2.7 Linear Equations:
Is it not interesting to
solve problems similar to?
1. If
sum of three consecutive even numbers is 252, which are they? (80,82,84 ?
70,72,74 ? . . . )
2. If
the length of a rectangle is 4cms more than breadth and the perimeter is 11cms
more than breadth. What is the length and breadth of the rectangle?
3. A
traveler on a pilgrimage
spends half the amount in Prayaga,
2/9 th of the remaining amount in Kashi, 1/4 th of the
remaining amount towards taxes, 6/10 th of the
remaining amount in
Definition : Equation
is statement of equality of two algebraic expressions involving one or more
unknown quantities (called Variables).
The expression on Left Hand
Side is called (LHS) and expression on Right Hand Side is called RHS.
Observations:
Is it not true that 6=6 ? ======è
(1)
Here Left hand side (LHS)
is 6 and Right hand side is also 6 and both sides are equal
We thus say LHS=RHS (as
both sides are equal)
Let us add the same number
2 to both sides of (1)
We get LHS =6+2=8 and also RHS = 6+2
=8
We still notice that LHS=
RHS (as both sides are equal)
Let us subtract the same
number 3 from both sides of (1)
We get LHS= 6-3 =3 and RHS
= 6-3 =3
We still notice that
LHS=RHS (as both sides are equal)
Let us multiply both sides
of (1) by the same number 6
We get LHS= 6*6=36 and RHS=
6*6 =36
We still notice that
LHS=RHS (both sides are equal)
Let us divide both sides of (1) by the same number 3
We get LHS = 6/3=2 and RHS
= 6/3=2
We still notice that
LHS=RHS (both sides are equal)
Properties of Equality
(Axioms)
1. Equality does not change
when we add same quantity to both sides
2 Equality does not
change when we subtract same quantity from both sides
3. Equality does not change
when we multiply both sides by same quantity
4. Equality does not change
when we divide both sides by same non-zero quantity
If we carry out any of the
above mentioned operations on an equal statement (LHS=RHS), the equality
(LHS+RHS) still holds well even after the operation.
Definition :
An equation containing only linear
polynomial or variable in first degree
is called ‘linear equation’
Eaxmples : x+2 =5, 3*(a-5) =6, ½ x -4/5 = 3x+7 are all linear equations
x2-4
=0 is not a linear equation(
Its degree is 2)
Example 1:
Let us consider the
statement x-3 = 1 where x is a variable.
This statement can be
explained as “find the value of x, such that, when 3 is subtracted from that
number(x) we should get the result as 1”
Let us substitute the
following values for x as in the statement x-3 =1
1. Can x be 1? No, because
1-3 which is -2, is 
2. Can x be 2 ? No, because 2-3 which is -1, is
3. Can x be 5? No, because
5-3 which is 2, is
4. Can x be 4? Yes because
4-3=1.
What are we doing?
We are trying to find the
value of x by trial and error method which is time consuming and is not the
most effective way.
If so, then how do we find
out the value of x mathematically?
Let us add 3 to both sides
of the statement. Then we get x-3+3= 1+3 so we get x+0 = 4.
What did we do ?
We added the same number to
both sides of the statement as per 1st Axiom.
Why did we choose the
number 3 to be added?
We wanted only x to be
present on LHS and no other number along with x.
Example 2:Find
the value of x such that 6x+4 = 3x+10
Here LHS =6x+4 and RHS=
3x+10
Step1:
Subtract 3x
from both sides (Why? We want RHS to contain only the constant and not the
variable) then we get
RHS = 3x+10-3x= 10
LHS = 6x+4-3x = 3x+4
From 2nd Axiom, RHS=LHS
Step 2:
Subtract 4 from both sides
(Why? We want LHS to have only the variable part and not the constant) then we
get
LHS = 3x+4-4=3x
RHS = 10-4 = 6
From 2nd Axiom, LHS=RHS
Step
3
Divide
both sides by 3 (Why? We want LHS to have only variable with co-efficient=1)
then we get
LHS = 3x/3 =x
RHS = 6/3 =2
From 2nd Axiom, LHS=RHS
Thus
we have x=2
What
did we do in first two steps?
We
first subtracted 3x from both sides and then the
constant (number) 4.
This
is same as saying add additive inverse of 3x (i.e. -3x) and additive inverse of 4 (i.e. -4) to
both LHS and RHS
We
transposed (changed) 3x from one side
to another side of the equation with change in sign and then in 2nd
step, transposed 4 from
one side to another side of equation with change in sign
Let
us summarise our steps
|
Step |
Statement |
Explanation |
|
1 |
6x+4= 3x+10 |
Given equation: |
|
2 |
6x+4-3x =10 i.e. 3x+4 =10 |
Transposition of 3x from right side to left side with sign changed |
|
3 |
3x= 10-4 i.e. 3x =16 |
Transposition of 4 from left side to right side with change in sign |
|
4 |
x=2 |
Simplification(Divide both sides by 3) |
Verification:
Let
us replace x by 2 in the equation (1)
LHS
=6*2+4 = 16 and
RHS
= 3*2+10 =16
Since
LHS = RHS=16, x=2 is the correct answer.
