Class 10

### Topic Covered

♦ Equations Reducible to a Pair of Linear Equations in Two Variables

### Equations Reducible to a Pair of Linear Equations in Two Variables

=> In this section, we shall discuss the solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable substitutions. We now explain this process through some examples..
Q 3109078818

Solve the pair of equations:

2/x + 3/y = 13

5/x - 4/y = -2
Class 10 Chapter 3 Example 17
Solution:

Let us write the given pair of equations as

2 (1/x) + 3 ( 1/y) = 13 (1)

5 (1/x) -4 ( 1/y) = -2 (2)

These equations are not in the form ax + by + c = 0. However, if we substitute

1/x = p and 1/y = q in Equations (1) and (2), we get

2p + 3q = 13 (3)

5p – 4q = – 2 (4)

So, we have expressed the equations as a pair of linear equations. Now, you can use
any method to solve these equations, and get p = 2, q = 3.

You know that p = 1/x and q = 1/y .

Substitute the values of p and q to get

1/x = 2 , i.e., x = 1/2 and 1/y =3 , i.e.,  y = 1/3 .

Verification : By substituting x = 1/2  and y = 1/3 in the given equations, we find that
both the equations are satisfied.
Q 3129178911

Solve the following pair of equations by reducing them to a pair of
linear equations :

5/(x-1) + 1/(y-2) = 2

6/(x-1) - 3/(y-2) = 1
Class 10 Chapter 3 Example 18
Solution:

Let us put  1/(x-1) = p and 1/(y-2) = q Then the given equations

 5(1/(x-1) ) + 1/(y-2) =2 (1)

6 (1/(x-1) ) -3 (1/(y-2) ) =1 (2 )

can be written as : 5p + q = 2 (3)

6p – 3q = 1 (4 )

Equations (3) and (4) form a pair of linear equations in the general form. Now,
you can use any method to solve these equations. We get  p = 1/3 and q = 1/3 .

Now, substituting 1/(x-1) for p, we have

1/(x-1)= 1/3 ,

i.e., x – 1 = 3, i.e., x = 4.

Similarly, substituting  1/(y-2) for q, we get

1/(y-2) = 1/3

i.e., 3 = y – 2, i.e., y = 5
Hence, x = 4, y = 5 is the required solution of the given pair of equations.

Verification : Substitute x = 4 and y = 5 in (1) and (2) to check whether they are
satisfied .
Q 3149178913

A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km
down-stream. Determine the speed of the stream and that of the boat in still water.
Class 10 Chapter 3 Example 19
Solution:

Let the speed of the boat in still water be x km/h and speed of the stream be y km/h. Then the
speed of the boat downstream

= (x + y) km/h,

and the speed of the boat upstream = (x – y) km/h

Also,  text (time) = (text ( distance ) )/( text ( speed ) )

In the first case, when the boat goes 30 km upstream, let the time taken, in hour,
be t_1. Then

t_1 = 30/(x-y)

Let t_2 be the time, in hours, taken by the boat to go 44 km downstream. Then

t_2 = 44/(x+y) . The total time taken, t_1 + t_2, is 10 hours. Therefore, we get the equation

30/(x-y) + 44/(x+y) = 10 (1)

In the second case, in 13 hours it can go 40 km upstream and 55 km downstream. We
get the equation

40/(x-y) + 55/(x+y) = 13 (2)

Put 1/(x-y) = u and  1/(x+ y ) = v (3)

On substituting these values in Equations (1) and (2), we get the pair of linear
equations:

30u + 44v = 10 or 30u + 44v – 10 = 0 (4)

40u + 55v = 13 or 40u + 55v – 13 = 0 (5)

Using Cross-multiplication method, we get

u/(44 (-13) -55 (-10 )) = v/(40 (-10) -30 ( -13 ) ) = 1/(30 (55) -44 (40 ) )

i.e., u/(-22) = v/(-10) = 1/(-110)

i.e.  u = 1/5 , v = 1/11

Now put these values of u and v in Equations (3), we get

1/(x-y) = 1/5 and  1/(x+y) = 1/11

i.e.,  x-y = 5 and x+ y = 11 (6)