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### Course: Class 10 (Old)>Unit 3

Lesson 4: Equations reducible to linear form

# Solving equations reducible to linear form

## Introduction:

So far, we've solved a bunch of system of linear equations with two variables. We've used methods like substitution and elimination to solve such equations.
But what happens when the given pair of equations isn't linear?
Well, mostly it gets a lot messier. But sometimes we can convert them to a new set of equations that are linear.
And then try to solve them.
Let's try to understand that by solving one such problem.

## The problem:

A boat goes $30\phantom{\rule{0.167em}{0ex}}\text{km}$ upstream and $44\phantom{\rule{0.167em}{0ex}}\text{km}$ downstream in $10$ hours. It can go $40\phantom{\rule{0.167em}{0ex}}\text{km}$ upstream and $55\phantom{\rule{0.167em}{0ex}}\text{km}$ downstream in $13$ hours.
The boat has a constant speed in still water and the stream has a constant speed as well.
Moving upstream means moving against the flow and downstream means moving with the flow of water.
Find the speed of the boat in still water.

## Strategy:

We are given two scenarios where the boat moves, partly upstream and partly downstream. The distances and total time taken by the boat are known.
What we don't know is the speed of the $\text{stream}$ and the speed of the $\text{boat in still water}$. These are our unknown variables.
Using the two scenarios, we can create two equations that involve our two unknowns.
Once we have the equations, we can try to solve them.
If the equations are not linear, we might have to convert them to linear form before solving.

## Introducing variables:

Let the speed of the $\text{stream}$ be $w\phantom{\rule{0.167em}{0ex}}\frac{\text{km}}{\text{hr}}$.
Let the speed of the $\text{boat in still water}$ be $b\phantom{\rule{0.167em}{0ex}}\frac{\text{km}}{\text{hr}}$.
When the boat moves upstream, which is against the flow of water.
${\text{Speed}}_{\text{up}}=\left(b-w\right)\frac{\text{km}}{\text{hr}}$
When the boat moves downstream, which is with the flow of water.
${\text{Speed}}_{\text{down}}=\left(b+w\right)\frac{\text{km}}{\text{hr}}$

## Framing first equation:

Let's look at the first scenario.
The boat goes $30\phantom{\rule{0.167em}{0ex}}\text{km}$ upstream and $44\phantom{\rule{0.167em}{0ex}}\text{km}$ downstream in $10$ hours.
We can use the relation between speed, distance and time.
$\text{Time}=\frac{\text{Distance}}{\text{Speed}}$
When boat moves upstream,
$\begin{array}{rl}{\text{Time}}_{\text{up}}& =\frac{{\text{Distance}}_{\text{up}}}{{\text{Speed}}_{\text{up}}}\\ \\ {\text{Time}}_{\text{up}}& =\frac{30}{b-w}\phantom{\rule{0.167em}{0ex}}\text{hours}\end{array}$
When boat moves downstream,
$\begin{array}{rl}{\text{Time}}_{\text{down}}& =\frac{{\text{Distance}}_{\text{down}}}{{\text{Speed}}_{\text{down}}}\\ \\ {\text{Time}}_{\text{down}}& =\frac{44}{b+w}\phantom{\rule{0.167em}{0ex}}\text{hours}\end{array}$
But the boat took a total time of $10$ hours.
$\begin{array}{rl}{\text{Time}}_{\text{up}}+{\text{Time}}_{\text{down}}& =10\\ \\ \frac{30}{b-w}+\frac{44}{b+w}& =10\end{array}$
Try to make another equation using the second scenario.

