Q48. - Algebra 1

Question

CHALLENGE If $\frac{a+1}{b-1}=\frac{5}{1}$ and $\frac{a-1}{b+1}=\frac{1}{1},$ find the value of $\frac{b}{a}$  (Hint: Choose different values of and for which the proportions are true and evaluate the expression $\frac{b}{a}$.)

Step-by-Step Solution

Answer

$\frac{b}{a}=\frac{1}{2}$

Step 1. Understanding the concept of ratio.

A ratio is a number, which expresses one quantity as a fraction of the other.

For example, the ratio of 3 to 4 is 3 : 4 or $\frac{3}{4}$ .

 

Step 2. Choosing possible sets.

According to the hint we can choose any number to see if it fits the ratio and make it true. Let us choose the most basic numbers first like set $\left(a,b\right)$ as $\left(1,1\right)\left(1,2\right),\left(2,1\right)\left(2,2\right)$ and so on…

Step 3. Plug values in the expression.

We are given the equations $\frac{a+1}{b-1}=\frac{5}{1}$ and $\frac{a-1}{b+1}=\frac{1}{1}$

We start by plugging the first value: $a=1,b=1$

 Thus we have:

$\begin{array}{l}\frac{a+1}{b-1}=\frac{5}{1}\\ \frac{1+1}{1-1}=\frac{5}{1}\\ \frac{1}{0}\ne \frac{5}{1}\end{array}$

Thus $a=1,b=1$ is not a valid solution.

Similarly plugging in next set $a=1,b=2$ we get,

 $\begin{array}{l}\frac{a+1}{b-1}=\frac{5}{1}\\ \frac{1+1}{2-1}=\frac{5}{1}\\ \frac{2}{1}\ne \frac{5}{1}\end{array}$

Thus $a=1,b=2$ is not a valid solution…

Going in this manner we get to the solution set  $a=4,b=2$

Plugging in this set  $a=4,b=2$ we get,

 $\begin{array}{l}\frac{a+1}{b-1}=\frac{5}{1}\\ \frac{4+1}{2-1}=\frac{5}{1}\\ \frac{5}{1}=\frac{5}{1}\end{array}$

Thus $a=4,b=2$ is a valid solution.

Step 4. Verify validity using the second equation

Plugging in the valid set $a=4,b=2$, in the second equation we get,

 $\begin{array}{l}\frac{a-1}{b+1}=\frac{1}{1},\\ \frac{4-1}{2+1}=\frac{1}{1}\\ \frac{3}{3}=\frac{1}{1}\\ \frac{1}{1}=\frac{1}{1}\end{array}$

Thus $a=4,b=2$ is indeed a valid solution.

Step 5. Find the value of the expression

Since $a=4,b=2$ is a valid solution. We have,

 $\begin{array}{l}\frac{b}{a}=\frac{2}{4}\\ =\frac{1}{2}\end{array}$