         2.14 Solving of Simultaneous Linear Equations:

We are going to find solutions to the puzzle which we mentioned in the topic 2.1 while introducing algebra.

“Sum of my age and my father’s age is 55 years. If after 16 years, my father’s age is twice as that of mine, then can you tell me my age as on today”?

We know how to solve equations of type x+1 = 5, 2a+6 =10, as they have only one variable. They are called linear equations of one variable.

Let us take an equation of type x+y = 5. This equation has two variables namely x and y.

Let us substitute some values for x and y in the above equation. We find that the  pair  of values such as (x=1,y=4), (x=2,y=3), (x=3,y=2), (x=0,y=5), (x= -2, y=7)  satisfy the condition x+y = 5. Thus, we have infinite values of x and y satisfying the given condition. Why is this so?

This is because, by transposition of given equation we get y = 5-x and for any value of x we have corresponding value of y satisfying the condition x+y =5.

Since a linear equation in two variables have infinite number of solutions, for solving such equations we need to have another relationship between the two variables.

Suppose your friend gives the below mentioned puzzle to you and says that he will give as many CDs of your choice (music, games, video) as his age, if you solve the puzzle. The number of CDs he is going to give will be of his age.

Will you accept the challenge to get those many CDs?

2.14 Problem 1 (Puzzle): Sum of my age and my father’s age is 55 years. If after 16 years, my father’s age is twice as that of mine, then can you tell me my age as on today?

Since ages have to be positive integers and your friend is not a kid you can solve the puzzle easily by trial and error method starting with 9 as your friend’s age as detailed below.

 Now (total age =55) After 16 years Friends age His father’s age Friends age His father’s age 9 46 25 62 10 45 26 61 11 44 27 60 12 43 28 59 13 42 29 58 14 41 30 57 15 40 31 56

From the above table, you find that, If your friend’s age is 13 and his father’s age 42 now, then  after 16 years his age will be 29 and his father’s will be 58 which is twice the age of his. Now you can demand from your friend 13 CDs of your choice for having solved the puzzle.

But in more complex cases you will not have time to solve these types of problems by trial and error method.

Solution:

Let us solve the problem systematically.

Let y be your friend’s age and x be his father’s age since sum of their age is 55 we have

x+y =55

After 16 years, your friend’s age will be y+16 and his fathers age will be x+16.

We are given that that after 16 years, fathers age will be twice that of your friend, thus we have

x+16 =2*(y+16)

I.e. x+16 = 2y+ 32    (On simplification)

I.e. x-2y = 32-16 =16 (By transposition)

Finally we have following two equations

(1) x+y =55

(2) x-2y = 16

To solve a linear equation in one variable we need to have an equation with only one variable.

We should find a method to convert two equations to equation in single variable.

The given equations are

x+y =55   ==è (1)

x-2y=16   ==è (2)

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Subtract (2) from (1)   we get    0+3y =39       ==è (3)

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So 3y = 39 and hence y=13

Since x+y =55 we have x = 55-y (By transposition)

Substituting 13 for y in the above equation we get x=55-13 =42

Thus your friend’s age is 13 and his father’s age is 42 which is exactly what we got by trial and error method!

Verification:

Note:

1.  Now, if your friend’ age is 13 and his father’s age is 42, then the sum of ages is 55

After 16 years your age friend’s age will be 29 and father’s will be 58 which will be twice of his age

This is exactly what is given in the problem and hence our solution is correct.

2. Substitute values of x and y in (1) we get x+y = 42+13 = 55 which is the first equation

Substitute values of x and y in (2) we get x-2y = 42-26 = 16 which is the second equation

2.14 Problem 2: The cost of a geometry box is 18 Rs more than that of a pen. If your class teacher pays Rs 240 for buying 5 geometry boxes and 10 pens, find out the cost of geometry box and pen

Solution:

Let the cost of geometry box be y and the cost of pen be x. We have

(1) y = x+18       ==è(1)

(2) 5y+10x = 240 ==è(2)

Transposition of (1) gives us

y-x =18 and  by  multiplying  this equation by 5  we get

5y-5x= 90         ===è(3)

So we have two equations

5y+10x =240 ------è(2)

5y -5X =   90 ------è(3)

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[Subtract (3) from (2) to eliminate y                               0+15x = 150 ------è(4)

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Thus x = 10 which is cost of the pen.

Since y = x+18, cost of geometry box(y) is Rs.28 (10+18)

Exercise: Verify that these values satisfy the equations (1) and (2).

2.14 Problem 3: Solve 2x+2y =4   and x+y =2

Solution:

2x+2y =4   ===è(1)

x+y =    2  ===è(2)

Multiply (2) by 2 we get

2x+2y=4     ===è(3)

Subtract (1) from (3) we get   0 =0 which is always true

This means that there are many values of x and y which satisfy the given equations and there is no unique solution.