## Framing second equation:

Let's look at the second scenario.
The boat goes $40\phantom{\rule{0.167em}{0ex}}\text{km}$ upstream and $55\phantom{\rule{0.167em}{0ex}}\text{km}$ downstream in $13$ hours.
When boat moves upstream,
$\begin{array}{rl}{\text{Time}}_{\text{up}}& =\frac{{\text{Distance}}_{\text{up}}}{{\text{Speed}}_{\text{up}}}\\ \\ {\text{Time}}_{\text{up}}& =\frac{40}{b-w}\phantom{\rule{0.167em}{0ex}}\text{hours}\end{array}$
When boat moves downstream,
$\begin{array}{rl}{\text{Time}}_{\text{down}}& =\frac{{\text{Distance}}_{\text{down}}}{{\text{Speed}}_{\text{down}}}\\ \\ {\text{Time}}_{\text{down}}& =\frac{55}{b+w}\phantom{\rule{0.167em}{0ex}}\text{hours}\end{array}$
But the boat took a total time of $13$ hours.
$\begin{array}{rl}{\text{Time}}_{\text{up}}+{\text{Time}}_{\text{down}}& =13\\ \\ \frac{40}{b-w}+\frac{55}{b+w}& =13\end{array}$
Now that we have both the equations, let's try to solve them.

## Converting to linear form:

Our equations are not linear. Our variables sit in the denominators right now.
But we're in luck as the terms $\frac{1}{b-w}$ and $\frac{1}{b+w}$ repeat in both the equations.
$\begin{array}{rl}\frac{30}{b-w}+\frac{44}{b+w}& =10\\ \\ \frac{40}{b-w}+\frac{55}{b+w}& =13\end{array}$
If we replace $\frac{1}{b-w}$ by $x$ and $\frac{1}{b+w}$ by $y$, we can get rid of things in the denominator.
$\begin{array}{rl}30x+44y& =10\\ \\ 40x+55y& =13\end{array}$
Now we're talking! We finally have two linear equations in two variables.
We can proceed from here using either the substitution method or the elimination method.

## Solving linear equations for new variables:

$\begin{array}{rl}30x+44y& =10\\ \\ 40x+55y& =13\end{array}$
We can eliminate $x$ if we multiply the first equation by $4$ and second equation by $3$, and then subtract one from another.
$\begin{array}{rl}4×\left(30x+44y& =10\right)\\ \\ 3×\left(40x+55y& =13\right)\end{array}$
This gives,
$\begin{array}{rl}120x+176y& =40\\ \\ 120x+165y& =39\end{array}$
Subtracting the second equation from the first,
$\begin{array}{rl}120x+176y& =40\\ \\ -\left(120x+165y& =39\right)\\ \\ 11y& =1\\ \\ y& =\frac{1}{11}\end{array}$
Plugging the value of $y$ in the first equation:
$\begin{array}{rl}30x+44×\frac{1}{11}& =10\\ \\ 30x+4& =10\\ \\ 30x& =6\\ \\ x& =\frac{6}{30}\\ \\ x& =\frac{1}{5}\end{array}$
Now that we have $x$ and $y$, we can find $b$ and $w$.

## Solving for initial variables:

Let's plug in the values of $x$ and $y$ in these equations:
$\begin{array}{rl}\frac{1}{b-w}& =x=\frac{1}{5}\\ \\ \frac{1}{b+w}& =y=\frac{1}{11}\end{array}$
This means that:
$\begin{array}{rl}b-w& =5\\ \\ b+w& =11\end{array}$
$\begin{array}{rl}2×b& =5+11\\ \\ 2×b& =16\\ \\ b& =8\end{array}$
Plugging back the value of $b$:
$\begin{array}{rl}b-w& =5\\ \\ 8-w& =5\\ \\ w& =3\end{array}$
Thus, $b$ or the speed of the $\text{boat in still water}$ is $8\phantom{\rule{0.167em}{0ex}}\frac{\text{km}}{\text{hr}}$.
And $w$ or the speed of the $\text{stream}$ is $3\phantom{\rule{0.167em}{0ex}}\frac{\text{km}}{\text{hr}}$.

## Summary:

To solve a pair of equations that are reducible to linear form:
• Find the expressions that repeat in both the equations. Give them a simpler form: say $x$ and $y$.
• Solve the new pair of linear equations for the new variables.
• Plug back the values of the new variables and solve for the initial variables.
• Celebrate.

## Want to join the conversation?

• ritu can row downstream 20km in 2 hours, upstream 4km in 2 hours.find her speed of rowing in still water and speed of current.