Equations
We need to frame
equations of one or two unknown in invariably in every problem. Sometimes we
get three equations in three unknowns. In general, we need as many equations as
the variables we will have to solve for. So, for solving for the values fo
three unknowns, we need three equations (and hence the problem should give
three conditions from which we can frame three equations). Solving the
equations by itself is not a difficult task. The important part of the problem
is framing the equation/equations. Once the equations are framed, solving them
is very easy.
One equation in one unknown
An equation like 2x +
4 = 26 is an equation in one variable. We have only one variable X whose value
we have to find out.
Step I: Keep
variables on L.H.S and constants on R.H.S i.e 2x=26-4=22.
Step II: Dividing
both sides of equation with coefficient of x i.e 2.
i.e 
= 
= x = 11
Two
equations in two unknowns
A set of equations
like
2x + 3y = 8 → (1)
5x + 4y = 13 → (2)
is called simultaneous
equations in two unknown. Here, we have two variables (or unknowns) x and y
whose values we have to find out.
Step I: Using
both the equations we first eliminate one variable (so that we can then have
one equation in one unknown).
For this purpose, we multiply equation
(1) with 5 (the co-efficient of x in the second equation) and multiply equation
(2) with 2 (the co-efficient of x in the first equation) to eliminate x. Thus
we have.
(1) x 5 = 10x + 15y = 40 → (3)
(2) x 2 = 10x + 8y = 26 →
Now, subtracting equation
(4) from eqn (3) we have, 7y = 14 → (4)
This is one equation in one
unknown.
Step II: Solve
for the value of one variable from the equation (In one unknown) obtained from
step I above.
Therefore, y = 2.
Step II: Substitute
this value in one of two equations to get the value of the second variable.
Substituting the value of y
in eqn (1) or equation (2), we get x = 1
Therefore the values of x and y that
satisfy the given set of equations are x = 1 and y = 2
Worked
out examples
Example 1
The cost of 4 chairs and 3
tables is Rs.1800. The cost of 4 chair and 4 tables is Rs.2300. Find the cost
of each chair.
Solution
Let the costs of one
there and one table be Rs.C and Rs.T respectively:.
4C + 3T = 1800 → (1)
5C + 4T = 2300 → (2)
{(2) x 3} - {(1) x 4}
we get C = 300
So Cost each chair = Rs.300.
Example 2
The sum of the digits
of a two-digit number is 12. If 36 is subtracted from it, the reversed number
is obtained. Find the number.
Solution
The sum of the
digits=12
x + y = 12 → (1)
&⇒ 10x + y – 36 = 10y +
x
&⇒ 10x + y - (10y + x) =
36
&⇒ 9(x - y) = 36
&⇒ x – y = 4
Adding (1) and (2) ,
we get
2x = 16
x = 8
From (1) y =12 – x = 4
The number is 84
Example 3
The cost of 3 pens, 4
erasers and 5 sharpeners is Rs.40. The cost of 5 pens, 7 erasers and 9
sharpeners is Rs.70. Find the total cost of one of each.
Solution
Let the cost of each pen, eraser and sharper be p, e and
s respectively.
3p + 4e + 5s = 40 → (1)
5p + 7e + 9s = 70 → (2)
Multiplying the equations (1) b 2 and subtracting the
equating (2) from it. We get. p + e + s = 10.
Hence the total cost
of one of each is Rs.10.
Note: In
this case, there are three unknowns and two equations containing these
variables. As the number of equations is less than the number of variables,
most what is asked for is not a complete solution. It is just a certain
combination of one or more variables which sometimes can be found out.
Example 4
Manoj and Anuj have some chocolates with each of them, if
Manoj gives 5 chocolates to Anuj gives the same number of chocolates to Manoj, Manoj
would have twice as many chocolates as Anuj. Find the number of chocolates Anuj
has.
Solution
Let the number of
chocolates with Manoj and Anuj be S and R respectively. If Manoj gives 5
chocolates to Anuj, He would have S-5 chocolates and Anuj would have R+5
chocolates.
So S-5=R+5
S-R=10
If Anuj Gives 5
chocolates to Manoj, he would have R-5 chocolates and Manoj would have S+5
chocolates.
S+5=2(R-5)
2R-S=15 → (2)
Adding (1) and (2) we
get R = 25.
Example 5
A,B,C,D have a total Rs.200
with them. The amount with B is Rs.10 more than that with A. The amount with C
is Rs.20 more than that with B. The amount with D equals the sum of the amounts
with B and C together. Find the amount with A.
Solution
Let the amounts with
A,B,C,D be a,b,c and d respectively.
a+b+c+d=200
b=a+10
c=b+20=a+30
d=b+c=2a+40
a+b+c+d=5a+80=200
a=24
Example 6
The Present age of
father is two years more than thrice his son’s age 10 years later. The age of
the father ten years later will be 2 years more than 7 times his son’s present
age. Find the present age of the father.
Solution
Let the present age
of the father and the son be f and s respectively.
f=3(s+10) +2=3s+32
f+10=7s+2
Substituting f=3s+32
in the above equation,
3s+32+10=7s+2
40=4s
10=s
f=3(10)+32=62 years
