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

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

Step 1

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

Replace all occurrences of $z$ in $x + 4 y + 2 z = 1$ with $- 3$ .

$x + 4 y + 2 (- 3) = 1$

$z = - 3$

$5 y + 3 z = 1$

Simplify the left side.

Multiply $2$ by $- 3$ .

$x + 4 y - 6 = 1$

$z = - 3$

$5 y + 3 z = 1$

$x + 4 y - 6 = 1$

$z = - 3$

$5 y + 3 z = 1$

Replace all occurrences of $z$ in $5 y + 3 z = 1$ with $- 3$ .

$5 y + 3 (- 3) = 1$

$x + 4 y - 6 = 1$

$z = - 3$

Simplify the left side.

Multiply $3$ by $- 3$ .

$5 y - 9 = 1$

$x + 4 y - 6 = 1$

$z = - 3$

$5 y - 9 = 1$

$x + 4 y - 6 = 1$

$z = - 3$

$5 y - 9 = 1$

$x + 4 y - 6 = 1$

$z = - 3$

Step 2

Solve for $y$ in $5 y - 9 = 1$ .

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

Add $9$ to both sides of the equation.

$5 y = 1 + 9$

$x + 4 y - 6 = 1$

$z = - 3$

Add $1$ and $9$ .

$5 y = 10$

$x + 4 y - 6 = 1$

$z = - 3$

$5 y = 10$

$x + 4 y - 6 = 1$

$z = - 3$

Divide each term in $5 y = 10$ by $5$ and simplify.

Divide each term in $5 y = 10$ by $5$ .

$\frac{5 y}{5} = \frac{10}{5}$

$x + 4 y - 6 = 1$

$z = - 3$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 y}{5} = \frac{10}{5}$

$x + 4 y - 6 = 1$

$z = - 3$

Divide $y$ by $1$ .

$y = \frac{10}{5}$

$x + 4 y - 6 = 1$

$z = - 3$

$y = \frac{10}{5}$

$x + 4 y - 6 = 1$

$z = - 3$

$y = \frac{10}{5}$

$x + 4 y - 6 = 1$

$z = - 3$

Simplify the right side.

Divide $10$ by $5$ .

$y = 2$

$x + 4 y - 6 = 1$

$z = - 3$

$y = 2$

$x + 4 y - 6 = 1$

$z = - 3$

$y = 2$

$x + 4 y - 6 = 1$

$z = - 3$

$y = 2$

$x + 4 y - 6 = 1$

$z = - 3$

Step 3

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

Replace all occurrences of $y$ in $x + 4 y - 6 = 1$ with $2$ .

$x + 4 (2) - 6 = 1$

$y = 2$

$z = - 3$

Simplify the left side.

Simplify $x + 4 (2) - 6$ .

Multiply $4$ by $2$ .

$x + 8 - 6 = 1$

$y = 2$

$z = - 3$

Subtract $6$ from $8$ .

$x + 2 = 1$

$y = 2$

$z = - 3$

$x + 2 = 1$

$y = 2$

$z = - 3$

$x + 2 = 1$

$y = 2$

$z = - 3$

$x + 2 = 1$

$y = 2$

$z = - 3$

Step 4

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

Subtract $2$ from both sides of the equation.

$x = 1 - 2$

$y = 2$

$z = - 3$

Subtract $2$ from $1$ .

$x = - 1$

$y = 2$

$z = - 3$

$x = - 1$

$y = 2$

$z = - 3$

Step 5

Solve the system of equations $x = - 1 y = 2 z = - 3$ .

$x = - 1 y = 2 z = - 3$

Step 6

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

$(- 1, 2, - 3)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(- 1, 2, - 3)$

Equation Form:

$x = - 1, y = 2, z = - 3$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.