Method of Solving Problems Based On Simultaneous Equations –
Step.1) We have to read the problem carefully and then have to try to know about unknowns, Take these unknowns as x & y.
Step.2) According to given conditions, we have to form linear equations in x & y.
Step.3) Now we have to solve obtained equations simultaneously.
There are some examples are given below for your better understanding.–
Example.1) The sum of two numbers is 35 and the difference is 10. Find the numbers.
Ans.) Let two numbers are x & y
So, now according to the given conditions x + y = 35 ……………….. (i)
x – y = 15 ………………. (ii)
adding (i) and (ii) we get –
x + y = 35
x – y = 15
----------------
2x = 50
=> x = 25
Now we substitute the value of x in (i), and we find –
x + y = 35 ……………….. (i)
=> 25 + y = 35
=> y = 35 – 25 = 10
Hence the required two numbers are 25 & 10 (Ans.)
Example.2) A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the original number. Find the original number.
Ans.) Let the required number contain x at ten’s place and y at unit’s place.
Then, we get the number = 10x + y
And, as per the condition => 10x + y = 7(x + y)
=> 10x + y = 7x + 7y
=> 10x – 7x = 7y – y
=> 3x = 6y
=> x = 2y ………………….(i)
As per the given condition, number formed on reversing the digits and we get = 10y + x
And, as per condition => (10x + y) – (10y + x) = 18
=> 10x + y - 10y - x = 18
=> 9x - 9y = 18
=> x - y = 2 .....................(ii)
now, we will substitute x in (ii), and we get -
=> 2y - y = 2
=> y = 2
now, we will substitute the value of y in (i), and we get -
x - y = 2
=> x - 2 = 2
=> x = 4
Hence the original number is 42 (Ans.)
Example.3) Seven times a two digit number is equal to four times the number obtained by reversing the order of it’s digits. If the difference between the digits is 3, find the number.
Ans.) Let the required number contain x at ten’s place and y at unit’s place.
Then, the obtained number would be => (10x + y)
Number obtained by reversing the number, we get 10y + x
As per the given condition => 7(10x + y) = 4(10y + x)
=> 70x + 7y = 40y + 4x
=> 66x - 33y = 0
=> 2x - y = 0 ......................(i)
Unit digit = 2 (tens digit) => y > x
So, y – x = 3 …………………(ii)
now we will add (i) with (ii), and we get -
2x - y = 0
– x + y = 3
----------------
x = 3
Now, we will substitute the value of x in (i), and we will get -
2x - y = 0
=> 6 - y = 0
=> y = 6
Hence, the number is 36 (Ans.)
Example.4) A two digit number is such that the product of it’s digits is 16. If 54 is added to the number, the digits interchange their places. Find the number.
Ans.) Let the required number contain x at ten’s place and y at unit’s place.
As per the given condition xy = 16
Required number is = (10x + y)
Number obtained on reversing the digits = (10y + x)
So, as per the given condition –
(10x + y) + 54 = (10y + x)
=> (10y + x – 10x – y) = 54
=> (9y – 9x) = 54
=> 9(y –x ) = 54
=> y – x = 6 ………………… (i)
Now, as per formulae –
[(y + x)² - (y – x)²] = 4xy
Now, we will substitute the value of xy and (y – x) which is 16 & 6 respectively, and we get –
[(y + x)² - (y – x)²] = 4xy
=> (y + x)² = 4xy + (y – x)²
=> (y + x)² = (4 X 16) + (6)²
=> (y + x)² = 64 + 36 = 100
=> (y + x)² = 10²
=> (y + x) = 10 ……………………. (ii)
Add (i) & (ii), and we find –
y – x = 6
y + x = 10
-----------------
2y = 16
=> y = 8
Now, we will substitute the value of y in (ii) and we find –
y + x = 10
=> 8 + x = 10
=> x = 2
So, we obtained here xy = 28
Hence the required number is 28 (Ans.)
Example.5) The sum of the numerator and denominator of a fraction is 8. If 4 is added to both of the numerator and the denominator, the fraction becomes 4/5. Find the fraction.
x
Ans.) Let the required fraction is ------
y
as per the give condition, x + y = 8 …………………..(i)
x + 4 4
and, ---------- = -------
y + 4 5
or, 5(x + 4) = 4(y + 4)
or, 5x + 20 = 4y + 16
or, 5x – 4y = - 4 …………………… (ii)
now, we will multiply 4 with (i), and we get –
4x + 4y = 32 ………………… (iii)
Now, we will add (iii) and (ii), and will find –
4x + 4y = 32
5x – 4y = - 4
------------------
9x = 28
=> x = 28/9
Now, we will substitute the value of x in (i) and we get –
x + y = 8
28
------- + y = 8
9
28 72 – 28 44
y = 8 – -------- = ------------ = --------
9 9 9
x 28/9
So, the required fraction is = ------- = ---------
y 44/9
28 44
= -------- ÷ --------
9 9
28 9 7
= -------- X -------- = -------
9 44 11
So, the obtained fraction is 7/11 (Ans.)
