Q 177. - Intermediate Algebra

Question

Solve the system of equations:

$\left\{\begin{array}{l}\frac{1}{3}x-y+\frac{1}{2}z=4\\ \frac{2}{3}x+\frac{5}{2}y-4z=0\\ x-\frac{1}{2}y+\frac{3}{2}z=2\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of equations is,

$$\begin{gathered} x=3 \\ y=-4 \\ z=-2 \end{gathered}$$

Step 1. Given the information

The system of equations is,

$\left\{\begin{array}{l}\frac{1}{3}x-y+\frac{1}{2}z=4..........\left(1\right)\\ \frac{2}{3}x+\frac{5}{2}y-4z=0........\left(2\right)\\ x-\frac{1}{2}y+\frac{3}{2}z=2........\left(3\right)\end{array}\right.$

Step 2. Simplifying the equation by removing the fractions.

Simplifying equation (1), $\frac{1}{3}x-y+\frac{1}{2}z=4$ by multiplying both sides by $6$,

$$\begin{gathered} 6(\frac{1}{3}x-y+\frac{1}{2}z)=6\left(4\right) \\ 2x-6y+3z=24...........\left(4\right) \end{gathered}$$

Simplifying equation (2), $\frac{2}{3}x+\frac{5}{2}y-4z=0$ by multiplying both sides by $6$,

$$\begin{gathered} 6(\frac{2}{3}x+\frac{5}{2}y-4z)=6\left(0\right) \\ 4x+15y-24z=0.........\left(5\right) \end{gathered}$$

Simplifying equation (3), $x-\frac{1}{2}y+\frac{3}{2}z=2$ by multiplying both sides by $2$,

$$\begin{gathered} 2(x-\frac{1}{2}y+\frac{3}{2}z)=2\left(2\right) \\ 2x-y+3z=4...........\left(6\right) \end{gathered}$$

Step 3. Eliminating x from the equations (4) and (5).

Eliminating $x$ from equations (4) and (5),.

Multiplying $-2$ on both sides of equation (4),

$-4x+12y-6z=-48$

$4x+15y-24z=0$

Solving the equation we get,

$27y-30z=-48$

$9y-10z=-16........\left(7\right)$

Step 4. Eliminating x from the equations (5) and (6)

Eliminating $x$ from the equations (5) and (6).

Multiplying $-2$ on both sides of equation (6),

$$\begin{gathered} -4x+2y-6z=-8 \\ 4x+15y-24z=0 \end{gathered}$$

Solving the equations we get,

$17y-30z=-8......\left(8\right)$

Step 5. Solving equations (7) and (8)

Solving the equations,

$$\begin{gathered} 27y-30z=-48.......\left(7\right)\times 10 \\ 17y-30z=-8........\left(8\right) \end{gathered}$$

Subtracting the equations we get,

$$\begin{gathered} 10y=-40 \\ y=-4 \end{gathered}$$

Substituting the value of $y$ in the equation $9y-10z=-16$

$$\begin{gathered} 9(-4)-10z=-16 \\ -36-10z=-16 \\ -10z=-16+36 \\ -10z=20 \\ z=-2 \end{gathered}$$

Step 6. Finding the value of x .

Substituting $$\begin{gathered} y=-4 \\ z=-2 \end{gathered}$$in the equation

$2x-6y+3z=24$,

$$\begin{gathered} 2x-6(-4)+3(-2)=24 \\ 2x+24-6=24 \\ 2x+18=24 \\ 2x=24-18 \\ 2x=6 \\ x=3 \end{gathered}$$

Step 7. Checking the solution for the equation 1 3 x - y + 1 2 z = 4 .

Substituting $$\begin{gathered} x=3 \\ y=-4 \\ z=-2 \end{gathered}$$ in the equation

$\frac{1}{3}x-y+\frac{1}{2}z=4$,

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

This is true.

Step 8. Checking the solution for the equation 2 3 x + 5 2 y - 4 z = 0 .

Substituting $$\begin{gathered} x=3 \\ y=-4 \\ z=-2 \end{gathered}$$ in the equation

$\frac{2}{3}x+\frac{5}{2}y-4z=0$

$$\begin{gathered} \frac{2}{3}\left(3\right)+\frac{5}{2}(-4)-4(-2)=0 \\ 2-10+8=0 \\ 10-10=0 \\ 0=0 \end{gathered}$$

This is true.

Step 9. Checking the solution for the equation x - 1 2 y + 3 2 z = 2 .

Substituting $$\begin{gathered} x=3 \\ y=-4 \\ z=-2 \end{gathered}$$ in the equation

$x-\frac{1}{2}y+\frac{3}{2}z=2$

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

This is true.