Application of Linear Equations

 

 

We begin with a simple example.

Sum of two numbers is 74. One of the numbers is 10 more than the other. What are the numbers?

We have a puzzle here. We do not know either of the two numbers, and we have to find them. We are given two conditions.

  • One of the numbers is 10 more than the other.
  • Their sum is 74.

We already know from Class VII how to proceed. If the smaller number is taken to be x, the larger number is 10 more than x, i.e., x + 10. The other condition says that the sum of these two numbers x and x + 10 is 74.

This means that  x + (x + 10) = 74.

or    2x + 10 = 74

Transposing 10 to RHS,     2x =74 - 10

or   2x =64

Dividing both sides by 2,    x = 32. This is one number.

The other number is   x + 10 = 32 + 10 = 42

The desired numbers are 32 and 42. (Their sum is indeed 74 as given and also one number is 10 more than the other.)

We shall now consider several examples to show how useful this method is.

Example 5

What should be added to twice the rational number «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mo»-«/mo»«mn»7«/mn»«/mrow»«mn»3«/mn»«/mfrac»«/math»  to get «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»3«/mn»«mn»7«/mn»«/mfrac»«/math» ?

Solution

Twice the rational number «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mo»-«/mo»«mn»7«/mn»«/mrow»«mn»3«/mn»«/mfrac»«mo»§nbsp;«/mo»«mi»i«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mn»2«/mn»«mo»§#215;«/mo»«mo»§nbsp;«/mo»«mfenced»«mfrac»«mrow»«mo»-«/mo»«mn»7«/mn»«/mrow»«mn»3«/mn»«/mfrac»«/mfenced»«mo»=«/mo»«mfrac»«mrow»«mo»-«/mo»«mn»14«/mn»«/mrow»«mn»3«/mn»«/mfrac»«/math» . Suppose x added to this number gives «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»3«/mn»«mn»7«/mn»«/mfrac»«mo»;«/mo»«mo»§nbsp;«/mo»«mi»i«/mi»«mi»e«/mi»«mo».«/mo»«mo»,«/mo»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mo»§nbsp;«/mo»«mfenced»«mfrac»«mrow»«mo»-«/mo»«mn»14«/mn»«/mrow»«mn»3«/mn»«/mfrac»«/mfenced»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»3«/mn»«mn»7«/mn»«/mfrac»«/math»

or   «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»-«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»14«/mn»«mn»3«/mn»«/mfrac»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»3«/mn»«mn»7«/mn»«/mfrac»«/math»

 or    «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»3«/mn»«mn»7«/mn»«/mfrac»«mo»§nbsp;«/mo»«mo»+«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»14«/mn»«mn»3«/mn»«/mfrac»«mo»§nbsp;«/mo»«mo»(«/mo»«mi»t«/mi»«mi»r«/mi»«mi»a«/mi»«mi»n«/mi»«mi»s«/mi»«mi»p«/mi»«mi»o«/mi»«mi mathvariant=¨normal¨»sin«/mi»«mi»g«/mi»«mo»§nbsp;«/mo»«mfrac»«mn»14«/mn»«mn»3«/mn»«/mfrac»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»o«/mi»«mo»§nbsp;«/mo»«mi»R«/mi»«mi»H«/mi»«mi»S«/mi»«mo»)«/mo»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»=«/mo»«mfrac»«mrow»«mfenced»«mrow»«mn»3«/mn»«mo»§#215;«/mo»«mn»3«/mn»«/mrow»«/mfenced»«mo»+«/mo»«mfenced»«mrow»«mn»14«/mn»«mo»§#215;«/mo»«mn»7«/mn»«/mrow»«/mfenced»«/mrow»«mn»21«/mn»«/mfrac»«mo»§nbsp;«/mo»«mo»=«/mo»«mfrac»«mrow»«mn»9«/mn»«mo»+«/mo»«mn»98«/mn»«/mrow»«mn»21«/mn»«/mfrac»«mo»§nbsp;«/mo»«mo»=«/mo»«mfrac»«mn»107«/mn»«mn»21«/mn»«/mfrac»«/math»

 Thus «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»107«/mn»«mn»21«/mn»«/mfrac»«/math» should be added to «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mo»§#215;«/mo»«mfenced»«mfrac»«mrow»«mo»-«/mo»«mn»7«/mn»«/mrow»«mn»3«/mn»«/mfrac»«/mfenced»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»o«/mi»«mo»§nbsp;«/mo»«mi»g«/mi»«mi»i«/mi»«mi»v«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mfrac»«mn»3«/mn»«mn»7«/mn»«/mfrac»«/math»

Example 6

The perimeter of a rectangle is 13 cm and its width is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mfrac»«mn»3«/mn»«mn»4«/mn»«/mfrac»«mo»§nbsp;«/mo»«mi»c«/mi»«mi»m«/mi»«mo».«/mo»«mo»§nbsp;«/mo»«mi»F«/mi»«mi»i«/mi»«mi»n«/mi»«mi»d«/mi»«mo»§nbsp;«/mo»«mi»i«/mi»«mi»t«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»l«/mi»«mi»e«/mi»«mi»n«/mi»«mi»g«/mi»«mi»t«/mi»«mi»h«/mi»«/math»

Solution

Assume the length of the rectangle to be x cm.

