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2.14 Solving of
Simultaneous Linear Equations:
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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.