Chapter 5.3, Problem 26E

To determine

To calculate:

The solution of the following linear equations by using any correct method:

  $\begin{array}{l}3x+y=\frac{1}{3}\\ 2x-3y=\frac{8}{3}\end{array}$

Answer to Problem 26E

The solution of the $\begin{array}{l}3x+y=\frac{1}{3}\\ 2x-3y=\frac{8}{3}\end{array}$ by elimination method is $x,y=\frac{17}{33},-\frac{6}{11}$ .

Explanation of Solution

Given information:

  $\begin{array}{l}3x+y=\frac{1}{3}\\ 2x-3y=\frac{8}{3}\end{array}$

Calculation:

The given equations are as:

  $3x+y=\frac{1}{3}$$\times 3$

  $9x+3y=3$................ (1)

  $2x-3y=\frac{8}{3}$................ (2)

From above listed two equations, the coefficient of variable $y$ is same. So by eliminating variable $y$ , the value of $x$ is obtainable. So, the correct method to get the solutions is elimination method.

Take addition of equation (1) and (2) to eliminate variable $x$ :

  $\begin{array}{l}9x+3y+2x-3y=3+\frac{8}{3}\\ 11x=\frac{17}{3}\\ x=\frac{17}{33}\end{array}$

Now calculate the value of variable $y$ by putting value of $x=\frac{17}{33}$ in eq. (1):

  $\begin{array}{l}9x+3y=3\\ 9\left(\frac{17}{33}\right)+3y=3\\ \frac{153}{33}+3y=3\\ 3y=3-\frac{153}{33}\\ y=-\frac{54}{33}\times \frac{1}{3}\\ y=-\frac{18}{33}\\ y=-\frac{6}{11}\end{array}$

Hence, by solving two equations by elimination method get variable values as $x,y=\frac{17}{33},-\frac{6}{11}$ .