Question

Solve each system of equations.

14

$\begin{array}{l}5x+2y=4\\ 3x+4y+2z=6\\ 7x+3y+4z=29\end{array}$

Short answer

The solution is $x=2,y=-3,z=6.$.

Step-by-step solution

Step-1 –Using elimination to obtain two equations with two variables

Given system of equations are

$\begin{array}{l}5x+2y=4\\ 3x+4y+2z=6\\ 7x+3y+4z=29\end{array}$

Multiplying the second equation by $2$and subtracting from the third equation, we get

$\begin{array}{l}7x-6x+3y-8y=29-12\\ x-5y=17\end{array}$

Step-2 –Solving the system of two equations with two variables

$\begin{array}{l}5x+2y=4\\ x-5y=17\end{array}$System of two equations with two variables are

Multiplying the second equation by $5$and subtracting from the first equation, we get

$\begin{array}{l}2y+25y=4-85\\ 27=-81\\ y=-3\end{array}$

Putting the value of$y$in the second equation, we get

$\begin{array}{l}x-5y=17\\ x-5(-3)=17\\ x+15=17\\ x=17-15\\ x=2\end{array}$

Step-3 –Substituting the value ofx and y in any equation with three variables

$\begin{array}{l}3x+4y+2z=6\\ 3\left(2\right)+4(-3)+2z=6\\ 6-12+2z=6\\ 2z=6+6\\ 2z=12\\ z=6.\end{array}$