Equations Reducible To A Pair Of Linear Equations Examples
Example 1: Solve the following system of equations
(frac { 1 }{ 2x }) – (frac { 1 }{ y }) = – 1; (frac { 1 }{ x }) + (frac { 1 }{ 2y }) = 8
Sol. We have,
(frac { 1 }{ 2x }) – (frac { 1 }{ y }) = – 1 ….(1)
(frac { 1 }{ x }) + (frac { 1 }{ 2y }) = 8 ….(2)
Let us consider 1/x = u and 1/y = v.
Putting 1/x = u and 1/y = v in the above equations, we get;
(frac { u }{ 2 }) – v = – 1 ….(3)
u + (frac { v }{ 2 }) = 8 ….(4)
Let us eliminate v from the system of equations. So, multiplying equation (3) with 1/2 and (4) with 1, we get
(frac { u }{ 4 }) – (frac { v }{ 2 }) = (-frac { 1 }{ 2 }) ….(5)
u + (frac { v }{ 2 }) = 8 ….(6)
Adding equation (5) and (6), we get ;
(frac { u }{ 4 }) + u = (-frac { 1 }{ 2 }) + 8
⇒ (frac { 5u }{ 4 }) = (frac { 15 }{ 2 })
⇒ u = (frac { 15 }{ 2 }) × (frac { 4 }{ 5 })
⇒ u = 6
We know,
(frac { 1 }{ x }) = u ⇒ (frac { 1 }{ x }) = 6
⇒ x = (frac { 1 }{ 6 })
Putting 1/x = 6 in equation (2), we get ;
6 + (frac { 1 }{ 2y }) = 8 ⇒ (frac { 1 }{ 2y }) = 2
⇒ (frac { 1 }{ y }) = 4 ⇒ y = (frac { 1 }{ 4 })
Hence, the solution of the system is,
x = (frac { 1 }{ 6 }) , y = (frac { 1 }{ 4 })
Example 2: Solve (frac { 2 }{ x }) + (frac { 1 }{ 3y }) = (frac { 1 }{ 5 }); (frac { 3 }{ x }) + (frac { 2 }{ 3y }) = 2 and also find ‘a’ for which y = ax – 2.
Sol. Considering 1/x = u and 1/y = v, the given system of equations becomes
2u + (frac { v }{ 3 }) = (frac { 1 }{ 5 })
⇒ (frac { 6u+v }{ 3 }) = (frac { 1 }{ 5 })
30u + 5v = 3 ….(1)
3u + (frac { 2v }{ 3 }) = 2 ⇒ 9u + 2v = 6 ….(2)
Multiplying equation (1) with 2 and equation (2) with 5, we get
60u + 10v = 6 ….(3)
45u + 10v = 30 ….(4)
Subtracting equation (4) from equation (3), we get
15u = – 24
u = (-frac { 24 }{ 15 }) = (-frac { 8 }{ 5 })
Putting u = (-frac { 8 }{ 5 }) in equation (2), we get;
9 × (frac { -8 }{ 5 }) + 2v = 6
⇒ (frac { -72 }{ 5 }) + 2v = 6
⇒ 2v = 6 + (frac { 72 }{ 5 }) = (frac { 102 }{ 5 })
⇒ v = (frac { 51 }{ 5 })
Here (frac { 1 }{ x }) = u = (frac { -8 }{ 5 })
⇒ x = (frac { -5 }{ 8 })
And, (frac { 1 }{ y }) = v = (frac { 51 }{ 5 }) ⇒ y = ⇒ (frac { 5 }{ 51 })
Putting x = (frac { -5 }{ 8 }) and y = (frac { 5 }{ 51 }) in y = ax – 2, we get;
(frac { 5 }{ 51 }) = (frac { -5a }{ 8 }) – 2
(frac { 5a }{ 8 }) = – 2 – (frac { 5 }{ 51 }) = (frac { -102-5 }{ 51 }) = (frac { -107 }{ 51 })
a = (frac { -107 }{ 51 }) × (frac { 8 }{ 5 }) = (frac { -856 }{ 255 })
