Question

Solve the system of equations.

$\left\{\begin{array}{l}2y+3z=-1\\ 5x+3y=-6\\ 7x+z=1\end{array}\right.$

Short answer

The solution for the system of equations is,

$$\begin{gathered} x=0 \\ y=-2 \\ z=1 \end{gathered}$$.

Step-by-step solution

Step 1. Given the information.

The system of equations is,

$\left\{\begin{array}{l}2y+3z=-1........\left(1\right)\\ 5x+3y=-6........\left(2\right)\\ 7x+z=1............\left(3\right)\end{array}\right.$

Step 2. Eliminating z from the equations (1) and (3).

Eliminating $z$from the equations (1) and (3).

$$\begin{gathered} 2y+3z=-1............\left(1\right) \\ -21x-3z=-3.......\left(3\right)\times -3 \end{gathered}$$

Solving the system of equations,

$-21x+2y=-4........\left(4\right)$

Step 3. Finding the value of x.

Solving the equations (2) and (4).

$$\begin{gathered} 10x+6y=-12.........\left(2\right)\times 2 \\ -63x+6y=-12......\left(4\right)\times 3 \end{gathered}$$

Solving the equations, we get,

$$\begin{gathered} 73x=0 \\ x=0 \end{gathered}$$

Step 4. Finding the value of y.

Substituting $x=0$in the equation

$5x+3y=-6$

$$\begin{gathered} 5\left(0\right)+3y=-6 \\ 0+3y=-6 \\ 3y=-6 \\ y=-2 \end{gathered}$$

Step 5. Finding the value of z.

Substituting $x=0$in the equation

role="math" localid="1644559173849" $7x+z=1$.

$$\begin{gathered} 7\left(0\right)+z=1 \\ 0+z=1 \\ z=1 \end{gathered}$$

Step 6. Checking the solution for the equation 2y+3z=-1.

Substituting $$\begin{gathered} y=-2 \\ z=1 \end{gathered}$$in the equation $2y+3z=-1$,

$$\begin{gathered} 2(-2)+3\left(1\right)=-1 \\ -4+3=-1 \\ -1=-1 \end{gathered}$$

This is true.

Step 7. Checking the solution for the equation 5x+3y=-6. 

Substituting $$\begin{gathered} x=0 \\ y=-2 \end{gathered}$$in thee eequation

$5x+3y=-6$

$$\begin{gathered} 5\left(0\right)+3(-2)=-6 \\ 0+(-6)=-6 \\ -6=-6 \end{gathered}$$

This is true.

Step 8. Checking the solution for the equation 7x+z=1.

Substituting $$\begin{gathered} x=0 \\ z=1 \end{gathered}$$n the equation

$7x+z=1$,

$$\begin{gathered} 7\left(0\right)+1=1 \\ 0+1=1 \\ 1=1 \end{gathered}$$

This is true.