To solve Equations with Variables and Numbers
in the denominator
1
2x
(i)
+
1
3y
=
2;
1
3x
+
1
2y
=
13
6
Soln.
Substituting
∴
p
3
12
–
2q
=
How is this sum
different from the
previous sums?
3p + 2q = 12
It has variables
as well as numbers
in the denominator
To get rid of the
variables from denominator,
we substitute…
1
2x
+
1
3y
=
2
To remove ‘2’ & ‘3’ from denominator multiply by LCM of 2 & 3
LCM of 2 & 3 is 6
Multiplying throughout by 6
3
2
3
x
+
2
y
=
12
3p + 2q = 12 …(iii)
1
3x
+
1
2y
=
13
6
Multiplying throughout by 6
2
3
2
x
+
3
y
=
13
2p + 3q = 13 …(iv)
Consider one of the two equations
Which equation is to be considered
You can consider either
of the two equation
It is better to consider simpler of the two equations
Lets us consider
equation no (iii)
Consider (iii)
Write their equation either
p = something
or
q = something
Number the equation as (v)
We need to substitute
something
Substitute what ?
Substitute (v)
Where ?
In the equation which was not considered
How to get rid of the
numbers from
Denominator?
By Multiplying throughout
By LCM
….. (i)
LCM of 2 , 3 & 6
is 6
….. (ii)
Look at (i) and (ii)
Are equations linear?
How to make the equations Linear?
By Substituting…
What to Substitute…
Common Term / Common Denominator in both equations
= p &
1
x
= q in (i) and (ii)
1
y
….. (v)
Substituting (v) in (iv)
12
- 2q
3
2
= 13
+ 3q
24
- 4q
3
= 13
+ 3q
∴
24
- 4q
3
+ 9q
∴
= 13
24
- 4q
= 39
+ 9q
∴
24
= 39
+ 5q
∴
= 15
5q
∴
q
=
3
∴
Substituting q = 3 in (v),
p
3
12
2(3)
=
–
p
=
2
∴
Re substituting p =
1
x
2 =
1
x =
2
∴
1
x
& q =
1
y
1
y
& 3 =
1
& y =
3