Equations Reducible to Linear Form

Equations Reducible to Linear Form" refers to equations that can be transformed or rewritten into a linear equation. These equations typically involve variables raised to powers other than 1, such as squared terms, cubed terms, or higher. By applying suitable substitutions or transformations, these equations can be simplified into linear equations, making them easier to solve using standard linear algebra techniques.

Standard Form of Two-Variable Linear Equation Pairs

A linear equation is a one-degree polynomial. It can have one or more than one variable in it. For example, 2x + 5, 5x + 10y ...etc are some examples of linear equation. Sometimes some equations are not linear in nature by default, but they can be brought into the linear form. 

For example: 

[Tex]\frac{7}{x} + \frac{3}{y} = 2[/Tex]

The equation given above is not a linear equation, but it can be brought into a linear form with some rearrangements and algebraic methods. Let's look at them in detail. 

Equations Reducible to Linear Form

Let's look at this procedure with an example. 

Suppose we take the equation given above [Tex]\frac{7}{x} + \frac{3}{y} = 2[/Tex]. This is equation is not linear in nature however if we decide to substitute 

[Tex]p = \frac{1}{x} \text{ and } q = \frac{1}{y}[/Tex]

Then this equation becomes, 

7p + 3q = 2

Now the solutions to this equation can be find out with “p” and “q” and then later we can get the real solutions with the previous relation, 

[Tex]p = \frac{1}{x} \text{ and } q = \frac{1}{y}[/Tex]

Determine the solution to the equation provided below:

Question 1: Reduce the following equation into the linear form. 

[Tex]\frac{2}{x - 3} + \frac{1}{y} = 1[/Tex]

Solution:

Like explained above, let's say.

[Tex]p = \frac{1}{x -3}       [/Tex] and [Tex]q = \frac{1}{y}[/Tex]

Substituting these values in the equation, we get

2p + q = 1

Now let's see how to solve a pair of such equations with an example, 

Question 2: Solve the given system of equations 

[Tex]\frac{2}{x - 4} + \frac{1}{y -1}  = 1 \\ \text{ and } \\ \frac{5}{x - 4} + \frac{2}{y-1} = 3[/Tex]

Solution: 

Let's say, 

p = [Tex]\frac{1}{x - 4}[/Tex] and q = [Tex] \frac{1}{y - 1}[/Tex]

Then both equations become, 

2p + q = 1  ......(1)

5p + 2q = 3 ........(2)

We now have to solve these equations, 

From equation (1), 

q = 1 - 2p

Plugging this value in the equation (2) 

5p + 2(1 - 2p) = 3 

⇒5p + 2 - 4p = 3 

⇒p + 2 = 3 

⇒p = 1 

Then q = -1. 

Now let's back the actual solutions by using the expressions we assumed in the beginning. 

p = [Tex]\frac{1}{x - 4}       [/Tex]  

⇒ 1 = [Tex]\frac{1}{x - 4}[/Tex]

⇒ x - 4 = 1 

⇒ x = 5

Similarly, 

q = [Tex] \frac{1}{y - 1}[/Tex]

⇒ -1 = [Tex] \frac{1}{y - 1}[/Tex]

⇒ -y + 1 = 1 

⇒ y = 0

Thus, the solution to this system of equation is x = 5 and y = 0. 

Word Problems

Question 1: A boat goes 60 Km upstream and 88Km downstream in 20 hours. In another trip, it went 80Km upstream and 110 Km downstream in 26 hours. Find out the speed of the stream and the speed of the boat in still water. 

Solution: 

Let's assume that speed of boat in still water is “x Km/h” and speed of stream is “y Km/h”. 

So, speed of boat in downstream = x + y 

And speed of boat in upstream = x - y

We know that, time = [Tex]\frac{distance}{speed}[/Tex]

So, for the first trip the equation will be, 

[Tex]20 = \frac{60}{x - y} + \frac{88}{x + y}       [/Tex] 

Similarly, the equation for the second trip will be, 

[Tex]26 = \frac{80}{x - y} + \frac{110}{x + y}[/Tex]

These equations now need to be reduced in a linear form, 

Let's say p = [Tex]\frac{1}{x - y}       [/Tex] and [Tex]q = \frac{1}{x + y} [/Tex]

Substituting this in the above equations, we get 

20 = 60p + 88q 

26 = 80p + 110q 

Solving this with cross multiplication method, 

[Tex]\frac{p}{88(-26) - (110)(-20)} = \frac{q}{80(-20)  - 60(-26)} = \frac{1}{(60(110) - (88)(80)} \\ \frac{p}{-22} = \frac{q}{-10} = \frac{1}{-110}[/Tex]

This gives us, 

p = [Tex]\frac{1}{5}      [/Tex] and q =[Tex] \frac{1}{11}[/Tex]

Now put these value of “p” and “q” in the equation, 

 [Tex]p = \frac{1}{x - y} \\ \frac{1}{5} = \frac{1}{x - y} \\ 5 = x - y      [/Tex] 

 [Tex]q = \frac{1}{x + y} \\ \frac{1}{11} = \frac{1}{x + y} \\ x + y = 11[/Tex]

To get the values of “x” and “y” we will solve these equations, 

x - y = 5 ..... (1)

x + y = 11 ..... (2)

Putting the value of x = y + 5 

y + 5 + y = 11 

⇒2y = 6 

⇒y = 3

So, x = 8

Thus, the solution is x = 8 and y = 3. 

