Question

These exercises refer to the following system:

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

To eliminate x from the third equation, we add_______ times the first equation to the third equation. The third equation becomes _______$=$________

Short answer

$-3$, $4y-5z$& $-4$

Step-by-step solution

Step 1. Given information

From the given question, we have been given 3 equations

$\begin{array}{l}\text{1.x-y+z=2}\\ \text{2.-x+2y+z=-3}\\ \text{3.3x+y-2z=2}\end{array}$

Step 2. Concept used

An equation is a statement in that the values of two mathematical expressions are equal (indicated by the sign $=$ ).

Step 3. Calculation

Lets we multiple the first equation by $-3$ and add it to the third equation, the coefficients of xcancel each other and x will be eliminated:

$\begin{array}{l}-3(x-y+z=2)\\ -3x+3y-3z=-6\end{array}$

Our first equation after multiplying by $-3$is $-3x+3y-3z=-6$

Adding the above equation and equation 3 (given in the question), we get

$4y-5z=-4$

So, the third equation becomes $4y-5z=-4$.