Equations Reducible To A Pair Of Linear Equations

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  $\frac{2}{x+2y}+\frac{6}{2x-y}=4\text{ ; }\frac{5}{2\left( x+2y \right)}+\frac{1}{3\left( 2x-y \right)}=1$   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 $\frac{1}{x+y}+\frac{2}{x-y}=2\text{  and  }\frac{2}{x+y}-\frac{1}{x-y}=3$    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 =  $\frac { -35 }{ 16 }$