# Applications of Linear Equations - Miscellaneous Solved Examples

In this section, we will discuss how to use the concept of pair of linear equations in applied situations.

**Example 1: **Five years ago, the age of Niraj was seven times that of his son Girish. Five years from, his age will be three times that of his son. What is the ratio of their ages today?

**Solution:** Let the ages of Niraj and Girish be *x* and *y* respectively. Then, their ages five years ago would have been:

\[x - 5,\;y - 5\]

And their ages five years from today will be:

\[x + 5,\;y + 5\]

From the information given in the problem, we have:

\[\begin{array}{l}\left\{ \begin{array}{l}x - 5 = 7\left( {y - 5} \right)\\x + 5 = 3\left( {y + 5} \right)\end{array} \right.\\ \Rightarrow \qquad x - 7y + 30 = 0\\ \qquad\quad x - 3y - 10 = 0\end{array}\]

Subtracting the two equations, we have:

\[\begin{array}{l} - 4y + 40 = 0\\ \Rightarrow y = 10\end{array}\]

Plugging this value of *y* back into the first equation, we have:

\[\begin{array}{l}x - 5 = 7\left( {y - 5} \right)\\ \Rightarrow \qquad x - 5 = 7\left( {10 - 5} \right) = 35\\ \Rightarrow \qquad x = 40\end{array}\]

Thus, their present ages are 40 years and 10 years.

**Example 2: **The cost of 2 pencils and 3 erasers is ₹ 9, while the cost of 5 pencils and 4 erasers is ₹19. What is the cost of each pencil and each eraser?

**Solution:** Let the cost of each pencil and each eraser be *p* and *e* respectively. Thus, we have:

\[\begin{array}{l}2p + 3e = 9\\5p + 4e = 19\end{array}\]

Let’s solve this pair by the cross-multiplication method. The coefficients grid is:

\[\begin{array}{l}2 && 3 && - 9\\5 && 4 && - 19\end{array}\]

Now, we calculate the terms which will come below *p*, negative *e* and 1 in the solution equality:RF

Thus, the solution is:

\[\begin{align}&\frac{p}{{ - 21}} = \frac{{ - e}}{7} = \frac{1}{{ - 7}}\\ &\Rightarrow \;\;\;p = 3,\;e = 1\end{align}\]

This means that each pencil costs ₹ 3, and each eraser costs ₹ 1.

**Example 3: **Solve the following pair of equations:

\[\begin{align}&\frac{2}{x} + \frac{3}{y} = 18\\&\frac{5}{x} - \frac{2}{y} = 7\end{align}\]

**Solution:** Clearly, this is not a pair of linear equations, but it can easily be reduced to one, using the following substitutions:

\[\frac{1}{x} \to p,\;\frac{1}{y} \to q\]

Thus, our pair of equations becomes:

\[\begin{array}{l}\left\{ \begin{array}{l}2p + 3q = 18\\5p - 2q = 7\end{array} \right.\\ \Rightarrow \;\;\;2p + 3q - 18 = 0\\\;\;\;\;\;\;5p - 2q - 7 = 0\end{array}\]

Let’s use cross-multiplication to solve this pair:

\[\begin{align}&\frac{p}{{\left( {3 \times - 7} \right) - \left( { - 2 \times - 18} \right)}} = \frac{{ - q}}{{\left( {2 \times - 7} \right) - \left( {5 \times - 18} \right)}}\\ &\qquad\qquad\qquad\qquad\qquad\;\;= \frac{1}{{\left( {2 \times - 2} \right) - \left( {5 \times 3} \right)}}\\ &\Rightarrow \;\;\;\frac{p}{{ - 21 - 36}} = \frac{{ - q}}{{ - 14 + 90}} = \frac{1}{{ - 4 - 15}}\\ &\Rightarrow \;\;\;\frac{p}{{ - 57}} = \frac{{ - q}}{{76}} = \frac{1}{{ - 19}}\\ &\Rightarrow \;\;\;p = 3,\;q = 4\end{align}\]

Note that these values correspond to *p* and *q*, whereas the original unknowns were *x* and *y*, so we still have one more step left:

\[\begin{align}{l}x = \frac{1}{p} = \frac{1}{3}\\y = \frac{1}{q} = \frac{1}{4}\end{align}\]

This completes our solution.

**Example 4: **Solve the following pair of equations:

\[\begin{align}{l}\frac{5}{{x - 1}} + \frac{1}{{y - 2}} = 2\\\frac{6}{{x - 1}} - \frac{3}{{y - 2}} = 1\end{align}\]

**Solution:** Once again, these equations are not linear, but we can easily reduce them to linear form:

\[\frac{1}{{x - 1}} \to p,\;\frac{1}{{y - 2}} \to q\]

Thus, our equations become:

\[\begin{array}{l}\left\{ \begin{array}{l}5p + q = 2\\6p - 3q = 1\end{array} \right.\\ \Rightarrow \;\;\;5p + q - 2 = 0\\\;\;\;\;\;\;6p - 3q - 1 = 0\end{array}\]

