divide $39$ by $13$ .

$y = 3$

$x = - 5 + 2 y$

$y = 3$

$x = - 5 + 2 y$

$y = 3$

$x = - 5 + 2 y$

$y = 3$

$x = - 5 + 2 y$

Replace all occurrences of $y$ with $3$ in each equation.

Replace all occurrences of $y$ in $x = - 5 + 2 y$ with $3$ .

$x = - 5 + 2 (3)$

$y = 3$

Simplify the right side.

Simplify $- 5 + 2 (3)$ .

Multiply $2$ by $3$ .

$x = - 5 + 6$

$y = 3$

Add $- 5$ and $6$ .

$x = 1$

$y = 3$

$x = 1$

$y = 3$

$x = 1$

$y = 3$

$x = 1$

$y = 3$

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

$(1, 3)$

The result can be shown in multiple forms.

Point Form:

$(1, 3)$

Equation Form:

$x = 1, y = 3$

Solve by Substitution 4y-x=5+2y 3x+7y=24

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