201920
L
INEAR
E
QUATIONS
IN
O
NE
V
ARIABLE
33
Did you observe how we
simplified the form of the given
equation? Here, we had to
multiply both sides of the
equation by the LCM of the
denominators of the terms in the
expressions of the equation.
RHS = 6
x
– 2 +
7
2
=
4
7
3
6
6
2
2
2
x
x

+
=
+
The equation is
x
+ 14 = 6
x
+
3
2
or
14 = 6
x
–
x
+
3
2
or
14 = 5
x
+
3
2
or
14 –
3
2
= 5
x
(transposing
3
2
)
or
28
3
2

= 5
x
or
25
2
= 5
x
or
x
=
25
1
5
5
5
2
5
2
5
2
×
×
=
=
×
Therefore, required solution is
x
=
5
2
.
Check
:
LHS =
=
25
25
25
2(5
7)
2( 2)
4
2
2
2


=


=
+
=
25
8
33
2
2
+
=
RHS =
=
26
7
33
2
2
+
=
= LHS.
(as required)
EXERCISE 2.5
Solve the following linear equations.
1.
1
1
2
5
3
4
x
x

=
+
2.
3
5
21
2
4
6
n
n
n

+
=
3.
8
17
5
7
3
6
2
x
x
x
+

=

Note, in this example we
brought the equation to a
simpler form by opening
brackets and combining like
terms on both sides of the
equation.
201920
34
M
ATHEMATICS
4.
5
3
3
5
x
x


=
5.
3
2
2
3
2
4
3
3
t
t
t

+

=

6.
1
2
1
2
3
m
m
m



=

Simplify and solve the following linear equations.
7.
3(
t
– 3) = 5(2
t
+ 1)
8.
15(
y
– 4) –2(
y
– 9) + 5(
y
+ 6) = 0
9.
3(5
z
– 7) – 2(9
z
– 11) = 4(8
z
– 13) – 17
10.
0.25(4
f
– 3) = 0.05(10
f
– 9)
2.7
Equations Reducible to the Linear Form
Example 18:
Solve
1
3
2
3
8
x
x
+
=
+
Solution:
Observe that the equation is not a linear equation, since the expression on its
LHS is not linear. But we can put it into the form of a linear equation. We multiply both
sides of the equation by (2
x
+ 3),
x
x
x
+
+
×
+
1
2
3
2
3
(
)
=
3
(2
3)
8
x
×
+
Notice that (2
x
+ 3) gets cancelled on the LHS We have then,
x
+ 1 =
3 (2
3)
8
x
+
We have now a linear equation which we know how to solve.
Multiplying both sides by 8
8 (
x
+ 1) = 3 (2
x
+ 3)
or
8
x
+ 8 = 6
x
+ 9
or
8
x
= 6
x
+ 9 – 8
or
8
x
= 6
x
+ 1
or
8
x
– 6
x
= 1
or
2
x
= 1
or
x
=
1
2
The solution is
x
=
1
2
.
Check :
Numerator of LHS =
1
2
+ 1 =
1
2
3
2
2
+
=
Denominator of LHS = 2
x
+ 3 =
1
2
2
×
+ 3 = 1 + 3 = 4
This step can be
directly obtained by
‘crossmultiplication’
Note that
2
x
+ 3
≠
0 (Why?)
201920
L
INEAR
E
QUATIONS
IN
O
NE
V
ARIABLE
35
LHS = numerator
÷
denominator =
3
4
2
÷
=
3
1
3
2
4
8
×
=
LHS = RHS.
Example 19:
Present ages of Anu and Raj are in the ratio 4:5. Eight years from now
the ratio of their ages will be 5:6. Find their present ages.
Solution:
Let the present ages of Anu and Raj be 4
x
years and 5
x
years respectively.
After eight years. Anu’s age = (4
x
+ 8) years;
After eight years, Raj’s age = (5
x
+ 8) years.
Therefore, the ratio of their ages after eight years =
4
8
5
8
x
x
+
+
This is given to be 5 : 6
Therefore,
4
8
5
8
x
x
+
+
=
5
6
Crossmultiplication gives
6
(4
x
+ 8) = 5 (5
x
+ 8)
or
24
x
+ 48 = 25
x
+ 40
or
24
x
+ 48 – 40 = 25
x
or
24
x
+ 8 = 25
x
or
8 = 25
x
– 24
x
or
8 =
x
Therefore,
Anu’s present age = 4
x
= 4 × 8 = 32 years
Raj’s present age = 5
x
= 5 × 8 = 40 years
EXERCISE 2.6
Solve the following equations.
1.
8
3
2
3
x
x

=
2.
9
15
7
6
x
x
=

3.
4
15
9
z
z
=
+
4.
3
4
2
2 – 6
5
y
y
+

=
5.
7
4
4
2
3
y
y
+

=
+
6.
The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of
their ages will be 3:4. Find their present ages.
7.
The denominator of a rational number is greater than its numerator by 8. If the
numerator is increased by 17 and the denominator is decreased by 1, the number
obtained is
3
2
. Find the rational number.
You've reached the end of your free preview.
Want to read all 16 pages?
 Fall '19
 Linear Equations, Numerical digit, RHS