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### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# in exercises 1 through 10, find all solutions of the linear systems using elimination.Then check your solutions$\left|x+2y=1\phantom{\rule{0ex}{0ex}}2x+3y=1\phantom{\rule{0ex}{0ex}}\right|$

The solution of system of equations is $x=-1,y=1$$x=-1,y=1$

See the step by step solution

## Step 1:Transforming the system

To get the solution, we will transform the value of $x,y\mathrm{ℤ}$.

$\left|x+2y=1\phantom{\rule{0ex}{0ex}}2x+3y=1\phantom{\rule{0ex}{0ex}}\right|$Into the form $\left|x=....\phantom{\rule{0ex}{0ex}}y=....\phantom{\rule{0ex}{0ex}}\right|$

## 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 xfrom first equation. First of all multiply the first equation by 2.

$\left|2x+4y=2\phantom{\rule{0ex}{0ex}}2x+3y=1\phantom{\rule{0ex}{0ex}}\right|$

Now subtract the first equation from the second equation.

$\left|2x-2x+4y-3y=2-1\phantom{\rule{0ex}{0ex}}2x+3y=1\phantom{\rule{0ex}{0ex}}\right|=\left|y=1\phantom{\rule{0ex}{0ex}}2x+3y=1\phantom{\rule{0ex}{0ex}}\right|$

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

$\left|y=1\phantom{\rule{0ex}{0ex}}2x+3\left(1\right)=1\phantom{\rule{0ex}{0ex}}\right|=\left|y=1\phantom{\rule{0ex}{0ex}}2x=1-3\phantom{\rule{0ex}{0ex}}\right|=\left|y=1\phantom{\rule{0ex}{0ex}}x=-1\phantom{\rule{0ex}{0ex}}\right|$

## Step 3: Checking the solution.

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

$\begin{array}{rcl}\left|2\left(-1\right)+4\left(1\right)\stackrel{?}{=2}\phantom{\rule{0ex}{0ex}}2\left(-1\right)+3\left(1\right)\stackrel{?}{=1}\right|& ⇒& \left|\begin{array}{c}-2+4\stackrel{?}{=2}\\ -2+3\stackrel{?}{=}1\end{array}\right|\\ & ⇒& \left|\begin{array}{c}2=2\\ 1=1\end{array}\right|\\ & & \end{array}$

Hence, the solution of the given system is $x=-1,y=1$