**Equations Reducible To A Pair Of Linear Equations Examples**

**Example 1:** Solve the following system of equations

– = – 1; + = 8

**Sol.** We have,

– = – 1 ….(1)

+ = 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;

– v = – 1 ….(3)

u + = 8 ….(4)

Let us eliminate v from the system of equations. So, multiplying equation (3) with 1/2 and (4) with 1, we get

– = ….(5)

u + = 8 ….(6)

Adding equation (5) and (6), we get ;

+ u = + 8

⇒ =

⇒ u = ×

⇒ u = 6

We know,

= u ⇒ = 6

⇒ x =

Putting 1/x = 6 in equation (2), we get ;

6 + = 8 ⇒ = 2

⇒ = 4 ⇒ y =

Hence, the solution of the system is,

x = , y =

**Example 2: **Solve + = ; + = 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 + =

⇒ =

30u + 5v = 3 ….(1)

3u + = 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 = =

Putting u = in equation (2), we get;

9 × + 2v = 6

⇒ + 2v = 6

⇒ 2v = 6 + =

⇒ v =

Here = u =

⇒ x =

And, = v = ⇒ y = ⇒

Putting x = and y = in y = ax – 2, we get;

= – 2

= – 2 – = =

a = × =

a =

**Example 3: **Solve where, x + 2y ≠ 0 and 2x – y ≠ 0

**Sol. **Taking = u and = v, the above system of equations becomes

2u + 6v = 4 ….(1)

+ = 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 =

Putting u = 14/43 in equation (1), we get

2 × + 6v = 4

⇒ 6v = 4 – = ⇒ v =

Now, u = =

⇒ 14x + 28y = 43 ….(4)

And, v = =

⇒ 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 where, x + y ≠ 0 and x – y ≠ 0

Taking = u and = 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 =

Putting v = 1/5 in equation (1), we get;

u + 2 × = 2 ⇒ u = 2 – =

Here, u = = ⇒ 8x + 8y = 5 ….(5)

And, v = = ⇒ 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 =

Now, putting the above value of x in equation (6), we get;

– y = 5 ⇒ y = – 5 =

Hence, solution of the system of the given equations is ;

x = , y =

Parneet says

Good explanation