Q 56 - Intermediate Algebra

Question

In the following exercise, solve the systems of equations by elimination.

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

Step-by-Step Solution

Answer

The solution is  $(-3,2)$.

Step 1. Given information

The given equations are

$\frac{1}{3}x-y=-3$

$x+\frac{5}{2}y=2$

Step 2. Calculating the value of y

To calculate the value of $y$ we have to eliminate the $x\text{'}s$ term.

Let us do some changes through which we can easily eliminate $x$.

1) multiply the first equation by $-3$

$-3\left(\frac{1}{3}x-y=-3\right)$

$-x+3y=9$

now, the new equations are 

$$\begin{gathered} -x+3y=9 \\ x+ \frac{5}{2}y=2 \end{gathered}$$

just add the equations

$-x+3y=9x+\frac{5}{2}y=2\frac{11}{2}y=11$

            $y=2$

the value of $y$=2

Step 3. Calculating the value of x

As we have the value of $y$, we can substitute it in any equation and hence calculate the value of $x$.

let us take the first equation

$\frac{1}{3}x-y=-3$

put $y$=2

$$\begin{gathered} \frac{1}{3}x-2=-3 \\ \frac{1}{3}x=-1 \\ x=-3 \end{gathered}$$

the value of x is $-3$.

Step 4. Checking the solution

Checking the solution by putting the value of $x,y$ in the equation, we get

$$\begin{gathered} \frac{1}{3}x-y=-3 \\ \frac{1}{3}(-3)-2=-3 \\ -1-2=-3 \\ -3=-3 \\ x+\frac{5}{2}y=2 \\ -3+\frac{5}{2}\left(2\right)=2 \\ -3+5=2 \\ 2=2 \end{gathered}$$

This is true, hence the solution is correct.