Q 175. - Intermediate Algebra

Question

Solve the system of equations.

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

Step-by-Step Solution

Answer

The solution for the system of equations is,

$$\begin{gathered} x=6 \\ y=-9 \\ z=-3 \end{gathered}$$.

Step 1. Given the information.

The system of equations is,

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

Step 2. Simplifying the equations by removing fractions.

Simplifying equation (1), $x+\frac{1}{2}y+\frac{1}{2}z=0$ by multiplying both sides by $2$,

$2(x+\frac{1}{2}y+\frac{1}{2}z)=2\left(0\right)$

$2x+y+z=0........\left(4\right)$

Simplifying equation (2), $\frac{1}{5}x-\frac{1}{5}y+z=0$by multiplying both sides by $5$,

$5(\frac{1}{5}x-\frac{1}{5}y+z)=5\left(0\right)$

$x-y+5z=0..........\left(5\right)$

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

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

$x-y+6z=-3...........\left(6\right)$

Step 3. Solving equations (4) and (5) for eliminating x .

Eliminating $x$ by solving the equations (4) and (5)

Multiplying equation (5) by $-2$,

$-2x+2y-10z=0$

$2x+y+z=0$

Solving we get, 

$3y-9z=0$

$y-3z=0......\left(7\right)$

Step 4. Solving equations (5) and (6) for eliminating x .

Eliminating $x$ by solving equations (5) and (6).

Multiplying $-1$ on both sides of the equation (5),

$-x+y-5z=0$

$x-y+6z=-3$

Solving we get,

$z=-3$.

Step 5. Substituting the value of z in equation (7) to find the value of y .

Substituting $z=-3$in the equation

$y-3z=0$

$y-3(-3)=0$

$y+9=0$

$y=-9$

Step 6. Finding the value of x .

Substituting $$\begin{gathered} y=-9 \\ z=-3 \end{gathered}$$in the equation

$2x+y+z=0$ to find the value of $x$.

$2x+(-9)+(-3)=0$

$2x-9-3=0$

$2x-12=0$

$2x=12$

$x=6$

Step 7. Checking the solution for the equation x + 1 2 y + 1 2 z = 0 .

Substituting $$\begin{gathered} x=6 \\ y=-9 \\ z=-3 \end{gathered}$$ in the equation

$x+\frac{1}{2}y+\frac{1}{2}z=0$

$6+\frac{1}{2}(-9)+\frac{1}{2}(-3)=0$

$6-\frac{9}{2}-\frac{3}{2}=0$

$6-\frac{12}{2}=0$

$6-6=0$

$0=0$

This s true.

Step 8. Checking the solution for the equation 1 5 x - 1 5 y + z = 0 .

Substituting $$\begin{gathered} x=6 \\ y=-9 \\ z=-3 \end{gathered}$$ in the equation

$\frac{1}{5}x-\frac{1}{5}y+z=0$

$\frac{1}{5}\left(6\right)-\frac{1}{5}(-9)+(-3)=0$

$\frac{6}{5}+\frac{9}{5}-3=0$

$\frac{15}{5}-3=0$

$3-3=0$

$0=0$

This is true.

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

Substituting $$\begin{gathered} x=6 \\ y=-9 \\ z=-3 \end{gathered}$$ in the equation

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

$\frac{1}{3}\left(6\right)-\frac{1}{3}(-9)+2(-3)=-1$

$2-(-3)-6=-1$

$2+3-6=-1$

$5-6=-1$

$-1=-1$

This is true.