Chapter 3.5, Problem 40WE

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 40WE

The system has a solution $\left(\frac{ac+bd}{a^2+b^2},\frac{ad-bc}{a^2+b^2}\right)$ such that $\frac{a}{b}\ne -\frac{b}{a}$ .

Explanation of Solution

Given information:

The system of the equation is $ax-by=c\text{ and }bx+ay=d$ where a , b, c, and d are constants.

Calculation:

Assume that the system $ax-by=c\text{ and }bx+ay=d$ has unique solution.

The values of a , b , c , and d are computed as follows,

Find the linear combination that eliminates the variable y .

Since the least common multiple of the y’s coefficients b and a is ab , multiply first equation by a and multiply second equation by b .

  $\begin{array}{l}{a}^{2}x-aby=ac\\ b^{2}x+aby=bd\end{array}$

Add these two equations to simplify,

  $\begin{array}{l}{a}^{2}x-aby=ac\\ b^{2}x+aby=bd\\ \overline{a^{2}x+b^{2}x+0=ac+bd}\end{array}$

Simplify further to find the values of x ,

  $\begin{array}{l}\left(a^2+b^2\right)x=ac+bd\\ x=\frac{ac+bd}{a^2+b^2}\left[\because \text{ Divide on both sides by }{a}^2+b^2\right]\end{array}$

Here, the denominator of $x=\frac{ac+bd}{a^2+b^2}$ cannot be zero.

That is,

  $\begin{array}{l}{a}^2+b^2\ne 0\\ a^2\ne -b^2\\ \frac{a^2}{ab}\ne -\frac{b^2}{ab}\left[\because \text{Divide on both sides by }ab\right]\\ \frac{a}{b}\ne -\frac{b}{a}\end{array}$

Thus, the value of $x=\frac{ac+bd}{a^2+b^2}$ such that $\frac{a}{b}\ne -\frac{b}{a}$ .

Substitute $x=\frac{ac+bd}{a^2+b^2}$ in $bx+ay=d$ and simplify,

  $\begin{array}{l}b\left(\frac{ac+bd}{a^2+b^2}\right)+ay=d\\ \frac{abc+b^{2}d}{a^2+b^2}+ay=d\\ ay=d-\frac{abc+b^{2}d}{a^2+b^2}\\ ay=\frac{d\left(a^2+b^2\right)}{a^2+b^2}-\frac{abc+b^{2}d}{a^2+b^2}\end{array}$

Simplify further to find the value of y ,

  $\begin{array}{l}ay=\frac{a^{2}d+b^{2}d}{a^2+b^2}-\frac{abc+b^{2}d}{a^2+b^2}\\ ay=\frac{a^{2}d+\cancel{b^{2}d}-abc-\cancel{b^{2}d}}{a^2+b^2}\\ ay=\frac{a^{2}d-abc}{a^2+b^2}\end{array}$

Divide on both sides by a ,

  $\begin{array}{l}\frac{ay}{a}=\frac{a^{2}d-abc}{a\left(a^2+b^2\right)}\\ y=\frac{a\left(ad-bc\right)}{a\left(a^2+b^2\right)}\\ y=\frac{ad-bc}{a^2+b^2}\end{array}$

Thus, the value of $y=\frac{ad-bc}{a^2+b^2}$ such that $\frac{a}{b}\ne -\frac{b}{a}$ .

Therefore, the system has a solution $\left(\frac{ac+bd}{a^2+b^2},\frac{ad-bc}{a^2+b^2}\right)$ such that $\frac{a}{b}\ne -\frac{b}{a}$ .