Example.6) If the numerator of a fraction is increased by 3 and its decreased by 2, then it becomes 4/5. If the numerator is decreased by 1 and denominator increased by 2, then it becomes 2/3. Find the fraction.
x
Ans.) Let the required fraction would be = -------
y
x + 3 4
As per the given condition, ----------- = ------
y – 2 5
or, 5(x + 3) = 4 (y – 2)
or, 5x + 15 = 4y – 8
or, 5x – 4y = - 15 – 8
or, 5x – 4y = – 23 …………………… (i)
4y – 23
or, x = -----------
5
x – 1 2
As per the given condition, --------- = ------
y + 2 3
or, 3(x – 1) = 2(y + 2)
or, 3x – 3 = 2y + 4
or, 3x – 2y = 7 …………….. (ii)
now, you will multiply – 2 with (ii), and we get –
or, - 6x + 4y = - 14 …………….. (iii)
now, we will add (i) & (iii), and we get –
5x – 4y = – 23
- 6x + 4y = – 14
--------------------
- x = - 37
Or, x = 37
Now we will substitute the value of ‘x’ in (i), and we get –
5x – 4y = – 23
=> 5x + 23 = 4y
=> 4y = (5 X 37) + 23
=> 4y = 208
=> y = 52
37
So, the required fraction would be ------ (Ans.)
52
Example.7) The students of a class are made to stand in complete rows. If one student is extra in each row, there would be 2 rows less, and if one student is less in each row, there would be 3 rows more. Find the number of students in the said class.
Ans.) Let, there would be ‘x’ rows, and each containing ‘y’ students. Then total number of students is xy
Case.1 – as per the given condition, if the number of students are (y + 1), then the numbers of rows would be (x – 2). So, now total number of students = (x – 2) (y + 1)
So, now xy = (x – 2) (y + 1)
=> xy = xy – 2y + x – 2
=> 2y – x = – 2
=> x = 2y + 2
=> x – 2y – 2 = 0 …………………(i)
Case.2 – as per the given condition, there are (x + 3) rows, then each rows would have (y – 1) students.
The total number of students = (x + 3) (y – 1)
Hence, xy = (x + 3) (y – 1)
=> xy = xy + 3y – x – 3
=> x – 3y + 3 = 0 …………………….(ii)
Now, we will substitute (ii) from (i)
x – 2y – 2 = 0
x – 3y + 3 = 0
- + -
--------------------
y – 5 = 0
=> y = 5
Now, we will substitute the value of y in (i), and we get –
x – 2y – 2 = 0
=> x = 2y + 2 = (2 X 5) + 2
=> x = 12
Hence, originally there are 12 rows, each containing 5 students, so the total number of students in the class = 12 X 5 = 60 students (Ans.)
Example.8) In an examination, the ratio of passes to failure was 2 : 1. If there 30 students less appeared and 20 less passed, the ratio of passes to failure would have been 4 : 1. How many students appeared for the examinations?
Ans.) Suppose, x no. of candidates are passed and y no. of candidates are failed.
x 2
Case.1) As per given condition, ------- = -------
y 1
=> x = 2y ………………….. (i)
total number of students would be - (x + y)
Case.2) When number of students appeared = (x + y – 30) and number of students passed = (x – 20)
Now, the number of failure students = {(x + y – 30) - (x – 20)}
= x + y – 30 – x + 20
= (y – 10)
Now, as per the given condition –
(x – 20) 4
------------ = -------
(y – 10) 1
Or, x – 20 = 4 (y – 10) …………………(ii)
Now, we will substitute the value of x in above obtained equation (ii), and we find -
Or, 2y – 20 = 4y – 40
Or, 4y – 2y = 40 – 20
Or, 2y = 20
Or, y = 10
Now, we will substitute the value of y in (i), and we find –
x = 2y
=> x = 20
Now, the number of students appeared in the examination =>
(x + y) = (20 + 10) = 30 students.
Hence the total number of students is 30 (Ans.)