Question :
Solve the following pairs of equations by reducing them to a pair of linear of linear equations :
(i) \(1\over 2x\) + \(1\over 3y\) = 2 and \(1\over 3x\) + \(1\over 2y\) = 2
(ii) \(2\over \sqrt{x}\) + \(3\over \sqrt{y}\) = 2 and \(4\over \sqrt{x}\) – \(9\over \sqrt{y}\) = 2
(iii) \(4\over x\) + 3y = 14 and \(3\over x\) – 4y = 23
(iv) \(4\over x\) + 3y = 14 and \(6\over x – 1\) – \(3\over y – 2\) = 1
(v) \(7x – 2y\over xy\) = 5 and \(8x + 7y\over xy\) = 15
(vi) 6x + 3y = 6xy and 2x + 4y = 5xy
(vii) \(10\over x + y\) + \(2\over x – y\) = 4 and \(15\over x + y\) – \(5\over x – y\) = -2
(viii) \(1\over 3x + y\) + \(1\over 3x – y\) = \(3\over 4\) and \(1\over 2(3x + y)\) – \(1\over 2(3x – y)\) = \(-1\over 8\)
Solution :
(i) Put \(1\over x\) = u and \(1\over y\) = v, then the equation becomes
\(1\over 2\)u + \(1\over 3\)v = 2 \(\implies\) 3u + 2v = 12 ………..(1)
\(1\over 3\)u + \(1\over 2\)v = \(13\over 6\) \(\implies\) 2u + 3v = 13 ………….(2)
Multiply the equation (1) by 3 and equation (2) by 2, we get
9u + 6v = 36 ……(3)
4u + 6v = 26 ……..(4)
Subtract equation (4) from equation ( 3), we get
5u = 10 \(\implies\) u = 2
Putting the value of u in equation (3), we get
18 + 6v = 36 \(\implies\) v = 3
Now, \(1\over x\) = u \(\implies\) x = \(1\over 2\)
and \(1\over y\) = v \(\implies\) y = \(1\over 3\)
(ii) Put \(1\over \sqrt{x}\) = u and \(1\over \sqrt{y}\) = v, then the equation becomes
2u + 3v = 2 ……(1)
and 4u – 9u = -1 ……(2)
Multiply equation (1) by 3 and adding it with equation (2), we get
10u = 5 \(\implies\) u = \(1\over 2\)
Put the value of u in equation (1), we get
3v = 1 \(\implies\) v = \(1\over 3\)
Now, \(1\over \sqrt{x}\) = u \(\implies\) x = 4
and \(1\over \sqrt{y}\) = v \(\implies\) y = 9
(iii) The given linear equations are
\(4\over x\) + 3y = 14 …….(1)
and \(3\over x\) – 4y = 23 …….(2)
Multiply equation (1) by 4 and equation (2) by 3, we get
\(16\over x\) + 12y = 56 …….(3)
\(9\over x\) – 12y = 69 …….(4)
Now, Add equation (3) and equation (3), we get
\(25\over x\) = 125 \(\implies\) x = \(1\over 5\)
Put the value x in equation (1), we get
3y = -6 \(\implies\) y = -2
Hence, x = \(1\over 5\) and y = -2.
(iv) Let u = \(1\over x – 1\) and v = \(1\over y – 2\). Then, the given linear equations becomes,
5u + v = 2 ……….(1)
and 6u – 3v = 1 ………..(2)
Multiply equation (1) by 3 and adding it to equation (2), we get
21 u = 7 \(\implies\) u = \(1\over 3\)
Put the value of u in equation (1), we get
\(5\over 3\) + v = 2 \(\implies\) v = \(1\over 3\)
Now, u = \(1\over x – 1\) \(\implies\) x = 4
and v = \(1\over y – 2\) \(\implies\) y = 5
Hence, x = 4 and y = 5.
(v) The given linear equations are :
\(7x – 2y\over xy\) = 5 \(\implies\) \(7\over y\) – \(2\over x\) = 5
and \(8x + 7y\over xy\) = 15 \(\implies\) \(8\over y\) + \(7\over x\) = 15
Let u = \(1\over x\) and v = \(1\over y\). Then, the above equations becomes
7v – 2u = 5 …….(1)
and 8v + 7u = 15 ………(2)
Multiply equation (1) by 7 and equation (2) by 2, we get
49 – 14u = 35 ……..(3)
and 16v + 14u = 30 ……..(4)
Add equation (3) and equation (4), we get
65v = 65 \(\implies\) v = 1
Put the value of v in equation (1), we get
7 – 2u = 5 \(\implies\) u = 1
Now, u = \(1\over x\) \(\implies\) x = 1
and v = \(1\over y\) \(\implies\) y = 1
Hence, x = 1, y = 1.
(vi) The given linear equations are
6x + 3y = 6xy
and 2x + 4y = 5xy
On dividing above equations by xy, we get
\(3\over x\) + \(6\over y\) = 6
and \(4\over x\) + \(2\over y\) = 5
Put \(1\over x\) = u and \(1\over y\) = u, the above equation becomes,
3u + 6v = 6 ……..(1)
and 4u + 2v = 5 ……….(2)
Multiply equation (2) by 3 and subtracting it from equation (1), we get
-9u = -9 \(\implies\) u = 1
Put the value of u in equation (1), we get
6v = 3 \(\implies\) v = \(1\over 2\)
Now, \(1\over x\) = u \(\implies\) x = 1
\(1\over y\) = v \(\implies\) y = 2
Hence, x = 1 and y = 2.
(vii) The given linear equations are
\(10\over x + y\) + \(2\over x – y\) = 4
and \(15\over x + y\) – \(5\over x – y\) = -2
Put u = \(1\over x + y\) and v = \(1\over x – y\), the given equations becomes
10u + 2v = 4 \(\implies\) 5u + v = 2 ………..(1)
and 15u – 5v = -2 ………(2)
Multiply equation (1) by 5 and adding it with equation (2), we get
40u = 8 \(\implies\) u = \(1\over 5\)
Put the value of u in equation (1), we get
v = 1
Now, u = \(1\over x + y\) \(\implies\) x + y = 5 …….(4)
and v = \(1\over x – y\) \(\implies\) x – y = 1 ……..(5)
Adding equation (4) and equation (5), we get
2x = 6 \(\implies\) x = 3
Put the value of x in equation (4), we get
3 + y = 5 \(\implies\) y = 2
Hence, x = 3 and y = 2.
(viii) The given equation are
\(1\over 3x + y\) + \(1\over 3x – y\) = \(3\over 4\)
and \(1\over 2(3x + y)\) – \(1\over 2(3x – y)\) = \(-1\over 8\)
Putting u = \(1\over 3x + y\) and v = \(1\over 3x – y\), then the equation becomes,
u + v = \(3\over 4\) ……….(1)
and \(1\over 2\)u – \(1\over 2\)v = \(-1\over 8\)
\(\implies\) u – v = \(-1\over 4\) ……..(2)
Add equation (1) and equation (2), we get
2u = \(1\over 2\) \(\implies\) u = \(1\over 4\)
Put the value of u in equation (1), we get
\(1\over 4\) + v = \(3\over 4\) \(\implies\) v = \(1\over 2\)
Now, u = \(1\over 3x + y\) \(\implies\) 3x + y = 4 ……..(3)
and v = \(1\over 3x – y\) \(\implies\) 3x – y = 2 ……..(4)
Adding equation (3) and equation (4), we get
6x = 6 or x = 1
Put the value x in equation (3), we get
3 + y = 4 or y = 1
Hence, x = 1, y = 1.