The perimeter of the rectangle = 2 × (length + width)

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»2«/mn»«mo»§#215;«/mo»«mo»§nbsp;«/mo»«mfenced»«mrow»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mn»2«/mn»«mfrac»«mn»3«/mn»«mn»4«/mn»«/mfrac»«/mrow»«/mfenced»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»2«/mn»«mfenced»«mrow»«mi»x«/mi»«mo»+«/mo»«mfrac»«mn»11«/mn»«mn»4«/mn»«/mfrac»«/mrow»«/mfenced»«/math»

The perimeter is given to be 13 cm. Therefore,

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mfenced»«mrow»«mi»x«/mi»«mo»+«/mo»«mfrac»«mn»11«/mn»«mn»4«/mn»«/mfrac»«/mrow»«/mfenced»«mo»=«/mo»«mn»13«/mn»«/math»

or  «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»+«/mo»«mfrac»«mn»11«/mn»«mn»4«/mn»«/mfrac»«mo»=«/mo»«mfrac»«mn»13«/mn»«mn»2«/mn»«/mfrac»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»(«/mo»«mi»d«/mi»«mi»i«/mi»«mi»v«/mi»«mi»i«/mi»«mi»d«/mi»«mi»i«/mi»«mi»n«/mi»«mi»g«/mi»«mo»§nbsp;«/mo»«mi»b«/mi»«mi»o«/mi»«mi»t«/mi»«mi»h«/mi»«mo»§nbsp;«/mo»«mi»s«/mi»«mi»i«/mi»«mi»d«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mi»b«/mi»«mi»y«/mi»«mn»2«/mn»«mo»)«/mo»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»=«/mo»«mfrac»«mn»13«/mn»«mn»2«/mn»«/mfrac»«mo»-«/mo»«mfrac»«mn»11«/mn»«mn»4«/mn»«/mfrac»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»26«/mn»«mn»4«/mn»«/mfrac»«mo»§nbsp;«/mo»«mo»-«/mo»«mfrac»«mn»11«/mn»«mn»4«/mn»«/mfrac»«mo»§nbsp;«/mo»«mo»=«/mo»«mfrac»«mn»15«/mn»«mn»4«/mn»«/mfrac»«mo»=«/mo»«mn»3«/mn»«mfrac»«mn»3«/mn»«mn»4«/mn»«/mfrac»«/math»

The length of the rectangle is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»3«/mn»«mfrac»«mn»3«/mn»«mn»4«/mn»«/mfrac»«mo»§nbsp;«/mo»«mi»c«/mi»«mi»m«/mi»«/math»

Example 7

The present age of Sahil’s mother is three times the present age of Sahil. After 5 years their ages will add to 66 years. Find their present ages.

Solution

Let Sahil’s present age be x years.

Therefore,   4x + 10 = 66

This equation determines Sahil’s present age which is x years. To solve the equation,

we transpose 10 to RHS,

4x = 66 – 10

or  4x = 56

or   «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»56«/mn»«mn»4«/mn»«/mfrac»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»14«/mn»«/math»   (solution)

Thus, Sahil’s present age is 14 years and his mother’s age is 42 years. (You may easily check that 5 years from now the sum of their ages will be 66 years.)

Example 8

 Bansi has 3 times as many two-rupee coins as he has five-rupee coins. If he has in all a sum of Rs 77, how many coins of each denomination does he have?

Solution

Let the number of five-rupee coins that Bansi has be x. Then the number of two-rupee coins he has is 3 times x or 3x.

The amount Bansi has:
 

  • from 5 rupee coins, Rs 5 × x = Rs 5x
  • from 2 rupee coins, Rs 2 × 3x = Rs 6x

Hence the total money he has = Rs 11x

But this is given to be Rs 77; therefore,

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»11«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»77«/mn»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mfrac»«mn»77«/mn»«mn»11«/mn»«/mfrac»«mo»=«/mo»«mn»7«/mn»«/math»

Thus, number of five-rupee coins = x = 7

and number of two-rupee coins = 3x = 21       (solution)

(You can check that the total money with Bansi is Rs 77.)

Example 9

The sum of three consecutive multiples of 11 is 363. Find these multiples.

Solution

If x is a multiple of 11, the next multiple is x + 11. The next to this is x + 11 + 11 or x + 22. So we can take three consecutive multiples of 11 as x, x + 11 and x + 22.

