2.14 Solving of Simultaneous Linear
Equations:
We are going to find solutions to
the puzzle which we mentioned in the topic 2.1 while introducing algebra.
“Sum of my age and my father’s age
is 55 years. If after 16 years, my father’s age is twice as that of mine, then
can you tell me my age as on today”?
We know how to solve equations of
type x+1 = 5, 2a+6 =10, as they have only one variable. They are called linear
equations of one variable.
Let us take an equation of type x+y = 5. This equation has two variables namely x and y.
Let us substitute some values for
x and y in the above equation. We find that the
pair of values such as (x=1,y=4),
(x=2,y=3), (x=3,y=2), (x=0,y=5), (x= -2, y=7)
satisfy the condition x+y = 5. Thus, we have
infinite values of x and y satisfying the given condition. Why is this so?
This is because, by transposition
of given equation we get y = 5-x and for any value of x we have corresponding
value of y satisfying the condition x+y =5.
Since a linear
equation in two variables have infinite number of solutions, for solving such
equations we need to have another relationship between the two variables.
Suppose your friend gives the
below mentioned puzzle to you and says that he will give as many CDs of your
choice (music, games, video) as his age, if you solve the puzzle. The number of
CDs he is going to give will be of his age.
Will you accept the challenge to
get those many CDs?
2.14 Problem 1 (Puzzle): Sum of my age and my father’s age
is 55 years. If after 16 years, my father’s age is twice as that of mine, then
can you tell me my age as on today?
Since ages have to be positive
integers and your friend is not a kid you can solve the puzzle easily by trial
and error method starting with 9 as your friend’s age as detailed below.
|
Now (total age =55) |
After 16 years |
||
|
Friends age |
His father’s age |
Friends age |
His father’s age |
|
9 |
46 |
25 |
62 |
|
10 |
45 |
26 |
61 |
|
11 |
44 |
27 |
60 |
|
12 |
43 |
28 |
59 |
|
13 |
42 |
29 |
58 |
|
14 |
41 |
30 |
57 |
|
15 |
40 |
31 |
56 |
From the above table, you find
that, If your friend’s age is 13
and his father’s age 42
now, then after 16 years his age will be 29 and his father’s will be 58 which is twice the age of
his. Now you can demand from your friend 13 CDs of your choice for having
solved the puzzle.
But in more complex cases you will
not have time to solve these types of problems by trial and error method.
Solution:
Let us solve the problem
systematically.
Let y be your friend’s age and x
be his father’s age since sum of their age is 55 we have
x+y =55
After 16 years, your friend’s age
will be y+16 and his fathers age will be x+16.
We are given that that after 16
years, fathers age will be twice that of your friend, thus we have
x+16 =2*(y+16)
I.e. x+16 = 2y+ 32 (On simplification)
I.e. x-2y = 32-16 =16 (By
transposition)
Finally we have following two
equations
(1) x+y =55
(2) x-2y = 16
To solve a linear equation in one
variable we need to have an equation with only one variable.
We should find a method to convert
two equations to equation in single variable.
The given equations are
x+y =55
==è (1)
x-2y=16
==è (2)
----------
Subtract (2) from (1) we get
0+3y =39 ==è (3)
-----------
So 3y = 39 and hence y=13
Since x+y
=55 we have x = 55-y (By transposition)
Substituting 13 for y in the above
equation we get x=55-13 =42
Thus your friend’s age is 13 and
his father’s age is 42 which is exactly what we got by trial and error method!
Verification:
Note:
1. Now, if your friend’ age is 13 and his father’s age is 42,
then the sum of ages is 55
After 16 years your age friend’s age will be 29 and father’s will be 58 which
will be twice of his age
This is exactly what is given in the problem and hence our solution is
correct.
2. Substitute values of x and y in
(1) we get x+y = 42+13 = 55 which is the first
equation
Substitute values of x and y in (2) we get x-2y = 42-26 = 16 which is the
second equation
2.14 Problem 2: The
cost of a geometry box is 18 Rs more than that of a pen. If your class teacher
pays Rs 240 for buying 5 geometry boxes and 10 pens, find out the cost of
geometry box and pen
Solution:
Let the cost of geometry box be y
and the cost of pen be x. We have
(1) y =
x+18 ==è(1)
(2) 5y+10x = 240 ==è(2)
Transposition of (1) gives us
y-x =18 and by
multiplying this equation by
5 we get
5y-5x= 90 ===è(3)
So we have two equations
5y+10x =240 ------è(2)
5y -5X =
90 ------è(3)
-----------------
[Subtract (3) from (2) to eliminate
y 0+15x = 150 ------è(4)
-----------------
Thus x = 10 which is cost of the
pen.
Since y = x+18, cost of geometry
box(y) is Rs.28 (10+18)
Exercise: Verify that these values satisfy
the equations (1) and (2).
