Question

Solve each system of equations

$\begin{array}{l}6x+2y+4z=2\\ 3x+4y-8z=-3\\ -3x-6y+12z=5\end{array}$

Short answer

The solution set of the given system of equations is $\left(\frac{1}{3},-\frac{1}{2},\frac{1}{4}\right)$.

Step-by-step solution

– Use the elimination method to get the system of equations in two variables.

Multiply the equation $6x+2y+4z=2$by $2$and add the new resultant equation to the equation $3x+4y-8z=-3$.

localid="1647997506005" role="math" $\begin{array}{l}\underset{}{\begin{array}{l}6x+2y+4z=2\\ 3x+4y-8z=-3\end{array}}\begin{array}{c}\text{multiplyby2}\\ \end{array}\underset{}{\begin{array}{l}12x+4y+8z=4\\ 3x+4y-8z=-3\end{array}}\\ 15x+8y+0=1\end{array}$

So, the resultant equation is $15x+8y=1$

Multiply the equation $3x+4y-8z=-3$by $3$and multiply the equation $-3x-6y+12z=5$by $2$.

Then add the two new resultant equations.

$\begin{array}{l}\underset{\_}{\begin{array}{l}3x+4y-8z=-3\\ -3x-6y+12z=5\end{array}}\begin{array}{c}\text{multiplyby3}\to \\ \text{multiplyby2}\to \end{array}\underset{\_}{\begin{array}{l}9x+12y-24z=-9\\ -6x-12y+24z=10\end{array}}\\ 3x-0+0=1\end{array}$

So, the resultant equation is $3x=1$.

– Use the elimination method to solve the system of two equations.

Solve $3x=1$for $x$:

$\begin{array}{l}3x=1\\ \frac{3x}{3}=\frac{1}{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Dividebothsidesby3}\\ x=\frac{1}{3}\end{array}$

– Find the values of yand z.

Substitute $x=\backslash frac\left\{1\right\}\left\{3\right\}\$in15x+8y=1$and find the value of $y$.

$\begin{array}{l}15x+8y=1\\ 15\left(\frac{1}{3}\right)+8y=1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Substitute}\frac{\text{1}}{3}\text{for}x\\ 5+8y=1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Simplify}\\ 8y=-4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract5frombothsides}\\ y=-\frac{1}{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Dividebothsidesby}8\end{array}$

Substitute localid="1647997601231" $x=\frac{1}{3}$,localid="1647997625663" $y=-\frac{1}{2}in3x+4y-8z=-3$and find the value of $z$

$\begin{array}{l}3x+4y-8z=-3\\ 3\left(\frac{1}{3}\right)+4(-\frac{1}{2})-8z=-3\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}substitute}\frac{\text{1}}{3}\text{for}x,-\frac{1}{2}\text{for}y\\ 1-2-8z=-3\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}simplify}\\ -1-8z=-3\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}simplify}\\ -8z=-2\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}add1onbothsides}\\ z=\frac{1}{4}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Dividebothsidesby}-8\end{array}$

Hence, the solution of the given system of equations is$\left(x,y,z\right)=\left(\frac{1}{3},-\frac{1}{2},\frac{1}{4}\right)$.