Solve the following pairs of equations by reducing them to a pair of linear equations: 1+1=2
2 3
1 + 1 = 13
3 2 6
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(i)
2x
1
+
3y
1
=2
3x
1
+
2y
1
=
6
13
Lets assume
x
1
=p and
y
1
=q, both equations will become
2
p
+
3
q
=2
⇒3p+2q=12 ... (A)
3
p
+
2
q
=
6
13
⇒2p+3q=13 ...(B)
Multiply (A) by 3 and (B) by 2, we get
9p+6q=36
4p+6q=26
Subtracting these both, we get
5p=10
⇒p=2
⇒x=
p
1
=
2
1
∴q=
6
26−8
=3
⇒y=
q
1
=
3
1
(ii)
x
2
+
y
3
=2
x
4
−
y
9
=−1
Lets assume
x
1
=p and
y
1
=q, both equations will become
2p+3q=2 ...(A)
4p−9q=−1 ...(B)
Multiply (A) by 3, we get
6p+9q=6
4p−9q=−1
Adding these both, we get
10p=5
⇒p=
2
1
⇒ x=
p
2
1
=4
∴q=
3
1
⇒y=
q
2
1
=9
(iii)
x
4
+3y=14
x
3
−4y=23
Lets assume
x
1
=p, both equations become
4p+3y=14 ...(A)
3p−4y=23 ...(B)
Multiply (A) by 4 and (B) by 3, we get
16p+12y=56
9p−12y=69
Adding these both, we get
25p=125
⇒p=5
⇒x=
p
1
=
5
1
⇒y=−2
(iv)
x−1
5
+
y−2
1
=2
x−1
6
−
y−2
3
=1
Lets assume
x−1
1
=p and
y−2
1
=q, both equations become
5p+q=2
6p−3q=1
Multiply (A) by 3, we get
15p+3q=6
6p−3q=1
Adding both of these, we get
21p=7
⇒p=
3
1
⇒ x=4
∴q=
3
1
⇒ y=5
(v)
xy
7x−2y
=5
⇒
y
7
−
x
2
=5
xy
8x+7y
=15
⇒
y
8
+
x
7
=15
Let t=
x
1
and r=
y
1
equation will be
7r−2t=5;8r+7t=15
On solving we get
t=1r=1
∴x=1andy=1
(vi)
Dividing both side by xy then we have
y
6
+
x
3
=6;
y
2
+
x
4
=5
Let t=
x
1
andr=
y
1
equation will be
6r+3t=6;2r+4t=5 on solving we get
t=1r=
2
1
∴x=1andy=2
(vii)
Let t=
x+y
1
andr=
x−y
1
then equation will be
10t+2r=4;15t−5r=−2 on solving we have
t=
5
1
andr=1
∴x+y=5andx−y=1
Hence, x=4,y=1
(viii)
Let t=
3x+y
1
andr=
3x−y
1
then equation will be
t+r=
4
3
;
2
t
−
2
r
=
8
−1
On solving we have
t=
4
1
andr=
2
1
∴3x+y=4and3x−y=2
Hence, x=1,y=1