2.14 Problem 3: Solve 2x+2y =4 and
x+y =2
Solution:
2x+2y
=4 ===è(1)
x+y = 2 ===è(2)
Multiply (2) by 2 we get
2x+2y=4 ===è(3)
Subtract (1) from (3) we get 0 =0 which is always true
This means that there are many
values of x and y which satisfy the given equations and there is no unique
solution.
2.14 Problem 4: Solve 2x+2y =4 and x+y = 3
Solution:
Given
equations are : 2x+2y =4 ====è(1)
x+y = 3 =====è(2)
Multiply(2) by 2 we get
2x+2y=6 =====è(3)
Given equation: 2x+2y =4 ===è(2)
Subtracting (1) from (3) we
get 0 =2 which is not true
Thus there are no values of x and
y which can satisfy the given equations
Definition: Two linear equations in two
variables taken together are called ‘simultaneous
linear equations’ and are of the form
a1 x+ b1
y = c1
a2 x+b2
y = c2
where a1, b1, a2,
b2, c1 and c2
are real numbers and x and y are
variables whose value we are asked to find
How did we solve these
simultaneous liner equations?
2.14.1 Method of Elimination by equating co –efficients:
Steps to be followed are:
1. Multiply the equations (one or
both) by suitable numbers such that the coefficients of either x or y are same
after multiplication, in both the equations.
2. Add or subtract these equations to
get a resulting equation whose coefficient of x or y is zero so that one of the
variable is not present in the resulting equation.
3. With only one variable present in
the resulting equation find the value of that variable.
4. Substituting this value in any one
of the equations find the value of other variable.
Observations:
It is not
true that it is always possible to find solutions to all the simultaneous
linear equations:
1. a1
x+ b1 y = c1
2. a2
x+b2 y = c2
1. They do not have solution if (a1
/ a2) = (b1 / b2) 
(c1 / c2)
2. They have infinite solutions if
(a1 / a2) = (b1 / b2) = (c1 /
c2)
3. They have unique solution only
if (a1 / a2) (b1 / b2)
2.14 Problem 5:
Solve x+y
=2xy ------à(1)
and
x-y = 6xy -----à(2)
Solution:
By
adding both the equations we get
2x = 8xy
I.e. 1 = 4y
y = 1/4
Substituting this in equation (1)
we get
x+ 1/4 = 2x/4 = x/2
On transposition we get
x-x/2 = - 1/4
x = -1/2
Verification:
x+y = -1/2+1/4 = -1/4
2xy = 2*(-1/2)*(1/4) = -1/4
x+y =2xy
x-y =
-1/2-1/4 = -3/4
6xy= 6*(-1/2)*(1/4) = -3/4
x-y = 6xy
2.14 Problem 6: In an examination
the ratio of passes to failures is 4:1.Had 30 less appeared for the examination
and had 20 less passed, the ratio of passes to failures would have been 5:1.
Find the number of students appeared for the examination.
Solution:
Let x be the number of pass
candidates and y be the number of failed candidates
It is given that x/y = 4/1. x=4y
If 30 had appeared less, and 20 less(x-20) passed then
-
the
total number of candidates = x+y-30
-
no
of passed candidates = (x+y-30) –(x-20)= y-10
it is given that in such a case
ratio of pass to failure is 5:1
(x-20)/(y-10)
= 5/1 (x-20) = 5(y-10)
Thus we need to solve the
equations
x=4y and
(x-20) = 5(y-10)
Exercise
: Solve
yourself to get the answer x = 120 and y= 30
and hence number of students appeared for examination = x+y= 150.
2.14 Problem 7: The sum of the digits of
a two digit number is 9. Nine times this
number is twice the number obtained by reversing the the
order of the digits of the given number. Find the number
Solution:
Let x be the number in tens place
and y be the number in digit place, so that the actual number is 10x+y. the
number got by reversing the digits is 10y+x.
The equations to be solved are :
x+y = 9
9(10x+y) = 2(10y+x)
Exercise: solve these two equations to find that x =1 and y=8 so
that 18 is the two digit number also verify to note that
1+8 = 9
9*18 =2*81
2.14 Problem 8: Suppose you intend traveling to your native
place from the place you are studying along with your mother.
Since you are student, you are
eligible to get the ticket at 50% of the cost of full ticket. However reservation charges are fixed and is common for you as well
as for your mother. If the cost of ticket for your mother(including
reservation) is Rs 2125 and for both of you it is Rs 3200 find out the full fare of the ticket and the
reservation charges
Solution:
Let x be the the
cost of full fare and y be the reservation charges
x+y = 2125 ----à(1)
Since you are traveling with half
the fare of your mother, the fare for you will be 1/2x
Note that you and your mother have
to pay same reservation charges and hence the total fare is {(1/2)x+y} + (x+y) which is given to be Rs 3200
{(1/2)x+y}+(x+y) = 3200
(3/2)x+2y
=3200; multiplying bothe sides by 2 we get
3x+4y =6400
-----à(2)
On solving equations (1) and (2)
we get {multiply (1) by 3 and then subtract the result from (2)}
We get y =25 and x= 2100
The full fare of the ticket is
Rs.2100 and reservation charge is Rs.25
Verification:
Note that In case of your mother,
full fare+ reservation charge=2100+25
In case of you and your mother
fare = half
fare + reservation +full fare +reservation = 1050+25+2100+25=3200
These are the values given in the
problem and hence our solution is correct
2.14 Summary of learning
|
No |
Points studied |
|
1 |
Simultaneous linear equations( a1 x+ b1
y = c1, a2 x+b2 y = c2)are
solved by reducing them to equations of single variable by
appropriate transpositions |
|
2 |
It is not always possible to solve all simultaneous linear equations |
2.14 Additional points:
2.14.2 Alternate method:
Method of elimination by substitution.
