Q. 189 - Intermediate Algebra

Question

Solve a System of Linear Equations with Three Variables:

$\left\{\begin{array}{l}x-2y+3z=1\\ x+y-3z=7\\ 3x-4y+5z=7\end{array}\right.$

Step-by-Step Solution

Answer

Solution of the system of equations is $(5+z,2+2z,z)$.

Step 1. Given Information.

We have been given system of equations :

$\left\{\begin{array}{l}x-2y+3z=1\\ x+y-3z=7\\ 3x-4y+5z=7\end{array}\right.$

Step 2. Using Elimination method to solve system of equations.

Multiply $-1$ in second equation to get opposite coefficients of $x$

$x+y-3z=7$

$-x-y+3z=-7$

Step 3. Add the equations to get two variable equation.

Add the equations : $\left\{\begin{array}{l}x-2y+3z=1\\ -x-y+3z=-7\end{array}\right.$

$-3y+6z=-6$

$-y+2z=-2$

Step 4. Multiply - 3 in first equation, to get opposite coefficients of x .

$x-2y+3z=1$

$-3(x-2y+3z)=-3\left(1\right)$

$-3x+6y-9z=-3$

Now, add the equations : $\left\{\begin{array}{l}-3x+6y-9z=-3\\ 3x-4y+5z=7\end{array}\right.$

$2y-4z=4$

$y-2z=2$

Step 5. Add the equations.

$\left\{\begin{array}{l}-y+2z=-2\\ y-2z=2\end{array}\right.$

$0=0$

This is True.

System has infinitely many solutions.

Step 6. Represent the solution showing how x , y are dependent on z .

So, we have $y-2z=2$

$y=2z+2$

And, equation $x-2y+3z=1$

Substitute -

$x-2(2z+2)+3z=1$

$x-4z-4+3z=1$

$x-z=5$

$x=5+z$

Solution of the system of equations is $(5+z,2+2z,z)$.