a = (frac { -856 }{ 255 })
Example 3: Solve (frac{2}{x+2y}+frac{6}{2x-y}=4text{ ; }frac{5}{2left( x+2y right)}+frac{1}{3left( 2x-y right)}=1) where, x + 2y ≠ 0 and 2x – y ≠ 0
Sol. Taking (frac { 1 }{ x+2y }) = u and (frac { 1 }{ 2x-y }) = v, the above system of equations becomes
2u + 6v = 4 ….(1)
(frac { 5u }{ 2 }) + (frac { v }{ 3 }) = 1 ….(2)
Multiplying equation (2) by 18, we have;
45u + 6v = 18 ….(3)
Now, subtracting equation (3) from equation (1), we get ;
–43u = – 14 ⇒ u = (frac { 14 }{ 43 })
Putting u = 14/43 in equation (1), we get
2 × (frac { 14 }{ 43 }) + 6v = 4
⇒ 6v = 4 – (frac { 28 }{ 43 }) = (frac { 172-28 }{ 43 }) ⇒ v = (frac { 144 }{ 43 })
Now, u = (frac { 14 }{ 43 }) = (frac { 1 }{ x+2y })
⇒ 14x + 28y = 43 ….(4)
And, v = (frac { 144 }{ 43 }) = (frac { 1 }{ 2x-y })
⇒ 288x – 144y = 43 ….(5)
Multiplying equation (4) by 288 and (5) by 14, the system of equations becomes
288 × 14x + 28y × 288 = 43 × 288
288x × 14 – 144y × 14 = 43 × 4
⇒ 4022x + 8064y = 12384 ….(6)
4022x – 2016y = 602 ….(7)
Subtracting equation (7) from (6), we get
10080y = 11782 ⇒ y = 1.6(approx)
Now, putting 1.6 in (4), we get,
14x + 28 × 1.6 = 63
⇒ 14x + 44.8 = 63 ⇒ 14x = 18.2
⇒ x = 1.3 (approx)
Thus, solution of the given system of equation is x = 1.3 (approx), y = 1.6 (approx).
Example 4: Solve (frac{1}{x+y}+frac{2}{x-y}=2text{ and }frac{2}{x+y}-frac{1}{x-y}=3 ) where, x + y ≠ 0 and x – y ≠ 0
Taking (frac { 1 }{ x+y }) = u and (frac { 1 }{ x-y }) = v the above system of equations becomes
u + 2v = 2 ….(1)
2u – v = 3 ….(2)
Multiplying equation (1) by 2, and (2) by 1, we get;
2u + 4v = 4 ….(3)
2u – v = 3 ….(4)
Subtracting equation (4) from (3), we get
5v = 1 ⇒ v = (frac { 1 }{ 5 })
Putting v = 1/5 in equation (1), we get;
u + 2 × (frac { 1 }{ 5 }) = 2 ⇒ u = 2 – (frac { 2 }{ 5 }) = (frac { 8 }{ 5 })
Here, u = (frac { 8 }{ 5 }) = (frac { 1 }{ x+y }) ⇒ 8x + 8y = 5 ….(5)
And, v = (frac { 1 }{ 5 }) = (frac { 1 }{ x-y }) ⇒ x – y = 5 ….(6)
Multiplying equation (5) with 1, and (6) with 8, we get;
8x + 8y = 5 ….(7)
8x – 8y = 40 ….(8)
Adding equation (7) and (8), we get;
16x = 45 ⇒ x = (frac { 45 }{ 16 })
Now, putting the above value of x in equation (6), we get;
(frac { 45 }{ 16 }) – y = 5 ⇒ y = (frac { 45 }{ 16 }) – 5 = (frac { -35 }{ 16 })
Hence, solution of the system of the given equations is ;
x = (frac { 45 }{ 16 }) , y = (frac { -35 }{ 16 })