Simultaneous linear equations can
also be solved using an alternate method:
1.
From
any one of the given equations, find the value of one variable(y) in terms of
the other variable(x).
2.
Substitute
the value of the variable(y) arrived in Step 1, in the other equation, to find
the value of the other variable(x).
3.
Substitute
the value of the variable(x) arrived in step 2, in one of the equations to get
the value of the remaining variable(y).
Let us solve problem 2.14.2 (previously
solved) using this alternate method.
Solve
5y+10x =240 ----à(1)
5y -5X = 90 ----à(2)
Let us take the first equation,
5y+10x = 240 and find the value of y in terms of x.
5y = 240-10x
y = 48-2x
Let us substitute this value of y
in the 2nd equation, 5y-5x = 90.
LHS = 5y - 5x = 5(48-2x) - 5x = 240-10x-5x =
240-15x
Since LHS = 90(RHS) we have
240-15x = 90
i.e. 240-90 = 15x
i.e. 150 = 15x
x = 10
Substituting this value of x in
the 1st equation we get
5y+10*10 = 240
i.e. 5y = 240-100=140
y = 28
This is exactly the same
solutions that were obtained in 2.14.Problem 2.
2.14.3 Solving linear equations in three variables.
We have learnt that, to solve
linear equations in two variables we need two equations. Extending this logic, to
solve linear equations in three variables we will need three equations.
Steps to be followed are:
1.
Take
any two equations to eliminate the third variable
2.
Repeat
step 1 by taking another pair of equations different from the one used in step
1
3.
Solve
the resulting two linear equations using any of the methods discussed above
4.
Substitute
the values of the two variables obtained in step 3, in any one of the 3
equations to arrive at the value of the third variable
2.14 Problem 8: Solve
2/x + 3/y - 4/z = -20
2/y - 4/x + 3/z = 45
3/x - 4/y + 2/z=5
Solution:
Let a=1/x, b=1/y and c=1/z
Then the give equations will
become
2a+3b-4c = -20 -----à(1)
-4a+2b+3c = 45 ------à(2)
3a-4b+2c = 5 -----à(3)
By multiplying equations (1) by 3
and (2) by 4 we get
6a+9b-12c = -60
-16a+8b+12c = 180
By adding the above two equations
we get
-10a+17b = 120 ----à(4)
By multiplying equations (1) by 1
and (3) by 2 we get
2a+3b-4c = -20
6a-8b+4c = 10
By adding the above two equations
we get
8a-5b = -10 ----à(5)
By multiplying equations (4) by 8
and (5) by 10 we get
-80a+136b = 960
80a-50b = -100
By adding the above two equations
we get
86b = 860
b = 10
Substituting b=10 in (4) we get
-10a+170 = 120
-10a = -50
a = 5
Substituting a=5 and b=10 in (1)
we get
10+30-4c =-20
-4c = -60
c = 15
Since a=5, b=10 and c=15 it
follows that
x=1/5, y=1/10 and z=1/15
Verification:
Substitute these values in the
given equations to confirm that our solution is correct.
2.14 Problem 8: What 3 carpenters earn in a day is earned by 4 male
workers in a day. The daily wages of 4 male workers is equal to the daily wages
of one carpenter and 4 female workers. If one carpenter, 2 male workers and 5
female workers are engaged for a day, their total wages is Rs 500. Find the
daily wages of carpenter, male worker and female worker.
Solution:
Let c, m and f be the daily wages
of carpenter, male worker and female worker respectively.
Since 3 carpenters’ wage is equal
to the wages of 4 male workers,
3c = 4m ----à(1)
Since daily wages of 4 male
workers is equal to daily wages of a carpenter and 4 female workers,
4m = c+4f ----à(2)
Since daily wages of 1 carpenter,
2 male workers and 5 female workers is 500,
c+2m+5f = 500 ---à(3)
By substituting 4m = 3c in (2) we
get
3c = c+4f
2c = 4f or c=2f
Substituting c=2f in (2) we get
4m =2f+4f
2m=3f
Substituting 2m=3f and c=2f in (3) we get
2f+3f+5f =500
10f = 500
f = 50
Substituting this value of f in 2m=3f and c=2f we get c = 100 and m =
75.
Hence the daily wages of a
carpenter, male worker and female worker is Rs 100, Rs 75 and Rs 50
respectively.