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Expert-verified Found in: Page 696 ### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508 # These exercises refer to the following system:$\left\{\begin{array}{l}x-y+z=2\\ -x+2y+z=-3\\ 3x+y-2z=2\end{array}\right\$To eliminate x from the third equation, we add_______ times the first equation to the third equation. The third equation becomes _______$=$________

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

See the step by step solution

## Step 1. Given information

From the given question, we have been given 3 equations

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

## 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 x cancel each other and x will be eliminated:

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

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

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

$4\text{}y-5\text{}z=-4$

So, the third equation becomes $4\text{}y-5\text{}z=-4$. ### Want to see more solutions like these? 