NCERT Class 10 Mathematics

64 MATHEMATICS

These equations are not in the form ax + by + c = 0. However, if we substitute

(^11) pand q
xy
in Equations (1) and (2), we get
2 p + 3q = 13 (3)
5 p – 4q = – 2 (4)
So, we have expressed the equations as a pair of linear equations. Now, you can use
any method to solve these equations, and get p = 2, q = 3.
You know that p =

1

x

and q =

1

y

✁

Substitute the values of p and q to get

(^11) 2, i.e., and (^11) 3, i.e.,
23
xy
xy

.

Verification : By substituting

11

and

23

xy✂ ✂ in the given equations, we find that

both the equations are satisfied.

Example 18 : Solve the following pair of equations by reducing them to a pair of
linear equations :

51

xy 12

✄

☎ ☎

=2

63

xy 12

✆

✆ ✆

=1

Solution : Let us put

11

and

12

p q

xy

✝ ✝

☎ ☎. Then the given equations

5 11

xy 12

✞ ✟✠

✡ ☛

✌ ☞ ✍ ☞ = 2 (1)

11

63

xy 12

✎ ✏✑ ✎ ✏

✒✔ ✑ ✓✕ ✒ ✑ ✓

✔ ✕

= 1 (2)

can be written as : 5 p + q = 2 (3)

6 p – 3q = 1 (4)