Question

Solve each system of equations

$\begin{array}{l}8x-6z=38\\ 2x-5y+3z=5\\ x+10y-4z=8\end{array}$

Short answer

The solution set of the given system of equations is $\left(4,0,-1\right)$.

Step-by-step solution

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

Multiply the equation $2x-5y+3z=5$by $2$and add the new resultant equation to $x+10y-4z=8$.

$\begin{array}{l}\underset{\_}{\begin{array}{l}2x-5y+3z=5\\ x+10y-4z=8\end{array}}\begin{array}{c}\text{multiplyby2}\\ \end{array}\underset{\_}{\begin{array}{l}4x-10y+6z=10\\ x+10y-4z=8\end{array}}\\ 5x+0+2z=18\end{array}$

So, the resultant equation is $5x+2z=18$.

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

Multiply $5x+2z=18$by 3 and add the new resultant equation to $8x-6z=38$.

$\begin{array}{l}\underset{\_}{\begin{array}{l}5x+2z=18\\ 8x-6z=38\end{array}}\begin{array}{c}\text{multiplyby3}\\ \end{array}\underset{\_}{\begin{array}{l}15x+6z=54\\ 8x-6z=38\end{array}}\\ 23x+0=92\end{array}$

Solve for :$\begin{array}{l}23x=92\\ \frac{23x}{23}=\frac{92}{23}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Dividebothsidesby23}\\ x=4\end{array}$

– Find the values of yand z.

Substitute $x=4in5x+2z=18$and find the value of $z$.

$\begin{array}{l}5x+2z=18\\ 5\left(4\right)+2z=18\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Substitute4for}x\\ 20+2z=18\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Simplify}\\ 2z=-2\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract20frombothsides}\\ z=-1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Dividebothsidesby2}\end{array}$

Substitute $x=4,z=-1in2x-5y+3z=5$and find the value of $y$

$\begin{array}{l}2x-5y+3z=5\\ 2\left(4\right)-5y+3(-1)=5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}substitute4for}x,-1\text{for}z\\ 8-5y-3=5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}simplify}\\ 5-5y=5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}simplify}\\ -5y=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}subtract5frombothsides}\\ y=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}dividebothsidesby}-\text{5}\end{array}$

Hence, the solution of the given system of equations is$\left(x,y,z\right)=\left(4,0,-1\right)$.

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