Solve by Substitution x-z+3y=4 , z=3y , y-x=5z

$x - z + 3 y = 4$ , $z = 3 y$ , $y - x = 5 z$

Step 1

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

Replace all occurrences of $z$ in $x - z + 3 y = 4$ with $3 y$ .

$x - (3 y) + 3 y = 4$

$z = 3 y$

$y - x = 5 z$

Simplify the left side.

Simplify $x - (3 y) + 3 y$ .

Multiply $3$ by $- 1$ .

$x - 3 y + 3 y = 4$

$z = 3 y$

$y - x = 5 z$

Combine the opposite terms in $x - 3 y + 3 y$ .

Add $- 3 y$ and $3 y$ .

$x + 0 = 4$

$z = 3 y$

$y - x = 5 z$

Add $x$ and $0$ .

$x = 4$

$z = 3 y$

$y - x = 5 z$

$x = 4$

$z = 3 y$

$y - x = 5 z$

$x = 4$

$z = 3 y$

$y - x = 5 z$

$x = 4$

$z = 3 y$

$y - x = 5 z$

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

$y - x = 5 (3 y)$

$x = 4$

$z = 3 y$

Simplify the right side.

Multiply $3$ by $5$ .

$y - x = 15 y$

$x = 4$

$z = 3 y$

$y - x = 15 y$

$x = 4$

$z = 3 y$

$y - x = 15 y$

$x = 4$

$z = 3 y$

Step 2

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

Replace all occurrences of $x$ in $y - x = 15 y$ with $4$ .

$y - (4) = 15 y$

$x = 4$

$z = 3 y$

Simplify the left side.

Multiply $- 1$ by $4$ .

$y - 4 = 15 y$

$x = 4$

$z = 3 y$

$y - 4 = 15 y$

$x = 4$

$z = 3 y$

$y - 4 = 15 y$

$x = 4$

$z = 3 y$

Step 3

Solve for $y$ in $y - 4 = 15 y$ .

Move all terms containing $y$ to the left side of the equation.

Subtract $15 y$ from both sides of the equation.

$y - 4 - 15 y = 0$

$x = 4$

$z = 3 y$

Subtract $15 y$ from $y$ .

$- 14 y - 4 = 0$

$x = 4$

$z = 3 y$

$- 14 y - 4 = 0$

$x = 4$

$z = 3 y$

Add $4$ to both sides of the equation.

$- 14 y = 4$

$x = 4$

$z = 3 y$

Divide each term in $- 14 y = 4$ by $- 14$ and simplify.

Divide each term in $- 14 y = 4$ by $- 14$ .

$\frac{- 14 y}{- 14} = \frac{4}{- 14}$

$x = 4$

$z = 3 y$

Simplify the left side.

Cancel the common factor of $- 14$ .

Cancel the common factor.

$\frac{- 14 y}{- 14} = \frac{4}{- 14}$

$x = 4$

$z = 3 y$

Divide $y$ by $1$ .

$y = \frac{4}{- 14}$

$x = 4$

$z = 3 y$

$y = \frac{4}{- 14}$

$x = 4$

$z = 3 y$

$y = \frac{4}{- 14}$

$x = 4$

$z = 3 y$

Simplify the right side.

Cancel the common factor of $4$ and $- 14$ .

Factor $2$ out of $4$ .

$y = \frac{2 (2)}{- 14}$

$x = 4$

$z = 3 y$

Cancel the common factors.

Factor $2$ out of $- 14$ .

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

$x = 4$

$z = 3 y$

Cancel the common factor.

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

$x = 4$

$z = 3 y$

Rewrite the expression.

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

$x = 4$

$z = 3 y$

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

$x = 4$

$z = 3 y$

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

$x = 4$

$z = 3 y$

Move the negative in front of the fraction.

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

$x = 4$

$z = 3 y$

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

$x = 4$

$z = 3 y$

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

$x = 4$

$z = 3 y$

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

$x = 4$

$z = 3 y$

Step 4

Replace all occurrences of $y$ with $- \frac{2}{7}$ in each equation.

Replace all occurrences of $y$ in $z = 3 y$ with $- \frac{2}{7}$ .

$z = 3 (- \frac{2}{7})$

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

$x = 4$

Simplify the right side.

Simplify $3 (- \frac{2}{7})$ .

Multiply $3 (- \frac{2}{7})$ .

Multiply $- 1$ by $3$ .

$z = - 3 (\frac{2}{7})$

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

$x = 4$

Combine $- 3$ and $\frac{2}{7}$ .

$z = \frac{- 3 \cdot 2}{7}$

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

$x = 4$

Multiply $- 3$ by $2$ .

$z = \frac{- 6}{7}$

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

$x = 4$

$z = \frac{- 6}{7}$

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

$x = 4$

Move the negative in front of the fraction.

$z = - \frac{6}{7}$

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

$x = 4$

$z = - \frac{6}{7}$

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

$x = 4$

$z = - \frac{6}{7}$

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

$x = 4$

$z = - \frac{6}{7}$

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

$x = 4$

Step 5

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

$(4, - \frac{2}{7}, - \frac{6}{7})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, - \frac{2}{7}, - \frac{6}{7})$

Equation Form:

$x = 4, y = - \frac{2}{7}, z = - \frac{6}{7}$

Solve by Substitution x-z+3y=4 , z=3y , y-x=5z

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