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 = 5x 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. x2y = 3216 =16
(By transposition)
Finally we have
following two equations
(1) x+y =55
(2) x2y = 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)
x2y=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 = 55y (By transposition)
Substituting 13 for y
in the above equation we get x=5513 =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
x2y = 4226 = 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
yx =18 and by
multiplying this equation by
5 we get
5y5x= 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
a_{1} x+ b_{1 }y = c_{1 }
a_{2}_{ }x+b_{2
}y _{= }c_{2 }
where a_{1, }b_{1,
}a_{2, }b_{2, }c_{1 } and c_{2} 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. a_{1} x+ b_{1 }y = c_{1}
2. a_{2}_{ }x+b_{2 }y _{= }c_{2 }
1. They do not have
solution if (a_{1 }/ a_{2}) = (b_{1 }/ b_{2}) _{} (c_{1 }/ c_{2})
2. They have infinite
solutions if (a_{1 }/ a_{2}) = (b_{1 }/ b_{2})
= (c_{1 }/ c_{2})
3. They have unique
solution only if (a_{1 }/ a_{2}) _{}(b_{1 }/ b_{2})
2.14 Problem 5:
Solve x+y =2xy à(1)
and
xy = 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
xx/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
xy = 1/21/4 = 3/4
6xy= 6*(1/2)*(1/4) = 3/4
_{} xy = 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(x20) passed then

the total number of candidates = x+y30

no of passed candidates = (x+y30) –(x20)=
y10
it is given that in such
a case ratio of pass to failure is 5:1
_{} (x20)/(y10)
= 5/1 _{} (x20) = 5(y10)
Thus we need to solve
the equations
x=4y and
(x20) = 5(y10)
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( a_{1}
x+ b_{1 }y = c_{1}, a_{2 }x+b_{2 }y _{= }c_{2})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 = 24010x
_{}y = 482x
Let us substitute this
value of y in the 2nd equation, 5y5x = 90.
LHS = 5y  5x =
5(482x)  5x =
24010x5x = 24015x
Since LHS = 90(RHS) we
have
24015x = 90
i.e. 24090 = 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 = 240100=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+3b4c = 20 à(1)
4a+2b+3c = 45 à(2)
3a4b+2c = 5 à(3)
By multiplying
equations (1) by 3 and (2) by 4 we get
6a+9b12c = 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+3b4c = 20
6a8b+4c = 10
By adding the above two
equations we get
8a5b = 10 à(5)
By multiplying
equations (4) by 8 and (5) by 10 we get
80a+136b = 960
80a50b = 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+304c =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.