It is given that the sum of these consecutive multiples of 11 is 363. This will give the following equation:

x + (x + 11) + (x + 22) = 363

or   x + x + 11 + x + 22 = 363

or  3x + 33 = 363

or  3x = 363 – 33

or  3x = 330

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»=«/mo»«mfrac»«mn»330«/mn»«mn»3«/mn»«/mfrac»«/math»

=110

Hence, the three consecutive multiples are 110, 121, 132 (answer).

Alternatively, we may think of the multiple of 11 immediately before x. This is (x – 11). Therefore, we may take three consecutive
multiples of 11 as x – 11, x, x + 11.

In this case we arrive at the equation

(x – 11) + x + (x + 11) = 363

or  3x = 363

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»=«/mo»«mfrac»«mn»363«/mn»«mn»3«/mn»«/mfrac»«mo»=«/mo»«mn»121«/mn»«mo».«/mo»«/math»  Therefore,

x = 121, x – 11 = 110, x + 11 = 132

Hence, the three consecutive multiples are 110, 121, 132.

We can see that we can adopt different ways to find a solution for the problem.

Example 10

The difference between two whole numbers is 66. The ratio of the two numbers is 2 : 5. What are the two numbers?

Solution

Since the ratio of the two numbers is 2 : 5, we may take one number to be 2x and the other to be 5x. (Note that 2x : 5x is same as 2 : 5.)

The difference between the two numbers is (5x – 2x). It is given that the difference is 66. Therefore,

5x – 2x = 66

or  3x = 66

or  x = 22

Since the numbers are 2x and 5x, they are 2 × 22 or 44 and 5 × 22 or 110, respectively.

The difference between the two numbers is 110 – 44 = 66 as desired.

Example 11

Deveshi has a total of Rs 590 as currency notes in the denominations of Rs 50, Rs 20 and Rs 10. The ratio of the number of Rs 50 notes and Rs 20 notes is 3:5. If she has a total of 25 notes, how many notes of each denomination she has?

Solution

Let the number of Rs 50 notes and Rs 20 notes be 3x and 5x, respectively. But she has 25 notes in total.

Therefore, the number of Rs 10 notes = 25 – (3x + 5x) = 25 – 8x

The amount she has

from Rs 50 notes : 3x × 50 = Rs 150x

from Rs 20 notes : 5x × 20 = Rs 100x

from Rs 10 notes : (25 – 8x) × 10 = Rs (250 – 80x)

Hence the total money she has =150x + 100x + (250 – 80x) = Rs (170x + 250)

But she has Rs 590. Therefore, 170x + 250 = 590

or  170x = 590 – 250 = 340

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mfrac»«mn»340«/mn»«mn»170«/mn»«/mfrac»«mo»=«/mo»«mn»2«/mn»«/math»

The number of Rs 50 notes she has = 3x

= 3 × 2 = 6

The number of Rs 20 notes she has = 5x = 5 × 2 = 10

The number of Rs 10 notes she has = 25 – 8x

= 25 – (8 × 2) = 25 – 16 = 9

Solving Equations having the Variable on both Sides

An equation is the equality of the values of two expressions. In the equation 2x – 3 = 7, the two expressions are 2x – 3 and 7. In most examples that we have come across so far, the RHS is just a number. But this need not always be so; both sides could have expressions with variables. For example, the equation 2x – 3 = x + 2 has expressions with a variable on both sides; the expression on the LHS is (2x – 3) and the expression on the RHS is (x + 2).

  • We now discuss how to solve such equations which have expressions with the variable on both sides.

Example 12

Solve 2x – 3 = x + 2

Solution

We have

2x = x + 2 + 3

or  2x = x + 5

or  2x – x = x + 5 – x (subtracting x from both sides)

or  x = 5 (solution)

Here we subtracted from both sides of the equation, not a number (constant), but a term involving the variable. We can do this as variables are also numbers. Also, note that subtracting x from both sides amounts to transposing x to LHS.

Example 13

Solve «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»5«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mfrac»«mn»7«/mn»«mn»2«/mn»«/mfrac»«mo»=«/mo»«mfrac»«mn»3«/mn»«mn»2«/mn»«/mfrac»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»-«/mo»«mn»14«/mn»«/math»

