Question

In the following exercises, solve the system of equations.

$\left\{\begin{array}{l}11x+9y+2z=-9\\ 7x+5y+3z=-7\\ 4x+3y+z=-3\end{array}\right.$

Short answer

The solution is $\left(2,-3,-2\right)$.

Step-by-step solution

Step 1. Given Information    

We are given system of linear equations,

$\left\{\begin{array}{l}11x+9y+2z=-9-\left(1\right)\\ 7x+5y+3z=-7-\left(2\right)\\ 4x+3y+z=-3-\left(3\right)\end{array}\right.$

Step 2. Eliminating the same variable 

Multiplying by $3$in third equation, we get

role="math" localid="1644665891276" $\begin{array}{l}3(4x+3y+z)=3(-3)\\ 12x+9y+3z=-9-\left(4\right)\end{array}$

Now, subtracting fourth and third equation, we get

role="math" localid="1644666068485" $$\begin{gathered} 12x-7x+9y-5y+3z-3z=-9+7 \\ 5x+4y=-2-\left(5\right) \end{gathered}$$

Now, multiplying by $2$in third equation, we get

$\begin{array}{l}2(4x+3y+z)=2(-3)\\ 8x+6y+2z=-6-\left(6\right)\end{array}$

Now, subtracting sixth and first equation, we get

$$\begin{gathered} 8x-11x+6y-9y+2z-2z=-6+9 \\ -3x-3y=3 \\ x+y=-1-\left(7\right) \end{gathered}$$

Step 3. Solving the new system of equation 

Now, multiplying by $4$in seventh equation, we get

$$\begin{gathered} 4(x+y)=4\left(-1\right) \\ 4x+4y=-4-\left(8\right) \end{gathered}$$

Subtracting eighth and fifth equation, we get

$$\begin{gathered} 5x-4x+4y-4y=-2+4 \\ x=2 \end{gathered}$$

Now, putting the value of $x$in seventh equation, we get

$$\begin{gathered} x+y=-1 \\ 2+y=-1 \\ y=-1-2 \\ y=-3 \end{gathered}$$

Now, putting the value of $x,y$in third equation, we get

$$\begin{gathered} 4x+3y+z=-3 \\ 4\left(2\right)+3(-3)+z=-3 \\ 8-9+z=-3 \\ z=-2 \end{gathered}$$

Step 4. Checking the solution 

Putting the value of $x,y,z$in the equations, we get

$$\begin{gathered} 11x+9y+2z=-9 \\ 11\left(2\right)+9(-3)+2(-2)=-9 \\ 22-27-4=-9 \\ -9=-9 \\ 7x+5y+3z=-7 \\ 7\left(2\right)+5(-3)+3(-2)=-7 \\ 14-15-6=-7 \\ -7=-7 \\ 4x+3y+z=-3 \\ 4\left(2\right)+3(-3)-2=-3 \\ 8-9-2=-3 \\ -3=-3 \end{gathered}$$

Hence the solution is correct.