Soln.
4
….(ii)
10p
+
2q
=
- 2
15p
-
5q
=
But what we
need to substitute….
Let us see
What is the difference between this sum and the sums done before this ?
Binomial terms in the denominator
So how to solve ?
By substituting
First let us number the equations.
Consider one of
the two equations
You can consider either of the two equations
It is better to consider simpler of the two equations
Let us consider equation (i)
Which equation is
to be considered ?
Consider (i) ,
2
5p
+
q
=
∴
q
=
2
-
5p
Shift one of the two variables to the L.H.S so that we get equation in
the form of either
p = something or
q = something…
… (iii)
What is the name of the method ?
SUBSTITUTION method
So we need to
substitute something
Substitute what ?
Where ?
In the equation which was not considered.
Substituting (iii) in (ii)
p
∴
=
8
40
p
∴
=
1
How to get the
value of q ?
We have to
substitute p = 1/5
Let us substitute in
equation (iii)
In either (i) , (ii) or (iii)
1
x + y
=
x + y =
x = 4
, y = 2
Re substituting p =
1
x – y
2x =
Re substituting q =
Equation (iii)
Substituting p = in (iii), we get
∴
Substituting
&
5
1
5
1
5
q
=
2
- 5 ( )
q
=
2
- 1
∴
q
∴
=
1
1
5
∴
5
1
=
6
Common Term / Common Denominator in both equations
10
(vii)
x
+
y
+
2
=
4
15
x
+
y
5
–
=
- 2
x
–
y
x
–
y
1
x + y
= p
1
x – y
= q
15p
-
5
(2 – 5p)
=
- 2
15p
=
- 10
-2
∴
+ 25 p
=
-2
∴
40 p
+ 10
=
8
∴
40 p
1
5
1
x + y
1
x - y
x – y =
∴
1
∴
2
….(i)
5p
+
q
=
… (iv)
… (v)
Adding (iv) and (v)
x + y =
5
x – y =
1
x =
3
Substituting x = 3 in (iv)
3 + y =
5
∴
y =
5 – 3
∴
y =
2
This equation can be simplified further….
By dividing throughout by 2
Even after
Re substitution, we aren’t getting the answer
We are getting 2 more equations….solve these equations to get the answer