Chapter 7, Problem 2PS

To determine

To find: the values of $k_1$ and $k_1$ such that the system of equations has an infinite number of solutions.

Answer to Problem 2PS

The required values are $k_1=-\frac{10}{3},k_2=\frac{16}{3}$

Explanation of Solution

Given:

The given equation is

  $\{\begin{array}{l}3x-5y=8\\ 2x+k_{1}y=k_2\end{array}$

Calculation:

Consider the system of equations

  $\begin{array}{l}3x-5y=8\\ 2x+k_{1}y=k_2\mathrm{...........}(1)\end{array}$

Consider the following two equations in two variables

  $\begin{array}{l}{a}_{1}x+b_{1}y=c_1\\ a_{2}x+b_{2}y=c_2\mathrm{..............}(2)\end{array}$

Then the equations are consistent and dependent with infinity many solutions, if and only if

  $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\mathrm{.................}(3)$

Compare the equations in (1) and (2) to get

  $a_1=3,a_2=2,b_1=-5,b_2=k_1,c_1=k_2$

Put these values in (3) to get

  $\frac{2}{3}=\frac{k_1}{-5}=\frac{k_2}{8}$

Solve the first equation

  $\begin{array}{c}\frac{2}{3}=\frac{k_1}{-5}\\ k_1=\frac{2}{3}\left(-5\right)\\ =-\frac{10}{3}\end{array}$

Solve the second equation

  $\begin{array}{c}\frac{2}{3}=\frac{k_2}{8}\\ k_2=\frac{2}{3}\left(8\right)\\ =\frac{16}{3}\end{array}$

Therefore, the required values are $k_1=-\frac{10}{3},k_2=\frac{16}{3}$