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EQUATIONS OF THE FORM
ax + by = c AND bx + ay = d, WHERE a ≠ b.
To solve the equations of the form :
ax + by = c.....(i)
and bx + ay = d ....(ii)
where a ¹ b, we follow the following steps :
StepI : Add (i) and (ii) and obtain (a + b)x + (b + a) y = c + d, i.e., x + y = ...(iii)
StepII : Subtract (ii) from (i) and obtain (a – b)x – (a – b) y = c – d, i.e., x – y = ...(iv)
StepIII : Solve (iii) and (iv) to get x and y.
Ex.12 Solve for x and y : 47x + 31y = 63, 31x + 47y = 15.
Sol. We have, 47x + 31y = 63 ...(i) and 31x + 47y = 15 ...(ii)
Adding (i) and (ii), we get : 78x + 78y = 78 ⇒ x + y = 1 ...(iii)
Subtracting (ii) from (i), we get : 16x – 16y = 48 ⇒ x – y = 3 ...(iv)
Now, adding (iii) and (iv), we get : 2x = 4 ⇒ x = 2
Putting x = 2 in (iii), we get : 2 + y = 1 ⇒ y = – 1
Hence, the solution is x = 2 and y = –1
EQUATIONS REDUCIBLE TO LINEAR EQUATIONS IN TWO VARIABLES
Equations which contain the variables, only in the denominators, are called reciprocal equations. These equations can be of the following types and can be solved by the under mentioned method :
TypeI : = c and c' a,b,c,a',b',c'∈ R
and find the value of x and y by any method described earlier.
Then
TypeII : au + bv = cuv and a'u + b'v = c'uv a,b,c,a',b',c'∈R
Divide both equations by uv and equations can be converted in the form explained in (I).
TypeIII : = k, = k a,b,k,a',b',k'∈R
Put 1
and
Then equations are au + bv = k and a'u + b'v = k'
Find the values of u and v and put in Ix + my
Again solve for x and y, by any method explained earlier.
Ex.13 Solve for x and y : + 5 = 0 and – 2 = 0 (x ¹ 0, y ¹ 0)
Sol. We have, + 5 = 0 and – 2 = 0
. Then, the given equations can be written as 3au – 2bv = –5 ...(i)
and au + 3bv = 2 ...(ii)
Multiplying (i) by 3 and (ii) by 2, we get
9au – 6bv = –15 ...(iii) and 2au + 6bv = 4 ...(iv)
Adding (iii) and (iv), we get 11au = –11 ⇒ u =1/a
Put u = 1/a in equation (ii), we get a + 3bv = 2 ⇒ 3bv = 3 ⇒ v =
Hence the solution is x = – a and y = b.
Ex.14 Solve = 5 and = 9.
Sol. We have
= 5 ⇒ – 5 = 0
and = 9 ⇒ – 9 = 0
Let = p and = q. Then, the given equations can be written as 57p + 6q – 5 = 0 and 38p + 21q – 9 = 0
By crossmultiplication method, we have
⇒ =
⇒
But = p and = q. Therefore
⇒ = 1/19 ⇒ x + y = 19...(i)
and = 1/3⇒ x – y = 3 ...(ii)
Adding (i) and (ii), we get 2x = 22 ⇒ x = 11
Put x = 11 in (i),
we get 11 + y = 19 ⇒ y = 8
Hence, the solution is x = 11 and y = 8.
Ex.15 Solve for x and y : = 5 , = 15,
Sol. = 5 ⇒ = 5 ...(i)
= 15 ⇒ = 15...(ii)
we get 7u – 2v = 5 ...(iii)
8u + 7v = 15 ...(iv)
Multiplying (iii) by 7 and (iv) by 2 and adding we get
49u – 14v = 35
and 16u + 14v = 30
65 u = 65 ⇒ u = 1 ⇒ 1/y = 1 or y = 1
Substituting u = 1 in (iii) we get : 7 – 2v = 5 ⇒ v = 1 ⇒ 1/x = 1 or x = 1
Hence, x = 1, y = 1.
APPLICATIONS OF LINEAR EQUATIONS IN TWO VARIABLES
In this section, we will study about some applications of simultaneous linear equations in solving variety of word problems related to our daytoday life situations. The following examples are selfexplanatory and will give some insight to the solution to such problems.
TypeI : Based on Articles And Their Costs/Quantities
Ex.16 7 audio cassettes and 3 video cassettes cost Rs.1110, while 5 audio cassettes and 4 video cassettes cost Rs. 1350.
Find the cost of an audio cassette and a video cassette.
Sol. Let the cost of an audio cassette and a video cassette be Rs. x and Rs. y respectively.
The cost of 7 audio cassettes and 3 video cassettes = Rs. 1110
⇒ 7x + 3y = 1110 ...(i)
The cost of 5 audio cassettes and 4 video cassettes = Rs. 1350
⇒ 5x + 4y = 1350 ...(ii)
Multiplying (i) by 4 and (ii) by 3, we get
28x + 12y = 4440 ...(iii)
15x + 12y = 4050 ...(iv)
Subtracting (iv) from (iii), we get 13x = 390 ⇒ x = 30
Putting x = 30 in (i), we get 7 × 30 + 3y = 1110
⇒ 210 + 3y = 1110 ⇒ 3y = 900 ⇒ y = 300
Hence, the cost of an audio cassette is Rs. 30 and that of a video casstte is Rs. 300.
