Q. 345 - Intermediate Algebra

Question

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

$\left\{\begin{array}{l}\frac{1}{3}x-\frac{1}{2}y=1\\ \frac{3}{4}x-y=\frac{5}{2}\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of linear equation is $\left(6,2\right)$.

Step 1. Given information.

The system of linear equations, 

$\left\{\begin{array}{l}\frac{1}{3}x-\frac{1}{2}y=1\\ \frac{3}{4}x-y=\frac{5}{2}\end{array}\right.$

Step 2. Calculating the value of x .

Now, we will make some changes in the equations to eliminate $y$ terms.

Multiplying the first equation with $2$ we get

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

Then the equations will be, 

$$\begin{gathered} \frac{2}{3}x-y=2 \\ \frac{3}{4}x-y=\frac{5}{2} \end{gathered}$$

Subtracting the above equations then 

$\frac{2}{3}x-y=2-\frac{3}{4}x+y=-\frac{5}{2}\left(\frac{2}{3}-\frac{3}{4}\right)x=2-\frac{5}{2}\frac{-1}{12}x=\frac{-1}{2}x=6$

Step 3. Calculating the value of y .

Substituting the value of $x$ in any equation.

$$\begin{gathered} \frac{1}{3}\left(6\right)-\frac{1}{2}y=1 \\ 2-\frac{1}{2}y=1 \\ -\frac{1}{2}y=-1 \\ y=2 \end{gathered}$$ 

Step 4. Checking the result.

Substituting the value of $x$ and $y$ in the first equation then

$$\begin{gathered} \frac{1}{3}x-\frac{1}{2}y=1 \\ \frac{1}{3}\left(6\right)-\frac{1}{2}\left(2\right)=1 \\ 2-1=1 \\ 1=1 \end{gathered}$$

Substituting the value of $x$ and $y$ in the second equation then

$$\begin{gathered} \frac{3}{4}x-y=\frac{5}{2} \\ \frac{3}{4}\left(6\right)-2=\frac{5}{2} \\ \frac{9}{2}-2=\frac{5}{2} \\ 9-4=5 \\ 5=5 \end{gathered}$$

Hence, this is true.