Definition:
Finding the value of variable which when substituted in the equation makes two
sides (LHS and RHS) equal, is called ‘solution’
to the equation.
In
the above case x =2 is the solution to the equation. You may notice that x=1 is
not a solution to the above equation (because when 1 is substituted for x, we
get LHS = 10 and RHS=13 and therefore LHS
)
2.7 Problem 1 :
Solve (Find value of x) in
5*(2x-3) =
2*(3x-7)
Solution:
|
Step |
Statement |
Explanation |
|
1 |
5*(2x-3) = 2*(3x-7) |
Given equation: |
|
2 |
10x -15 = 6x -14 |
Simplification |
|
3 |
10x -6x= -14+15 |
Transposition of 6x,15 from one side to another with sign change |
|
4 |
4x = 1:i,e x = ¼ |
Simplification |
Verification:
Substitute
1/4 for x in the given equation (1)
LHS
= 5*(2*1/4 -3) = 5*(1/2-3) = 5*(-5/2) = -25/2
RHS
= 2*(3*1/4-7) =
2*(3/4-7) = 2*(-25/4) = -25/2
Since
LHS= RHS =
-25/2, x =1/4 is the correct answer
2.7 Problem 2 :
Solve for x
= 1/2
Solution:
Get
the squares of both sides
(x-2)/(x+1) = 1/4
On
cross multiplication we get
4(x-2)
= x+1
I.e.
4x – 8 = x+1 (simplification)
I.e.
4x –x = 1+8 (
Transposition)
I.e. 3x = 9
x=3
Verification:
Substitute
x=3 in the given problem to note that 
= 1/2
2.7 Problem 3:
Find three consecutive even numbers whose sum is 252
Solution:
Step
1 : Let x be the first
even number
Step
2 : Then the next two consecutive even numbers are x+2
and x+4
Step
3 : Sum of these three consecutive even numbers are x+(x+2)+(x+4) = 3x+6
LHS
= 3x+6
RHS
= 252(Given data)
We
need to solve 3x+6 = 252
On
transposition we get 3x = 252-6=246 and thus x = 82
The
three numbers are 82(=x), 84(=x+2) and 86(=x+4)
Verification:
82,
84, 86 are three consecutive even numbers
and their sum is = 82+84+86
which is = 252
2.7 Problem 4: A
steamer goes downstream from one port to another in 9 hours. It covers the same
distance upstream in 10hours. If the speed of stream be 1km per hour, find out
the distance between ports
Solution:
|
Step1: Let the speed of steamer in still water be x km
per hour. The speed of downstream = (x+1) km per hour. The speed
in upstream = (x-1) km per hour. The distance covered in 9hours while going down =9(x+1).
The distance covered upstream in 10hours = 10(x-1) Step 2: Since distance between 2 ports is same (going
upstream or going downstream) 9(x+1) = 10(x-1) Step3: We need to solve the equation 9x+9 =10x-10i.e.
9+10 =10x-9x(By transposition) Therefore 19=x .Substitute this value in 9(x+1) The distance covered while in downstream =9(x+1) =
9*(19+1) =9*20= 180km |
|
Verification:
Speed
of steamer while going downstream = (distance/time) – (speed of stream) =
(180/9)-1 = (20-1) km/hr=19km/hr
Speed
of steamer while going upstream =
(distance/time) + (speed of stream) = (180/10) +1 = (18+1) km/hr=19km/hr
Since
both are same, our solution is correct.
2.7 Problem 5: A
number consists of 2 digits. The digit in tens place is 2 times the digit at
the units place. The number formed by reversing the digits is 27 less than the
original number. Find the original number
Solution:
|
1)Let x be the digit in unit place. Since the digit in 10’s place is 2 times
the digit in units place, it has to be2x. Since the number consists of 2
digits, its value = 10*digit in 10’s place + digit in units place Thus
the number = 10*2x+x. =20x+x -------------è (1) When we
reverse the digits of this number we get the reverse number x 2x(x in tens
place and 2x in units place) whose value= 10* digit in 10’s place + digit in
units place =
10*x+2x =10x+2x
--------------è reversed number It is
given that reversed number = oldnumber-27 ie 27 = 21x-12x (By transposition of 12x and 27) i,e 27 =9x
Hence
the original number is 63(digit in tens place is twice the digit in units
place) |
|
Verification:
Since the original number
is 63, the reversed number is 36. 36 is 27 less than
63.
Thus reversed number is 27
less than original number (which is as given in the problem).
Hence we have solved the
problem correctly
2.7 Problem 6: The length of a rectangle is 4cms more than
breadth and the perimeter is 11cms more than breadth. Find the length and
breadth of the rectangle.
Solution:
Step 1: let x be the
breadth. Then length = x+4.