2.14 Problem 4: Solve 2x+2y =4 and x+y = 3

Solution:

Given equations are :                                  2x+2y =4   ====è(1)

x+y = 3  =====è(2)

Multiply(2) by 2 we get

2x+2y=6  =====è(3)

Given   equation:                 2x+2y =4 ===è(2)

Subtracting (1) from (3) we get   0 =2 which is not true

Thus there are no values of x and y which can satisfy the given equations

Definition: Two linear equations in two variables taken together are called ‘simultaneous linear equations’ and are of the form

a1 x+ b1 y = c1

a2 x+b2 y = c2

where a1, b1, a2, b2, c1  and c2 are real numbers and x and y  are variables whose value we are asked to find

How did we solve these simultaneous liner equations?

2.14.1 Method of Elimination by equating co –efficients:

Steps to be followed are:

1.      Multiply the equations (one or both) by suitable numbers such that the coefficients of either x or y are same after multiplication, in both the equations.

2.      Add or subtract these equations to get a resulting equation whose coefficient of x or y is zero so that one of the variable is not present in the resulting equation.

3.      With only one variable present in the resulting equation find the value of that variable.

4.      Substituting this value in any one of the equations find the value of other variable.

Observations:

It is not true that it is always possible to find solutions to all the simultaneous linear equations:

1. a1 x+ b1 y = c1

2. a2 x+b2 y = c2

1. They do not have solution if (a1 / a2) = (b1 / b2) (c1 / c2)

2. They have infinite solutions if (a1 / a2) = (b1 / b2) = (c1 / c2)

3. They have unique solution only if (a1 / a2) (b1 / b2)

2.14 Problem 5:

Solve x+y =2xy   ------à(1)

and

x-y = 6xy           -----à(2)

Solution:

By adding both the equations we get

2x = 8xy

I.e.  1 = 4y y = 1/4

Substituting this in equation (1) we get

x+ 1/4 =  2x/4 = x/2

On transposition we get

x-x/2 = - 1/4 x = -1/2

Verification:

x+y =  -1/2+1/4 = -1/4

2xy =  2*(-1/2)*(1/4) = -1/4 x+y =2xy

x-y =  -1/2-1/4 = -3/4

6xy=  6*(-1/2)*(1/4) = -3/4 x-y = 6xy

2.14 Problem 6:  In an examination the ratio of passes to failures is 4:1.Had 30 less appeared for the examination and had 20 less passed, the ratio of passes to failures would have been 5:1. Find the number of students appeared for the examination.

Solution:

Let x be the number of pass candidates and y be the number of failed candidates

It is given that x/y = 4/1. x=4y

If 30 had appeared less,  and 20 less(x-20) passed then

-         the total number of candidates = x+y-30

-         no of passed candidates =  (x+y-30)(x-20)= y-10

it is given that in such a case ratio of pass to failure is 5:1 (x-20)/(y-10) = 5/1 (x-20) = 5(y-10)

Thus we need to solve the equations

x=4y and

(x-20) = 5(y-10)

Exercise : Solve yourself to get the answer x = 120 and y= 30  and hence number of students appeared for examination = x+y= 150.

2.14 Problem 7:   The sum  of the digits of a two digit number is  9. Nine times this number is twice the number obtained by reversing the the order of the digits of the given number. Find the number

Solution:

Let x be the number in tens place and y be the number in digit place, so that the actual number is 10x+y. the number got by reversing the digits is 10y+x.

The equations to be solved are :

x+y = 9

9(10x+y) =  2(10y+x)

Exercise: solve these two equations to find that x =1 and y=8 so that 18 is the two digit number also verify to note that

1+8 = 9

9*18 =2*81

2.14 Problem 8:    Suppose you intend traveling to your native place from the place you are studying along with your mother.

Since you are student, you are eligible to get the ticket at 50% of the cost of full ticket. However reservation charges are fixed and is common for you as well as for your mother. If the cost of ticket for your mother(including reservation) is Rs 2125 and for both of you it is Rs 3200  find out the full fare of the ticket and the reservation charges

Solution:

Let x be the the cost of full fare and y be the reservation charges x+y = 2125    ----à(1)

Since you are traveling with half the fare of your mother, the fare for you will be 1/2x

Note that you and your mother have to pay same reservation charges and hence the total fare is {(1/2)x+y} + (x+y) which is given to be Rs 3200 {(1/2)x+y}+(x+y) = 3200 (3/2)x+2y =3200; multiplying bothe sides by 2 we get

3x+4y =6400  -----à(2)

On solving equations (1) and (2) we get {multiply (1) by 3 and then subtract the result from (2)}

We get y =25 and x= 2100

The full fare of the ticket is Rs.2100 and reservation charge is Rs.25

Verification:

Note that In case of your mother, full fare+ reservation charge=2100+25

In case of you and your mother fare =  half fare + reservation +full fare +reservation = 1050+25+2100+25=3200

These are the values given in the problem and hence our solution is correct

2.14 Summary of learning

 No Points studied 1 Simultaneous linear equations( a1 x+ b1 y = c1, a2 x+b2 y = c2)are solved by reducing them  to  equations of single variable by appropriate  transpositions 2 It is not always possible to solve  all simultaneous linear equations

2.14 Additional points:

2.14.2 Alternate method: Method of elimination by substitution.

Simultaneous linear equations can also be solved using an alternate method:

1.      From any one of the given equations, find the value of one variable(y) in terms of the other variable(x).

2.      Substitute the value of the variable(y) arrived in Step 1, in the other equation, to find the value of the other variable(x).

3.      Substitute the value of the variable(x) arrived in step 2, in one of the equations to get the value of the remaining variable(y).

Let us solve problem 2.14.2 (previously solved) using this alternate method.

Solve

5y+10x =240     ----à(1)

5y -5X =   90    ----à(2)

Let us take the first equation, 5y+10x = 240 and find the value of y in terms of x.

5y = 240-10x y = 48-2x

Let us substitute this value of y in the 2nd equation, 5y-5x = 90.

LHS = 5y - 5x = 5(48-2x) - 5x  = 240-10x-5x = 240-15x

Since LHS = 90(RHS) we have

240-15x = 90

i.e. 240-90 = 15x

i.e. 150 = 15x x = 10

Substituting this value of x in the 1st equation we get

5y+10*10 = 240

i.e. 5y = 240-100=140 y = 28

This is exactly the same solutions that were obtained in 2.14.Problem 2.

2.14.3 Solving linear equations in three variables.

We have learnt that, to solve linear equations in two variables we need two equations. Extending this logic, to solve linear equations in three variables we will need three equations.

Steps to be followed are:

1.      Take any two equations to eliminate the third variable

2.      Repeat step 1 by taking another pair of equations different from the one used in step 1

3.      Solve the resulting two linear equations using any of the methods discussed above

4.      Substitute the values of the two variables obtained in step 3, in any one of the 3 equations to arrive at the value of the third variable

2.14 Problem 8:    Solve

2/x + 3/y - 4/z = -20

2/y - 4/x + 3/z = 45

3/x - 4/y + 2/z=5

Solution:

Let a=1/x, b=1/y and c=1/z

Then the give equations will become

2a+3b-4c = -20     -----à(1)

-4a+2b+3c = 45    ------à(2)

3a-4b+2c = 5        -----à(3)

By multiplying equations (1) by 3 and (2) by 4 we get

6a+9b-12c = -60

-16a+8b+12c = 180

By adding the above two equations we get

-10a+17b = 120     ----à(4)

By multiplying equations (1) by 1 and (3) by 2 we get

2a+3b-4c = -20

6a-8b+4c =   10

By adding the above two equations we get

8a-5b = -10          ----à(5)

By multiplying equations (4) by 8 and (5) by 10 we get

-80a+136b = 960

80a-50b = -100

By adding the above two equations we get

86b = 860 b = 10

Substituting b=10 in (4) we get

-10a+170 = 120 -10a = -50 a = 5

Substituting a=5 and b=10 in (1) we get

10+30-4c =-20 -4c = -60 c = 15

Since a=5, b=10 and c=15 it follows that

x=1/5, y=1/10 and z=1/15

Verification:

Substitute these values in the given equations to confirm that our solution is correct.

2.14 Problem 8: What 3 carpenters earn in a day is earned by 4 male workers in a day. The daily wages of 4 male workers is equal to the daily wages of one carpenter and 4 female workers. If one carpenter, 2 male workers and 5 female workers are engaged for a day, their total wages is Rs 500. Find the daily wages of carpenter, male worker and female worker.

Solution:

Let c, m and f be the daily wages of carpenter, male worker and female worker respectively.

Since 3 carpenters’ wage is equal to the wages of 4 male workers,

3c = 4m        ----à(1)

Since daily wages of 4 male workers is equal to daily wages of a carpenter and 4 female workers,

4m = c+4f   ----à(2)

Since daily wages of 1 carpenter, 2 male workers and 5 female workers is 500,

c+2m+5f = 500  ---à(3)

By substituting 4m = 3c in (2) we get

3c = c+4f 2c = 4f or c=2f

Substituting c=2f in (2) we get

4m =2f+4f 2m=3f

Substituting 2m=3f and c=2f in (3) we get

2f+3f+5f =500

10f = 500 f = 50

Substituting this value of f in 2m=3f and c=2f we get c = 100 and m = 75.

Hence the daily wages of a carpenter, male worker and female worker is Rs 100, Rs 75 and Rs 50 respectively.         