Solution

Multiply both sides of the equation by 2. We get

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mo»§nbsp;«/mo»«mo»§#215;«/mo»«mo»§nbsp;«/mo»«mfenced»«mrow»«mn»5«/mn»«mi»x«/mi»«mo»+«/mo»«mfrac»«mn»7«/mn»«mn»2«/mn»«/mfrac»«/mrow»«/mfenced»«mo»=«/mo»«mn»2«/mn»«mo»§#215;«/mo»«mfenced»«mrow»«mfrac»«mn»3«/mn»«mn»2«/mn»«/mfrac»«mi»x«/mi»«mo»-«/mo»«mn»14«/mn»«/mrow»«/mfenced»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfenced»«mrow»«mn»2«/mn»«mo»§nbsp;«/mo»«mo»§#215;«/mo»«mo»§nbsp;«/mo»«mn»5«/mn»«mi»x«/mi»«/mrow»«/mfenced»«mo»+«/mo»«mfenced»«mrow»«mn»2«/mn»«mi»x«/mi»«mfrac»«mn»7«/mn»«mn»2«/mn»«/mfrac»«/mrow»«/mfenced»«mo»=«/mo»«mfenced»«mrow»«mn»2«/mn»«mo»§#215;«/mo»«mfrac»«mn»3«/mn»«mn»2«/mn»«/mfrac»«mi»x«/mi»«/mrow»«/mfenced»«mo»-«/mo»«mfenced»«mrow»«mn»2«/mn»«mo»§#215;«/mo»«mn»14«/mn»«/mrow»«/mfenced»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»10«/mn»«mi»x«/mi»«mo»+«/mo»«mn»7«/mn»«mo»=«/mo»«mn»3«/mn»«mi»x«/mi»«mo»-«/mo»«mn»28«/mn»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»10«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»-«/mo»«mo»§nbsp;«/mo»«mn»3«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mn»7«/mn»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»-«/mo»«mn»28«/mn»«/math»     (transposing 3x to LHS)

or  «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»7«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mo»§nbsp;«/mo»«mn»7«/mn»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mo»-«/mo»«mn»28«/mn»«/math»

or  «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»7«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»-«/mo»«mn»28«/mn»«mo»-«/mo»«mn»7«/mn»«/math»

or  «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»7«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mo»-«/mo»«mn»35«/mn»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mfrac»«mrow»«mo»-«/mo»«mn»35«/mn»«/mrow»«mn»7«/mn»«/mfrac»«/math»

or  «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»-«/mo»«mn»5«/mn»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»(«/mo»«mi»s«/mi»«mi»o«/mi»«mi»l«/mi»«mi»u«/mi»«mi»t«/mi»«mi»i«/mi»«mi»o«/mi»«mi»n«/mi»«mo»)«/mo»«/math»

Some More Applications

Example 14

The digits of a two-digit number differ by 3. If the digits are interchanged,and the resulting number is added to the original number, we get 143. What can be the original number?

Solution

Take, for example, a two-digit number, say, 56. It can be written as 56 = (10 × 5) + 6.

If the digits in 56 are interchanged, we get 65, which can be written as (10 × 6 ) + 5.

Let us take the two digit number such that the digit in the units place is b. The digit in the tens place differs from b by 3. Let us take it as b + 3. So the two-digit number is 10 (b + 3) + b = 10b + 30 + b = 11b + 30.

With interchange of digits, the resulting two-digit number will be

10b + (b + 3) = 11b + 3

If we add these two two-digit numbers, their sum is

(11b + 30) + (11b + 3) = 11b + 11b + 30 + 3 = 22b + 33

It is given that the sum is 143. Therefore, 22b + 33 = 143

or  22b = 143 – 33

or  22b = 110

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»b«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mfrac»«mn»110«/mn»«mn»22«/mn»«/mfrac»«/math»

or  b = 5

The units digit is 5 and therefore the tens digit is 5 + 3 which is 8. The number is 85.

Check: On interchange of digits the number we get is 58. The sum of 85 and 58 is 143 as given.

Example 15

Arjun is twice as old as Shriya. Five years ago his age was three times Shriya’s age. Find their present ages.

Solution

Let us take Shriya’s present age to be x years.

Then Arjun’s present age would be 2x years.

Shriya’s age five years ago was (x – 5) years.

Arjun’s age five years ago was (2x – 5) years.

It is given that Arjun’s age five years ago was three times Shriya’s age.

Thus, 2x – 5 = 3(x – 5)

or  2x – 5 = 3x – 15

or  15 – 5 = 3x – 2x

or  10 = x

So, Shriya’s present age = x = 10 years.

Therefore, Arjun’s present age = 2x = 2 × 10 = 20 years.

Reducing Equations to Simpler Form

Example 16

Solve  «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mn»6«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mn»1«/mn»«/mrow»«mn»3«/mn»«/mfrac»«mo»+«/mo»«mn»1«/mn»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mrow»«mi»x«/mi»«mo»-«/mo»«mn»3«/mn»«/mrow»«mn»6«/mn»«/mfrac»«/math»

