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

Here we discuss the solution of such pair of equations which is not linear but can be reduced to linear form by making suitable substitutions. We will explain the with the following examples.

Â

Solve for x and y: , (x â‰ 0, y â‰ 0).

Let = u and = v

148u + 231v = 527uvÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i)

231u + 148v = 610uvÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â€¦â€¦â€¦â€¦.â€¦â€¦â€¦â€¦â€¦(ii)

Adding (i) and (ii),

379u + 379v = 1137uv

Dividing by 379,

u + v = 3uvÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(iii)

Subtracting (i) from (ii),

83u â€“ 83v = 83uv

Dividing by 83,

â‡’ u â€“ v = uvÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(iv)

(iii) + (iv),

2u = 4uv âˆ´ 2v = 1 âˆ´ v =

Substituting the value of v = in (iii),

u + = 3 Ã— u Ã—

u â€“ u =

â‡’

u = 1 âˆ´ u =âˆ´

x = 1Since v =

âˆ´

y = 2 âˆ´ x = 1, y = 2.

Solve the following equations algebraically:

Â + =

Â - =

+ =

- =

Let u = and v =

Then + =

LCM of 2 and 5 is 10.

â‡’ 20u + 8v = 10

Dividing by 2 we get,

10u + 4v = 5 â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)

Then 5u â€“ 2v =

50u - 20v = 1 â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)

Multiplying (i) by 5 and adding with eqn (ii)

u = = 0.26

Substituting u = 0.26 in (i)

10(0.26) + 4v = 5

Â Â Â Â Â Â Â Â 2.6 + 4v = 5

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 4v = 5 â€“ 2.6

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 4v = 2.4

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â v = 0.6

Substituting u = 0.26 in

= 0.26

3x + 4y = â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(iii)

Substituting v = 0.6 in

2x - 3y = â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(iv)

Multiplying (iii) by 3 and (iv) by 4

Â Â Â Â Â Â Â Â Â

Substituting in (iv)

Â Â - 3y =

Â Â Â Â Â Â Â Â Â Â Â Â -3y =

-3y =

Â Â Â Â

Â y =

Â Â Â

The solution set is .

A boat travels 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can travel 40 kmÂ upstream andÂ 55 km downstream. Determine theÂ speed of the stream and that of the boatÂ in stillÂ water?

Let the speed of the boat in still water be x km/hr

and the speed of the stream be y km/hr.

Speed of the boat while travelling downstream = (x + y) km/hr.

and speed of the boat while travellingÂ upstream = (x - y) km/hr

Condition 1: A boat travels 30 km upstream and 44 km downstream in 10 hrs.

Time taken to travel upstream by the boat =

and time taken to travel downstream by the boat =Â

â€¦â€¦â€¦..(i)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Condition 2: A boat travels 40 km upstream and 55 km downstream in 13 hrs.

Time taken toÂ travel upstream by the boat =

and time taken to travel downstream by the boat =Â

â€¦â€¦â€¦â€¦..(ii)

__Â __

The L.C.M. of 44 and 55 is 220.

âˆ´ Multiplying (i) by 5 and (ii) by 4

-10 = -2(x - y)

Â Â 5 = x - y

.^{.}. x - y = 5Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â€¦â€¦â€¦â€¦â€¦..(v)

Substituting x - y = 5 in (i)

44 = 10(x + y) â€“ 6(x + y)

44 = 4(x + y)

11 = x + y

âˆ´ x + y = 11Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â€¦â€¦â€¦â€¦â€¦..(vi)

Adding (v) and (vi)

Â Â Â x = 8

Substituting x = 8 in (v)

8 - y = 5

Â Â Â -y = 5 - 8Â

Â Â Â -y = -3

Â Â Â Â y = 3

âˆ´ The speed of the stream is 3 km/hr and the speed of the boat in still water is 8 km/hr.

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