To solve Equations with “xy” terms
2
7
(vi)
7x – 2y
xy
=
5 ,
xy
=
15
Soln.
Substituting
1
y
=
p
&
1
x
=
q
∴
7p
–
2q
=
5
8p
7q
=
15
+
... (ii)
∴
... (i)
8x + 7y
7x
2y
xy
xy
–
=
5 ,
8x
xy
xy
+
=
15
7y
y
x
–
8
y
7
x
+
=
5 ,
=
15 ,
7
y
2
x
×
1
1
×
–
Solution is x = 1,y = 1
What is the difference in this sum and other sum ?
There is xy term in the denominator
To remove the ‘xy’ term from the denominator we split the numerator and give ‘xy’ term to both the parts of numerator
Now let us number the equations as (i) and (ii)
Now this sum has become like previous sums and we solve it in the same way
After solving it we get the final answer as
(vi)
3y
6x
+
6xy ;
=
4y
2x
+
5xy
=
Soln.
What is the difference in this sum and previous sum
xy is in the numerator
Bring xy to denominator by cross multiplication
6x + 3y
xy
=
6 ,
xy
=
5
2x + 4y
Now the sum is similar to the previous sum
Q.] Solve the following pair of equations by reducing them to a pair
of linear equations:
6x + 3y = 6xy ; 2x + 4y = 5xy
Sol :
Divide both eqn by xy
6
y
+
3
x
=
6
;
2
y
+
4
x
=
5
Lets do the substitution to reduce it to linear equation
Substituting
p
=
1
y
&
q
=
1
x
6p
+
3q
=
6
...... (i)
2p
+
4q
=
5
...... (ii)
Multiplying (ii) by 3, we get
6p
+
12q
=
15
...... (iii)
Subtracting (i) from (iii), we get
6p
+
12q
=
15
6p
+
3q
=
6
(–)
(–)
(–)
9q
=
9
q
=
1
∴
Substituting q = 1 in (ii)
2p
+
4(1)
=
5
∴
2p
+
4
=
5
∴
2p
=
5
–
4
∴
2p
=
1
∴
p
=
1
2
Resubstituting the values
of p and q
1
y
=
1
2
∴
y
=
2
1
x
=
1
∴
x
=
1
Solution is
x
=
1,
y
=
2