Solution

Multiplying both sides of the equation by 6,

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mn»6«/mn»«mfenced»«mrow»«mn»6«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mn»1«/mn»«/mrow»«/mfenced»«/mrow»«mn»3«/mn»«/mfrac»«mo»+«/mo»«mn»6«/mn»«mo»§nbsp;«/mo»«mo»§#215;«/mo»«mn»1«/mn»«mo»=«/mo»«mfrac»«mrow»«mn»6«/mn»«mfenced»«mrow»«mi»x«/mi»«mo»-«/mo»«mn»3«/mn»«/mrow»«/mfenced»«/mrow»«mn»6«/mn»«/mfrac»«/math»

or    «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mfenced»«mrow»«mn»6«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mn»1«/mn»«/mrow»«/mfenced»«mo»+«/mo»«mn»6«/mn»«mo»§nbsp;«/mo»«mo»=«/mo»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»-«/mo»«mn»3«/mn»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»12«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mn»2«/mn»«mo»+«/mo»«mn»6«/mn»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»-«/mo»«mn»3«/mn»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»(«/mo»«mi»o«/mi»«mi»p«/mi»«mi»e«/mi»«mi»n«/mi»«mi»i«/mi»«mi»n«/mi»«mi»g«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»h«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mi»b«/mi»«mi»r«/mi»«mi»a«/mi»«mi»c«/mi»«mi»k«/mi»«mi»e«/mi»«mi»t«/mi»«mi»s«/mi»«mo»)«/mo»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»12«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mn»8«/mn»«mo»§nbsp;«/mo»«mo»=«/mo»«mi»x«/mi»«mo»-«/mo»«mn»3«/mn»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»12«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»-«/mo»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mn»8«/mn»«mo»=«/mo»«mo»-«/mo»«mn»3«/mn»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»11«/mn»«mi»x«/mi»«mo»+«/mo»«mn»8«/mn»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»-«/mo»«mn»3«/mn»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»11«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»-«/mo»«mn»3«/mn»«mo»-«/mo»«mn»8«/mn»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»11«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mo»-«/mo»«mn»11«/mn»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mo»-«/mo»«mn»1«/mn»«mo»§nbsp;«/mo»«mo»(«/mo»«mi»r«/mi»«mi»e«/mi»«mi»q«/mi»«mi»u«/mi»«mi»i«/mi»«mi»r«/mi»«mi»e«/mi»«mi»d«/mi»«mo»§nbsp;«/mo»«mi»s«/mi»«mi»o«/mi»«mi»l«/mi»«mi»u«/mi»«mi»t«/mi»«mi»i«/mi»«mi»o«/mi»«mi»n«/mi»«mo»)«/mo»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»c«/mi»«mi»h«/mi»«mi»e«/mi»«mi»c«/mi»«mi»k«/mi»«mo»:«/mo»«mo»§nbsp;«/mo»«mi»L«/mi»«mi»H«/mi»«mi»S«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mrow»«mn»6«/mn»«mfenced»«mrow»«mo»-«/mo»«mn»1«/mn»«/mrow»«/mfenced»«mo»+«/mo»«mn»1«/mn»«/mrow»«mn»3«/mn»«/mfrac»«mo»+«/mo»«mn»1«/mn»«mo»§nbsp;«/mo»«mo»=«/mo»«mfrac»«mrow»«mo»-«/mo»«mn»6«/mn»«mo»+«/mo»«mn»1«/mn»«/mrow»«mn»3«/mn»«/mfrac»«mo»+«/mo»«mn»1«/mn»«mo»=«/mo»«mfrac»«mrow»«mo»-«/mo»«mn»5«/mn»«/mrow»«mn»3«/mn»«/mfrac»«mo»+«/mo»«mfrac»«mn»3«/mn»«mn»3«/mn»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«mo»-«/mo»«mn»5«/mn»«mo»+«/mo»«mn»3«/mn»«/mrow»«mn»3«/mn»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«mo»-«/mo»«mn»2«/mn»«/mrow»«mn»3«/mn»«/mfrac»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»R«/mi»«mi»H«/mi»«mi»S«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mrow»«mfenced»«mrow»«mo»-«/mo»«mn»1«/mn»«/mrow»«/mfenced»«mo»-«/mo»«mn»3«/mn»«/mrow»«mn»6«/mn»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«mo»-«/mo»«mn»4«/mn»«/mrow»«mn»6«/mn»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«mo»-«/mo»«mn»2«/mn»«/mrow»«mn»3«/mn»«/mfrac»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»L«/mi»«mi»H«/mi»«mi»S«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mi»R«/mi»«mi»H«/mi»«mi»S«/mi»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»(«/mo»«mi»a«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»r«/mi»«mi»e«/mi»«mi»q«/mi»«mi»u«/mi»«mi»i«/mi»«mi»r«/mi»«mi»e«/mi»«mi»d«/mi»«mo»)«/mo»«/math»

Example 17

Solve  «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»5«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»-«/mo»«mn»2«/mn»«mo»§nbsp;«/mo»«mfenced»«mrow»«mn»2«/mn»«mi»x«/mi»«mo»-«/mo»«mn»7«/mn»«/mrow»«/mfenced»«mo»=«/mo»«mn»2«/mn»«mo»(«/mo»«mn»3«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»-«/mo»«mn»1«/mn»«mo»)«/mo»«mo»+«/mo»«mfrac»«mn»7«/mn»«mn»2«/mn»«/mfrac»«/math»

