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Expert-verified Found in: Page 5 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # In exercises, 1 through 10, findall solutions of the linear systems using elimination.T hen check your solutions.10.$\begin{array}{rcl}& & |\begin{array}{c}x+2y+3z=1\\ 2x+4y+7z=2\\ 3x+7y+11z=8\end{array}|\end{array}$

The solution of system of equation is $x=-9,y=5,z=0$

See the step by step solution

## Step 1:Transforming the system

To get the solution, we will transform the value of x,y,z.

$\left|x+2y+3z=1\text{}......\left(1\right)\phantom{\rule{0ex}{0ex}}2x+4y+7z=2\text{}......\left(2\right)\phantom{\rule{0ex}{0ex}}3x+7y+11z=8\text{}......\left(3\right)\right|$Into the form $\left|x=....\phantom{\rule{0ex}{0ex}}y=....\phantom{\rule{0ex}{0ex}}z=...\phantom{\rule{0ex}{0ex}}\right|$

## Step 2: Eliminating the variables

In the given system of equations, we can eliminate the variables by adding or subtracting the equations.

In this system, we can eliminate the variable x from equation 2 and 3. First of all subtract the 2 times of equation 1 from equation 2 and then subtract the 3 times of first equation from equation 3.

$\begin{array}{rcl}\left|x+2y+3z=1\phantom{\rule{0ex}{0ex}}2x-2x+4y-4y+7z-6z=0\phantom{\rule{0ex}{0ex}}3x-3x+7y-6y+11z-9z=5\right|& =& \left|x+2y+3z=1\phantom{\rule{0ex}{0ex}}z=0\phantom{\rule{0ex}{0ex}}y+2z=5\right|\end{array}$

Now put the value of z which is in second equation to the first and second equation.

$\begin{array}{rcl}\left|x+2\left(y\right)+3\left(0\right)=1\phantom{\rule{0ex}{0ex}}z=0\phantom{\rule{0ex}{0ex}}y+2\left(0\right)=5\right|& =& \left|x=1-2y\phantom{\rule{0ex}{0ex}}z=0\phantom{\rule{0ex}{0ex}}y=5\right|=\left|x=-9\phantom{\rule{0ex}{0ex}}z=0\phantom{\rule{0ex}{0ex}}y=5\right|\end{array}$

Hence, we get value of $x=-9,y=5,z=0$

## Step 3: Checking the solution

Now check the solution by putting the value of x, y and z in the given system of equation.

$-9+2\left(5\right)+3\left(0\right)=1\phantom{\rule{0ex}{0ex}}2\left(-9\right)+4\left(5\right)+7\left(0\right)=2\phantom{\rule{0ex}{0ex}}3\left(-9\right)+7\left(5\right)+11\left(0\right)=8\phantom{\rule{0ex}{0ex}}$

Which is true.

Hence, the solution of system of equation is $x=-9,y=5,z=0$ ### Want to see more solutions like these? 