Question 2: Distance between two cities is 300Km. A person has to travel from one to another using Bus and Train both. It takes 4 hours to travel if we go 60Km by train and the remaining distance is covered by bus. If 100km is travelled by train and the remaining by bus, it takes 10 minutes longer. Find the speed of the train and bus that's been running between two cities. 

Solution: 

Let the speed of bus be “x” and that of train be “y”. 

So total time taken, 

 [Tex]4 = \frac{60}{x} + \frac{240}{y}[/Tex]

[Tex]4 + \frac{1}{6} = \frac{100}{x} + \frac{200}{y}[/Tex]

Let's say [Tex]p = \frac{1}{x} \text{ and } q = \frac{1}{y} [/Tex]

Now the two equations become, 

4  = 60p + 240q 

[Tex]\frac{25}{6}      [/Tex] = 100p + 200q

Solving these equations, 

The solution comes out to be 

[Tex]p = \frac{1}{60} \text{ and } q = \frac{1}{80}[/Tex]

Computing the values of x and y from this, 

x = 60 and y = 80 

Thus, speed of the bus is 60Km/h and that of train is 80Km/h. 

Question 3: Ramesh's age is twice his son's age, Four years ago, his son's age was 20 years less than his father. Find their current age.

Solution: 

Let the age of Ramesh be x and the age of his son is y

The equations obtained are,

y= 2x ⇢ 1

y-4 = (x-4) +20 ⇢ 2

2x- 4 = x-4 + 20

x = 20 years

Question 4: Solve the following equation and obtain the value of p and q,

1/(2p) - 1/q= 2

1/p + 1/(2q)= 10

Solution: 

As it is clearly seen that the above two equations are not in the form of linear equation as the reciprocal of p and q are given.

Simplifying,

(1/2)(1/p) - 1/q = 2

1/p + (1/2)(1/q) = 10

Let 1/p= x and 1/q = y

Now, the equations shall look like,

1/2(x) - y = 2

x + 1/2(y) = 10

Simplifying further to find the values of x and y,

x- 2y = 4     (1)

2x+ y= 20        (2)

Multiplying equation 2 with 2 in order to equate both equations

x- 2y= 4

4x+ 2y= 40

5x = 44

x= 44/5

y= 12/5

Therefore, p= 1/x= 5/44

q= 1/y= 5/12

Question 5: There are 2 friends, Aditya and Aman, they both decided to buy some candies and chocolates, they both together spent Rs. 200 on buying items but Aman spent 4 more than half the amount Aditya spent. Calculate the precise amount spent by both the friends.

Solution: 

Let the amount spent by both Aditya and Aman be a and b respectively,

Now, they both collectively spent 200 rupees

a+ b= 200  (1)

Aman spent 4 rupees more than half the aditya's amount

a/2 +4 = b  (2)

Solving equation (2)

a+ 8= 2b

a= 2b-8

Putting the value of a in equation (1)

2b- 8 +b = 200

3b = 208

b= 208/3 = Rs. 69.7 ∼ Rs. 70

b= Rs. 70

Putting the value of b in equation (1)

a= Rs. 130

Related Articles:

• Linear Equations Formula
• Linear Equations
• Linear Equations in One Variable

Practice Problems - Equations Reducible to Linear Form

Problem 1: [Tex]\frac{x^2 + 3x}{x + 2} = 2[/Tex] Reduce this equation to a linear form and solve for x.

Problem 2: [Tex]\frac{2x^2 - 5x}{x - 1} = 3[/Tex] Transform this equation into a linear form and solve for x.

Problem 3: [Tex]\frac{3x - 2}{x + 4} = 1[/Tex] Reduce to a linear form and solve for x.

Problem 4: [Tex]\frac{4x^2 - 9}{x^2 - 1} = 2[/Tex] Simplify and solve the equation for x.

Problem 5: [Tex]\frac{5x + 7}{2x - 3} = 4[/Tex] Reduce this to a linear equation and find x.

Problem 6: [Tex]\frac{x^2 - 2}{x + 3} = 3[/Tex] Transform and solve for x.

Problem 7: [Tex]\frac{6x^2 - x - 2}{3x + 1} = 2[/Tex] Convert to a linear equation and solve.

Problem 8: [Tex]\frac{7x - 5}{x - 2} = 3[/Tex] Simplify to linear form and solve for x.

Problem 9: [Tex]\frac{2x^2 + x - 3}{x^2 - 4} = 1[/Tex] Reduce and solve for x.

Problem 10: [Tex]\frac{8x + 1}{x + 5} = 2[/Tex] Simplify this equation to linear form and solve.