Solution

Let us open the brackets,

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»L«/mi»«mi»H«/mi»«mi»S«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»5«/mn»«mi»x«/mi»«mo»-«/mo»«mn»4«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mn»14«/mn»«mo»§nbsp;«/mo»«mo»=«/mo»«mi»x«/mi»«mo»+«/mo»«mn»14«/mn»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»R«/mi»«mi»H«/mi»«mi»S«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»6«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»-«/mo»«mn»2«/mn»«mo»+«/mo»«mfrac»«mn»7«/mn»«mn»2«/mn»«/mfrac»«mo»=«/mo»«mn»6«/mn»«mi»x«/mi»«mo»-«/mo»«mfrac»«mn»4«/mn»«mn»2«/mn»«/mfrac»«mo»+«/mo»«mfrac»«mn»7«/mn»«mn»2«/mn»«/mfrac»«mo»=«/mo»«mn»6«/mn»«mi»x«/mi»«mo»+«/mo»«mfrac»«mn»3«/mn»«mn»2«/mn»«/mfrac»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»T«/mi»«mi»h«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mi»e«/mi»«mi»q«/mi»«mi»u«/mi»«mi»a«/mi»«mi»t«/mi»«mi»i«/mi»«mi»o«/mi»«mi»n«/mi»«mo»§nbsp;«/mo»«mi»i«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»x«/mi»«mo»+«/mo»«mn»14«/mn»«mo»§nbsp;«/mo»«mo»=«/mo»«mn»6«/mn»«mi»x«/mi»«mo»+«/mo»«mfrac»«mn»3«/mn»«mn»2«/mn»«/mfrac»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»14«/mn»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»6«/mn»«mi»x«/mi»«mo»-«/mo»«mi»x«/mi»«mo»+«/mo»«mfrac»«mn»3«/mn»«mrow»«mn»2«/mn»«mo»§nbsp;«/mo»«/mrow»«/mfrac»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»14«/mn»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»5«/mn»«mi»x«/mi»«mo»+«/mo»«mfrac»«mn»3«/mn»«mn»2«/mn»«/mfrac»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»14«/mn»«mo»§nbsp;«/mo»«mo»-«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»3«/mn»«mn»2«/mn»«/mfrac»«mo»=«/mo»«mn»5«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»(«/mo»«mi»t«/mi»«mi»r«/mi»«mi»a«/mi»«mi»n«/mi»«mi»s«/mi»«mi»p«/mi»«mi»o«/mi»«mi mathvariant=¨normal¨»sin«/mi»«mi»g«/mi»«mfrac»«mn»3«/mn»«mn»2«/mn»«/mfrac»«mo»)«/mo»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mn»28«/mn»«mo»-«/mo»«mn»3«/mn»«/mrow»«mn»2«/mn»«/mfrac»«mo»=«/mo»«mn»5«/mn»«mi»x«/mi»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»25«/mn»«mn»2«/mn»«/mfrac»«mo»=«/mo»«mn»5«/mn»«mi»x«/mi»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»=«/mo»«mfrac»«mn»25«/mn»«mn»2«/mn»«/mfrac»«mo»§#215;«/mo»«mfrac»«mn»1«/mn»«mn»5«/mn»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«mn»5«/mn»«mo»§#215;«/mo»«mn»5«/mn»«/mrow»«mrow»«mn»2«/mn»«mo»§#215;«/mo»«mn»5«/mn»«/mrow»«/mfrac»«mo»=«/mo»«mfrac»«mn»5«/mn»«mn»2«/mn»«/mfrac»«/math»

