Chapter 3.5, Problem 39WE

To determine

To find:the solution of the system and the conditions should be placed on the constants in the system to be certain that the system has a unique solution.

Answer to Problem 39WE

The system has a solution $\left(-1,b+a\right)$ such that $a\ne b$ .

Explanation of Solution

Given information:

The system of the equation is $ax+y=b\text{ and }bx+y=a$ where a and b are constant

Calculation:

Assume that the system $ax+y=b\text{ and }bx+y=a$ has unique solution.

The values of a and b are computed as follows,

Subtract the equation $bx+y=a$ from the equation $ax+y=b$ to eliminate the variable y ,

  $\begin{array}{l}ax+y=b\\ bx+y=a\\ \overline{ax-bx+y-y=b-a}\end{array}$

Simplify further to find the values of x ,

  $\begin{array}{l}ax-bx+0=b-a\\ x\left(a-b\right)=-\left(a-b\right)\\ x=\frac{-\left(a-b\right)}{\left(a-b\right)}\end{array}$

Since the system has unique solution, the values $a\ne b$ ,

  $x=-1\left[\because \text{ Divide on both sides by }a-b\right]$

Thus, the value of $x=-1$ .

Substitute $x=-1$ in $ax+y=b$ ,

  $\begin{array}{l}a\left(-1\right)+y=b\\ -a+y=b\\ y=b+a\end{array}$

Therefore, the system has a solution $\left(-1,b+a\right)$ such that $a\ne b$ .