By cross-multiplication, the solution is:

\[\begin{align}&\frac{p}{{\left( {1 \times - 1} \right) - \left( { - 3 \times - 2} \right)}} = \frac{{ - q}}{{\left( {5 \times - 1} \right) - \left( {6 \times - 2} \right)}}\\ &\qquad\qquad\qquad\qquad\qquad= \frac{1}{{\left( {5 \times - 3} \right) - \left( {6 \times 1} \right)}}\\ &\Rightarrow \;\;\;\frac{p}{{ - 7}} = \frac{{ - q}}{7} = \frac{1}{{ - 21}}\\ &\Rightarrow \;\;\;p = \frac{1}{3},\;q = \frac{1}{3}\end{align}\]

Now, we evaluate the original unknowns:

\[\begin{align}&\frac{1}{{x - 1}} = p = \frac{1}{3}\\ &\Rightarrow \;\;\;x - 1 = 3\;\; \Rightarrow \;\;x = 4\\&\frac{1}{{y - 2}} = q = \frac{1}{3}\\ &\Rightarrow \;\;\;y - 2 = 3\;\; \Rightarrow \;\;y = 5\end{align}\]

Thus, the final solution is:

\[x = 4,\;y = 5\]

**Example 5: **A boat is travelling in a river in which the water is flowing with a certain speed. In 10 hours, the boat can go 30 km upstream and then 52 km downstream. In 15 hours, it can go 50 km upstream and 65 km downstream. Determine the speed of the boat in still water, and the speed of the flowing water.

**Solution: **We suppose that the speed of the boat in still water is *x* km/hr, while the speed of the water is *y* km/hr. Clearly, when the boat is traveling upstream, its speed *with respect to the ground* will reduce compared to *x*, while when it is traveling downstream, its speed with respect to the ground will increase compared to *x*. In fact, when traveling downstream, the speed of the boat with respect to the ground will be *x* + *y* km/hr, while when traveling upstream, its speed with respect to the ground will be *x* – *y* km/hr.

Now, we know that time taken by the boat in covering a certain distance will be that distance divided by the boat’s speed with respect to the ground. Thus, when the boat goes 30 km upstream, it will take a time of

\[\frac{{30}}{{x - y}}\;{\rm{hours}}\]

And when it goes 52 km downstream, it will take a time of

\[\frac{{52}}{{x + y}}\;{\rm{hours}}\]

The total time in going 30 km upstream and 52 km downstream is given to be 10 hours, and so:

\[\frac{{30}}{{x - y}} + \frac{{52}}{{x + y}} = 10\]

Similarly, we have:

\[\frac{{50}}{{x - y}} + \frac{{65}}{{x + y}} = 15\]

We thus have two equations, which are not linear, but can easily be reduced to linear form using the following substitutions:

\[\frac{1}{{x - y}} \to p,\;\frac{1}{{x + y}} \to q\]

Thus, our pair of equations becomes:

\[\begin{array}{l}30p + 52q = 10\\50p + 65q = 15\end{array}\]

We can simplify this pair a bit, by taking out the common factors:

\[\begin{array}{l}15p + 26q = 5 & ...({\rm{I}})\\10p + 13q = 3 & ...({\rm{II}})\end{array}\]

We can easily eliminate *q* by doing (I) – 2 × (II):

\[\begin{array}{l}\left\{ \begin{array}{l}\,\,\,\,\,15p + 26q = 5\\\,\,\,\,\,20p + 26q = 6\\\underline { - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ - 5p\,\,\,\, = \,\,\,\,\,\,\,\,\,\,\,\, - 1\end{array} \right.\\ \Rightarrow \;\;\;p = \frac{1}{5}\\ \Rightarrow \;\;\;q = \frac{1}{{13}} & ({\rm{from}}\;{\rm{I}}\;{\rm{or}}\;{\rm{II}})\end{array}\]

Now that we have *p* and *q*, we go on to find out the original unknowns, *x* and *y*:

\[\begin{array}{l}x - y = \frac{1}{5} = 5\\x + y = \frac{1}{q} = 13\\ \Rightarrow \;\;\;2x = 18\;\; \Rightarrow \;\;x = 9\\ \Rightarrow \;\;\;y = x - 5 = 4\end{array}\]

Thus, the speed of the boat in still water is 9 km/hr, while the speed of the flowing water is 4 km/hr.

**Example 6: **A train moves between two stations A and B at a uniform speed, and takes a certain amount of time. If the train had gone 10 km/hr faster, it would have taken 1 hour less than the scheduled duration. Had it gone 10 km/hr slower, it would have taken 2 hours more than the scheduled duration. What is the distance between A and B?