Therefore, required solution is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»=«/mo»«mfrac»«mn»5«/mn»«mn»2«/mn»«/mfrac»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»c«/mi»«mi»h«/mi»«mi»e«/mi»«mi»c«/mi»«mi»k«/mi»«mo»:«/mo»«mo»§nbsp;«/mo»«mi»L«/mi»«mi»H«/mi»«mi»S«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»5«/mn»«mo»§#215;«/mo»«mfrac»«mn»5«/mn»«mn»2«/mn»«/mfrac»«mo»-«/mo»«mn»2«/mn»«mfenced»«mrow»«mfrac»«mn»5«/mn»«mn»2«/mn»«/mfrac»«mo»§#215;«/mo»«mn»2«/mn»«mo»-«/mo»«mn»7«/mn»«/mrow»«/mfenced»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»=«/mo»«mfrac»«mn»25«/mn»«mn»2«/mn»«/mfrac»«mo»-«/mo»«mn»2«/mn»«mfenced»«mrow»«mn»5«/mn»«mo»-«/mo»«mn»7«/mn»«/mrow»«/mfenced»«mo»=«/mo»«mfrac»«mn»25«/mn»«mn»2«/mn»«/mfrac»«mo»-«/mo»«mn»2«/mn»«mfenced»«mrow»«mo»-«/mo»«mn»2«/mn»«/mrow»«/mfenced»«mo»=«/mo»«mfrac»«mn»25«/mn»«mn»2«/mn»«/mfrac»«mo»+«/mo»«mn»4«/mn»«mo»=«/mo»«mfrac»«mrow»«mn»25«/mn»«mo»+«/mo»«mn»8«/mn»«/mrow»«mn»2«/mn»«/mfrac»«mo»=«/mo»«mfrac»«mn»33«/mn»«mn»2«/mn»«/mfrac»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»R«/mi»«mi»H«/mi»«mi»S«/mi»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»2«/mn»«mo»§nbsp;«/mo»«mfenced»«mrow»«mfrac»«mn»5«/mn»«mn»2«/mn»«/mfrac»«mo»§#215;«/mo»«mn»3«/mn»«mo»-«/mo»«mn»1«/mn»«/mrow»«/mfenced»«mo»+«/mo»«mfrac»«mn»7«/mn»«mn»2«/mn»«/mfrac»«mo»=«/mo»«mn»2«/mn»«mfenced»«mrow»«mfrac»«mn»15«/mn»«mn»2«/mn»«/mfrac»«mo»-«/mo»«mfrac»«mn»2«/mn»«mn»2«/mn»«/mfrac»«/mrow»«/mfenced»«mo»+«/mo»«mfrac»«mn»7«/mn»«mn»2«/mn»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«mn»2«/mn»«mo»§#215;«/mo»«mn»13«/mn»«/mrow»«mn»2«/mn»«/mfrac»«mo»+«/mo»«mfrac»«mn»7«/mn»«mn»2«/mn»«/mfrac»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»=«/mo»«mfrac»«mrow»«mn»26«/mn»«mo»+«/mo»«mn»7«/mn»«/mrow»«mn»2«/mn»«/mfrac»«mo»=«/mo»«mfrac»«mn»33«/mn»«mn»2«/mn»«/mfrac»«mo»=«/mo»«mi»L«/mi»«mi»H«/mi»«mi»S«/mi»«mo».«/mo»«mo»§nbsp;«/mo»«mo»(«/mo»«mi»a«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»r«/mi»«mi»e«/mi»«mi»q«/mi»«mi»u«/mi»«mi»i«/mi»«mi»r«/mi»«mi»e«/mi»«mi»d«/mi»«mo»)«/mo»«/math»

Equations Reducible to the Linear Form

Example 18

Solve «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»x«/mi»«mo»+«/mo»«mn»1«/mn»«/mrow»«mrow»«mn»2«/mn»«mi»x«/mi»«mo»+«/mo»«mn»3«/mn»«/mrow»«/mfrac»«mo»=«/mo»«mfrac»«mn»3«/mn»«mn»8«/mn»«/mfrac»«/math»

Solution

Observe that the equation is not a linear equation, since the expression on its LHS is not linear. But we can put it into the form of a linear equation. We multiply both sides of the equation by (2x + 3),

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfenced»«mfrac»«mrow»«mi»x«/mi»«mo»+«/mo»«mn»1«/mn»«/mrow»«mrow»«mn»2«/mn»«mi»x«/mi»«mo»+«/mo»«mn»3«/mn»«/mrow»«/mfrac»«/mfenced»«mo»§#215;«/mo»«mfenced»«mrow»«mn»2«/mn»«mi»x«/mi»«mo»+«/mo»«mn»3«/mn»«/mrow»«/mfenced»«mo»=«/mo»«mfrac»«mn»3«/mn»«mn»8«/mn»«/mfrac»«mo»§#215;«/mo»«mfenced»«mrow»«mn»2«/mn»«mi»x«/mi»«mo»+«/mo»«mn»3«/mn»«/mrow»«/mfenced»«/math»

Notice that «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfenced»«mrow»«mn»2«/mn»«mi»x«/mi»«mo»+«/mo»«mn»3«/mn»«/mrow»«/mfenced»«/math» gets cancelled on the LHS We have then,

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»+«/mo»«mn»1«/mn»«mo»§nbsp;«/mo»«mo»=«/mo»«mfrac»«mrow»«mn»3«/mn»«mfenced»«mrow»«mn»2«/mn»«mi»x«/mi»«mo»+«/mo»«mn»3«/mn»«/mrow»«/mfenced»«/mrow»«mn»8«/mn»«/mfrac»«/math»

We have now a linear equation which we know how to solve.Multiplying both sides by 8