We know perimeter P =
2*length + 2*breadth = 2(x+4)+2x =2x+8+2x --------------èEquation (1)
We are also given that
perimeter is 11cm more than breadth ( ie
P = x+11) --------------è Equation (2)
Step 2 :
Both are perimeters of the same rectangle and hence they have to be equal:
So Equation (1) = Equation
(2)
I.e. 2x+8+2x
= x+11:
I.e.4x+8 = x+11
I.e. 4x-x
= 11-8(By transposition of x and 8):
I.e.3x = 3:
x = 1.
Therefore breadth =1cm and
length=5cms(x+4)
Verification:
The perimeter of rectangle
= 2*length + 2*breadth = 2*5+2*1 = 10+2 = 12cms = 11cms +1cm = 11cms +breadth.
Hence we have solved the
problem correctly.
2.7 Problem 7: In a fraction, twice the numerator is 2 more
than the denominator. If 3 is added to each numerator
and denominator, the new fraction is 2/3. Find the original fraction
Solution:
Step1:
Let x be the numerator
It is given that twice the
numerator is two more than the denominator ( That
is to say 2*numerator = denominator+2)
Hence denominator =2x-2.
Thus the original fraction
is x/2x-2
When 3 is added to
denominator the new denominator= (2x-2) +3=2x+1
When 3 is added to
numerator the new numerator = x+3
Thus the new fraction is
(new numerator)/ (new denominator) = (x+3)/ (2x+1)
Since the new fraction is
2/3 we have
Step2 :
2/3 = (x+3)/(2x+1) --------------è(1)
By Multiplying both sides
of (1) by (2x+1), we get
2*(2x+1)/3 = x+3 --------------è(2)
By multiplying both sides
of (2) by 3, we get
2*(2x+1) =3(x+3)
I.e. 4x+2 =3x+9 and on
transposition of 3x and 2 we get
4x-3x= 9-2
I.e. x= 7
The original denominator = 2x-2 =14-2=12
Thus the original fraction
was 7/12
Verification:
To get the new fraction,
add 3 to both numerator and denominator so we get the new fraction as 10/15
which is 2/3
This shows that we solved
the problem correctly
2.7 Problem 8: Divide 32 in to two parts such that if the
larger is divided by the smaller, the quotient is 2 and the reminder is 5
Solution:
|
Let x be the larger number and
hence 32-x will be the smaller number We know dividend =
quotient*divisor + reminder Therfore we
need to solve the equation: x/(32-x) = 2+ 5(as
reminder) Exercise : Solve x = 2(32-x) +5 to get the answer x=23 and
other number as 9 |
|
2.7 Problem 9: Find a positive value of variable x for which
the given equation x2-9/( x2+5) =
-5/9 is satisfied
Solution:
Given
equation is x2-9/( x2+5) = -5/9
On
cross multiplication we get 9(x2-9) = -5(x2+5)
On
simplification we get 9x2-81 = -5x2 -25
On
transposition 14x2 = 56
x2
= 4
x = +2 or -2
Verification:
Substitute
x=2 in the given equation to arrive at LHS = -5/9 which is RHS and hence x=2 is
correct
2.7 Problem 10: Of the group of butter flies, 1/5 th proceeded towards Kadamba
tree, 1/3 of the group
proceeded towards Shilindhra
tree, thrice the differences of the
above two groups proceeded towards Kutaja
tree. If the remaining only one was flying around attracted by aroma of ketakamalati flowers, Lilavati
tell me the count of butter flies.
( Lilavati Shloka 56)
Solution:
Let the number of butter flies be x.
|
Step |
To where |
How many |
|
1 |
Kadamba tree |
(x/5) |
|
2 |
Shilindhra tree |
(x/3) |
|
3 |
differences of the above |
(x/3) – (x/5) = (2x/15) |
|
4 |
Kutaja tree |
3*(2x/15)=(2x/15) |
|
5 |
Remaining |
1 |
x- {(x/5)+(x/3)+(2x/5) =1
{15x-(3x+5x+6x)/15} =1
x=15
Verification:
15=3+5+6+1
2.7 Problem 11: A traveler on pilgrimage spends half the amount in Prayaga,
2/9 th of the remaining amount in Kashi, 1/4 th of the remaining
amount towards taxes, 6/10 th of the remaining amount
in
Solution:
Let
the number of
butter flies be x.
|
Step |
To where Towards |
How much |
Balance |
|
1 |
Prayaga |
(x/2) |
x-(x/2) = (x/2) |
|
2 |
Kashi |
(2/9)*(x/2)=(x/9) |
(x/2)-(x/9) = (7x/18) |
|
3 |
Taxes |
(1/4)*(7x/18)
=(7x/72) |
(7x/18) - (7x/72)= (21x/72) =(7x/24) |
|
4 |
|
(6/10)*(7x/24)=(7x/40) |
(7x/24)- (7x/40) ={(35x-21x)/120}=(7x/60) |
|
5 |
Remaining |
63 |
|
(7x/60)
=63
x=540
Verification:
Try
yourself
2.7 Summary of
learning
|
No |
Points studied |
|
1 |
Transposition
of terms from LHS to RHS with change
in sign |
|
2 |
Simplify
LHS and RHS such that variables are on one side and constants are on the
other side. |