TypeII : Based on Numbers
Ex.17 The sum of the digits of a twodigit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.
Sol. Let the digit at ten's place be x and that at unit's place be y. Then,
x + y = 12 ...(i)
And , the two digits number = 10x + y
Now, according to the question, (10y + x) = (10x + y) + 18 ⇒ 9y – 9x = 18 ⇒ y – x = 2 ...(ii)
Adding (i) and (ii), we get
2y = 14 ⇒ y = 7 Put y = 7 in (i), we get
x + 7 = 12 ⇒ x = 5
Hence, the required number is (10 × 5 + 7), i.e., 57.
TypeIII : Based on Fractions
Ex.18 If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It become 1/2. if we only add 1 to the denominator. What is the fraction?
Sol. Let the required fraction be x y . Then,
= 1 ⇒ x + 1 = y – 1 ⇒ x – y = –2 ...(i)
and
= 1/2 ⇒ 2x = y + 1 ⇒ 2x – y = 1 ...(ii)
Subtracting (i) from (ii), we get x = 3
Put x = 3 in (i), we get 3 – y = –2 ⇒ y = 5
Hence, the fraction is 3/5
TypeIV : Based on Ages
Ex.19 Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. Find the present ages of father and son.
Sol. Let the present ages of the father and the son be x years and y years respectively.
Two years ago, Father's age = (x – 2) years and son's age = (y – 2) years
∴ (x – 2) = 5 (y – 2) ⇒ x – 5y = – 8....(i)
Two years later, Father's age = (x + 2) years and son's age = (y + 2) years
∴ (x + 2) = 3 (y + 2) + 8 ⇒ x + 2 = 3y + 6 + 8 ⇒ x – 3y = 12 ....(ii)
Subtracting (i) from (ii), we get 2y = 20 ⇒ y = 10
Putting y = 10 in (ii), we get x – 3 × 10 = 12 ⇒ x – 30 = 12 ⇒ x = 42
Hence, the present ages of father and son are 42 years and 10 years respectively.
TypeV : Based on Geometrical Applications
Ex.20 The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Sol. Let the larger angle be x° and the smaller angle by y°.
Then, x + y = 180 ...(i)
and x = y + 18 ⇒ x – y = 18 ...(ii)
Adding (i) and (ii), we get2x = 198 ⇒ x = 99
Putting x = 99 in (i), we get 99 + y = 180 ⇒ y = 81
Hence the required angles are 99° and 81°.
TypeVI : Based on Time, Distance and Speed.
Formulae to be used:
1. (a) Speed =
(b) Distance = Speed × Time
(c)Time =
2. Let speed of a boat in still water = u km/h and speed of the current = v km/h. Then,
(a) Speed of a boat downstream = (u + v)km/h
(b) Speed of a boat upstream = (u – v)km/h.
Ex.21 A man travels 370 km partly by train and partly by car. If he covers 250 km by train and the rest by car it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18 minutes, longer. Find the speed of the train and that of the car.
Sol. Let the speeds of the train and that of the car be x km/h and y km/h respectively.
If he covers 250 km by train and 120 km by car it takes 4 hours. Therefore,
...(i)
And if he covers 130 km by train and 240 km by car it takes 4 hours and 18 minutes.
Therefore,
...(ii)
Let 1/x = u and 1/y = v. Then, the equations (i) and (ii), can be written as
250u + 120 v = 4 ...(iii)
and 130u + 240v =43/10...(iv)
Multiplying (iii) by 2, we get 500u + 240v = 8 ...(v)
Subtracting (iv) from (v), we get 370u = 8  (43/10)
⇒ 370u = 37/10
⇒ u =1/100
Putting u =1/100 in (iii), we get 250 × 1/100+ 120 v = 4 ⇒5/2
+ 120 v = 4
⇒ 120v = 4  (5/2)
⇒ v =
but u = 1/x and v =1/x.
Therefore, 1/x = 1/100
⇒ x = 100 and 1/y = 1/80
⇒ y = 80
Hence the speeds of the train and that of the car are 100 km/h and 80 km/h respectively.
TypeVII : Miscellaneous
Ex.22 8 men and 12 boys can finish a piece of work in 10 days while 6 men and 8 boys can finish it in 14 days. Find the time taken by one men alone and that by one boy alone to finish the work.
Sol. Let one man alone can finish the work in x days and one boy alone can finish the work in y days. Then, the work done by one man in one day = 1/x and the work done by one boy in one day = 1/y
According to the question, ...(i)
Also,
...(ii)
Multiplying (i) by 4 and (ii) by 3, we get :
...(iii)
and . ..(iv)
Subtracting (iii) from (iv), we get ⇒ x = 140
Putting x = 140 in (ii), we get
⇒ y = 280
Hence, one man alone can finish the work in 140 days and one boy alone can finish the work in 280 days.
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