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»8«/mn»«mfenced»«mrow»«mi»x«/mi»«mo»+«/mo»«mn»1«/mn»«mo»)«/mo»«/mrow»«/mfenced»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»3«/mn»«mfenced»«mrow»«mn»2«/mn»«mi»x«/mi»«mo»+«/mo»«mn»3«/mn»«/mrow»«/mfenced»«/math»

or  «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»8«/mn»«mi»x«/mi»«mo»+«/mo»«mn»8«/mn»«mo»=«/mo»«mn»6«/mn»«mi»x«/mi»«mo»+«/mo»«mn»9«/mn»«/math»

or  «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»8«/mn»«mi»x«/mi»«mo»=«/mo»«mn»6«/mn»«mi»x«/mi»«mo»+«/mo»«mn»9«/mn»«mo»-«/mo»«mn»8«/mn»«/math»

or  «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»8«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»6«/mn»«mi»x«/mi»«mo»+«/mo»«mn»1«/mn»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»8«/mn»«mi»x«/mi»«mo»-«/mo»«mn»6«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mn»1«/mn»«/math»

or «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mn»1«/mn»«/math»

or  «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»T«/mi»«mi»h«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mi»s«/mi»«mi»o«/mi»«mi»l«/mi»«mi»u«/mi»«mi»t«/mi»«mi»i«/mi»«mi»o«/mi»«mi»n«/mi»«mo»§nbsp;«/mo»«mi»i«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«mi»h«/mi»«mi»e«/mi»«mi»c«/mi»«mi»k«/mi»«mo»:«/mo»«mo»§nbsp;«/mo»«mi»N«/mi»«mi»u«/mi»«mi»m«/mi»«mi»e«/mi»«mi»r«/mi»«mi»a«/mi»«mi»t«/mi»«mi»o«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mi»O«/mi»«mi»f«/mi»«mo»§nbsp;«/mo»«mi»L«/mi»«mi»H«/mi»«mi»S«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mo»+«/mo»«mn»1«/mn»«mo»=«/mo»«mfrac»«mrow»«mn»1«/mn»«mo»+«/mo»«mn»2«/mn»«/mrow»«mn»2«/mn»«/mfrac»«mo»=«/mo»«mfrac»«mn»3«/mn»«mn»2«/mn»«/mfrac»«/math»

 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»D«/mi»«mi»e«/mi»«mi»n«/mi»«mi»o«/mi»«mi»m«/mi»«mi»i«/mi»«mi»n«/mi»«mi»a«/mi»«mi»t«/mi»«mi»o«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mi»o«/mi»«mi»f«/mi»«mo»§nbsp;«/mo»«mi»L«/mi»«mi»H«/mi»«mi»S«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»2«/mn»«mi»x«/mi»«mo»+«/mo»«mn»3«/mn»«mo»=«/mo»«mn»2«/mn»«mo»§nbsp;«/mo»«mo»§#215;«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mo»+«/mo»«mn»3«/mn»«mo»=«/mo»«mn»1«/mn»«mo»+«/mo»«mn»3«/mn»«mo»=«/mo»«mn»4«/mn»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»L«/mi»«mi»H«/mi»«mi»S«/mi»«mo»=«/mo»«mo»§nbsp;«/mo»«mi»n«/mi»«mi»u«/mi»«mi»m«/mi»«mi»e«/mi»«mi»r«/mi»«mi»a«/mi»«mi»t«/mi»«mi»o«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mo»§#247;«/mo»«mo»§nbsp;«/mo»«mi»d«/mi»«mi»e«/mi»«mi»n«/mi»«mi»o«/mi»«mi»m«/mi»«mi»i«/mi»«mi»n«/mi»«mi»a«/mi»«mi»t«/mi»«mi»o«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»3«/mn»«mn»2«/mn»«/mfrac»«mo»§#247;«/mo»«mn»4«/mn»«mo»=«/mo»«mfrac»«mn»3«/mn»«mn»2«/mn»«/mfrac»«mo»§#215;«/mo»«mfrac»«mn»1«/mn»«mn»4«/mn»«/mfrac»«mo»=«/mo»«mfrac»«mn»3«/mn»«mn»8«/mn»«/mfrac»«/math»

 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»L«/mi»«mi»H«/mi»«mi»S«/mi»«mo»=«/mo»«mo»§nbsp;«/mo»«mi»R«/mi»«mi»H«/mi»«mi»S«/mi»«/math»

 Example 19

Present ages of Anu and Raj are in the ratio 4:5. Eight years from now the ratio of their ages will be 5:6. Find their present ages.

Solution

Let the present ages of Anu and Raj be 4x years and 5x years respectively.

After eight years. Anu’s age = (4x + 8) years;

After eight years, Raj’s age = (5x + 8) years.

Therefore, the ratio of their ages after eight years = «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mn»4«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mn»8«/mn»«/mrow»«mrow»«mn»5«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mn»8«/mn»«/mrow»«/mfrac»«/math»

This is given to be 5 : 6

Therefore,        «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mn»4«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mn»8«/mn»«/mrow»«mrow»«mn»5«/mn»«mi»x«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mn»8«/mn»«/mrow»«/mfrac»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»5«/mn»«mn»6«/mn»«/mfrac»«/math»

Cross-multiplication gives 6 (4x + 8) = 5 (5x + 8)

or   24x + 48 = 25x + 40

or   24x + 48 – 40 = 25x

or  8 = 25x – 24x

or   8 = x

Therefore, Anu’s present age = 4x = 4 × 8 = 32 years

Raj’s present age = 5x = 5 × 8 = 40 years

 

 

 

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