**Solution:** Let the distance between A and B be *x* km, and let the normal speed of the train be *y* km/hr. The scheduled duration of the journey will be:

\[\frac{x}{y}\;{\rm{hours}}\]

However, if the train had gone 10 km/hr faster, the time taken would have been:

\[\frac{x}{{y + 10}}\;{\rm{hours}}\]

Since this is 1 hour less than the scheduled duration, we have:

\[\frac{x}{{y + 10}} = \frac{x}{y} - 1 ...({\rm{I}})\]

Similarly, from the other part of the problem statement, we have:

\[\frac{x}{{y - 10}} = \frac{x}{y} + 2 ...({\rm{II}})\]

Now, we have two equations in *x* and *y*, which are not linear. But we already know a very useful technique from having solved linear equations: the technique of elimination. Somehow, we eliminate the constant terms from the pair of equations. For that purpose, we do 2 × (I) + (II):

\[\begin{align}&2 \times ({\rm{I}}):\;\;\;\frac{{2x}}{{y + 10}} = \frac{{2x}}{y} - 2\\&({\rm{II}}):\;\;\;\frac{x}{{y - 10}} = \frac{x}{y} + 2\\&2 \times ({\rm{I}}) + ({\rm{II}}):\\&\frac{{2x}}{{y + 10}} + \frac{x}{{y - 10}} = \frac{{3x}}{y}\\ &\Rightarrow \;\;\;x\left( {\frac{2}{{y + 10}} + \frac{1}{{y - 10}}} \right) = x\left( {\frac{3}{y}} \right)\\ &\Rightarrow \;\;\;\frac{{2\left( {y - 10} \right) + \left( {y + 10} \right)}}{{{y^2} - 100}} = \frac{3}{y}\left\{ \begin{array}{l}x\;{\rm{gets}}\\{\rm{eliminated!}}\end{array} \right\}\\ &\Rightarrow \;\;\;\frac{{3y - 10}}{{{y^2} - 100}} = \frac{3}{y}\\ &\Rightarrow \;\;\;3{y^2} - 10y = 3{y^2} - 300\\ &\Rightarrow \;\;\;10y = 300\\ &\Rightarrow \;\;\;y = 30\end{align}\]

Once we have *y*, we plug this value into any of the two original equations, say (I):

\[\begin{align}&\frac{x}{{30 + 10}} = \frac{x}{{30}} - 1\\ &\Rightarrow \;\;\;\frac{x}{{40}} = \frac{{x - 30}}{{30}}\\ &\Rightarrow \;\;\;x = \frac{4}{3}x - 40\\ &\Rightarrow \;\;\;\frac{1}{3}x = 40\\ &\Rightarrow \;\;\;x = 120\end{align}\]

Thus, the distance between A and B is 120 km.

Note that the key to solving this problem was the technique of elimination. This example should show that elimination is a general technique which applies not only to solving linear equations but non-linear equations as well.

**Example 7: **There is a certain task which a boy and a girl will take different times to complete. 4 boys and 6 girls working together can complete this task in one day, and so can 2 boys and 9 girls working together. How many days will it take 2 boys and 3 girls working together to complete the task?

**Solution:** Suppose that a boy working alone will take *b* days to complete the task, and a girl working alone will take *g* days to complete it. Then, in one day, a single boy can do what fraction of the complete task? Obviously, he can complete:

\[\frac{1}{b}\;{\rm{of}}\;{\rm{the}}\;{\rm{task}}\]

Similarly, in one day, a girl working alone can complete:

\[\frac{1}{g}\;{\rm{of}}\;{\rm{the}}\;{\rm{task}}\]

4 boys and 6 girls working together can complete:

\[\frac{1}{b} + \frac{6}{g}\;{\rm{of}}\;{\rm{the}}\;{\rm{task}}\]

But it is also given that 4 boys and 6 girls working together can complete this task in one day, which means that:

\[\begin{align}&\frac{4}{b} + \frac{6}{g} \equiv {\rm{Complete}}\;{\rm{task}}\\ &\Rightarrow \;\;\;\frac{4}{b} + \frac{6}{g} = 1\end{align}\]

Similarly,

\[\frac{2}{b} + \frac{9}{g} = 1\]

Thus, we have a pair of (non-linear) equations in *b* and *g*, which can easily be reduced to linear form using the following substitutions:

\[\frac{1}{b} \to x,\;\frac{1}{g} \to y\]

Our pair of equations thus becomes:

\[\begin{array}{l}4x + 6y = 1\\2x + 9y = 1\end{array}\]

This can be solved easily to yield:

\[\begin{align}&x = \frac{1}{8},\;y = \frac{1}{{12}}\\ &\Rightarrow \;\;\;b = \frac{1}{x} = 8,\;g = \frac{1}{y} = 12\end{align}\]

Thus, a boy working alone can complete the task in 8 days, and a girl working alone can complete the task in 12 days. Now, if 2 boys and 3 girls work together, what fraction of the task can they complete in one day? This is straightforward:

\[\begin{array}{l}\left( {2 \times \frac{1}{b} + 3 \times \frac{1}{g}} \right)\;{\rm{of}}\;{\rm{the}}\;{\rm{task,}}\;{\rm{or}}\\\left( {\frac{2}{8} + \frac{3}{{12}}} \right)\;{\rm{of}}\;{\rm{the}}\;{\rm{task,}}\;{\rm{or}}\\ \qquad\quad \frac{1}{2}\;{\rm{of}}{\rm{the}}\;{\rm{task}}\end{array}\]

Since this group (2 boys, 3 girls) can complete half of the task in one day, they will take 2 days to complete the whole task. This completes our solution.