Linear equation Part 2

  
                                       
To reduce a given equation to linear form, you need to cross- multiply its terms on both sides of equality. After
cross-multiplication, you will get an equation involving variables having power more than one. But on simplifying, the terms with power more than one like x², cancel out and you are left with a linear equation which you can solve in the usual manner. Observe the following example to learn how to reduce a given equation to a linear form and then solve it.

Example: Solve the following equation: 2x - 5/8x + 3 = x - 5/4x


Solution: Cross multiplying the equation, we get
               2x - 5/8x + 3 = x - 5/ 4x
                                         (2x - 5)(4x) = (x - 5)(8x + 3)
                                              8x - 20x = 8x + 3x - 40x - 15
                                             8x -  20x = 8x - 37x - 15


Transposing the terms, we get


                                        8x² - 8²x - 2x + 37x = - 15
or         17x = -15
or            x  = - 15/17


Note: Sometimes due to a mistake in calculation, terms like x² do not get cancelled out and you are left with an
equation which is not a linear equation. In such cases, recheck all the steps to fing a error, since all the terms involving x² have to cancel out


Example 2: Find a positive value of tthe variable x for which the follwong equations is satisfied.
x² + 13/3x² + 10 = 1/2


Solution: Let us cross multiplying the equation, we get


2(x² + 13) = 1(3x²+ 10)
2x² + 26   = 3x² + 10
Transposing the terms, we get
3x² -2x² = 26 - 10
       x²  = 16
It implies x has two values.
x = + 4 or - 4
since only the positive value of x is required


So, x = + 4


               
Application of Linear Equations in solving Problems





Many problems in almost all topics of Math's involve relationships among known quantities ( numbers or
constant) and unknown quantities (variables). These problems are often stated in words and hence also known as word problems. To solve a word problem, first a suitable equation containing both variables and constants is framed. Then, a suitable solution to this equation is found that satisfies the given word problem.


You need to follow the steps mentioned below to solve a word problem.


Step 1: Read the problem carefully and note down what is given and what is required to be found out.


Step 2: Denote the unknown quantity by some English letter like x, y, z, a, b, c, p, q, r etc.


Step 3: Translate the statements of the problems into mathematical statements.


Step 4: Understand the relationship between variables and constants, and accordingly frame an
equation containing variables and constants.


Step 5: Solve the equation using any one of the methods discussed in this chapter so far.


Step 6: Check whether the solution obtained in Step 5 satisfies the conditions of the word problem.


Checking Solution: The solution of a linear equation may be rational number (negative, positive, fractions etc.) But the solution of an equation, framed for a word problem, has certain limitations depending upon the nature of a variable, say x, whose value you are trying to find to satisfy the given conditions.


If x is number of persons : - It cannot be negative nor a fraction; it has to be whole number.


If x is the age of a person :- it can not be negative


If x is length - it can not be negative.


If x is a digit - It has to be a digit form 0 to 9, it can neither be negative nor a fraction.


Now, observe the following examples to learn how to solve word problems by framing and solving equations


Example: The sum of three consecutive multiples of 5 of 555. Find the numbers.


Solution: There  is need to assume a variable for the unknown quantity. Here we have three unknows but
they are related to each other.


If we assume that first multiple of 5 to be x, th other tow multiple can be written as
x + 5 and x + 10 because they are consecutive multiples of 5 as 555.
Therefore                      x + (x + 5) + (x +10) = 555
⇒                                                     3x + 15 = 555
⇒                                                              3x = 555 – 15                       (transposing 15)
 ⇒                                                                 = 540
⇒                                                              x  = 540/3
⇒                                                               x = 180

Hence, the required multiples of 5 are 180, 185 and 190




Check:-  Adding these three numbers, we get 180 + 185 + 190 = 555.