Chapter 10.1, Problem 54E

To determine

For the given system of equation, the values of $x$ and $y$ in terms of $a$ and $b$ are to be estimated:

  $\{\begin{array}{l}ax+by=0\\ x+y=1\end{array}\left(a\ne b\right)$

Answer to Problem 54E

The solutions $x$ and $y$ of set of eqns. $\{\begin{array}{l}ax+by=0\\ x+y=1\end{array}\left(a\ne b\right)$ in terms of $a$ and $b$ are $x=\frac{b}{\left(b-a\right)},y=\frac{a}{\left(a-b\right)}$

Explanation of Solution

Given:

  $\{\begin{array}{l}ax+by=0\\ x+y=1\end{array}\left(a\ne b\right)$

Calculations:

The given eqns. are:

  $ax+by=0$.......(1)

  $x+y=1$.......(2)

Eq. (2)can be written as multiply by $a$ :

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

  $ax+ay=a$.......(3)

Now, subtract (3) from eq. (1) :

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

Put the value of $y$ into eq. (2):

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

So, the solutions of given set of eqns. is $x=\frac{b}{\left(b-a\right)},y=\frac{a}{\left(a-b\right)}$ .

Conclusion:

Hence, the solutions of given set of eqns. is:

  $x=\frac{b}{\left(b-a\right)},y=\frac{a}{\left(a-b\right)}$