Question 1.
Solve the following pairs of linear equations by the elimination method and the substitution method:
Solution:
By the method of elimination.
x + y = 5 ……………………………….. (i)
2x – 3y = 4 ……………………………..(ii)
When the equation (i) is multiplied by 2, we get
2x + 2y = 10 ……………………………(iii)
When the equation (ii) is subtracted from (iii) we get,
5y = 6
y = 6/5 ………………………………………(iv)
Substituting the value of y in eq. (i) we get,
x = 5 – 6 / 5 = 19/5
∴ x = 19/5 , y = 6/5
By the method of substitution.
From the equation (i), we get:
x = 5 – y………………………………….. (v)
When the value is put in equation (ii) we get,
2 (5 – y) – 3y = 4
–5y = -6
y = 6/5
When the values are substituted in equation (v), we get:
x =5 – 6/5 = 19/5
∴ x = 19/5, y = 6/5
(ii) 3x + 4y = 10 and 2x – 2y = 2
By the method of elimination.
3x + 4y = 10……………………….(i)
2x – 2y = 2 ………………………. (ii)
When the equation (i) and (ii) is multiplied by 2, we get:
4x – 4y = 4 ………………………..(iii)
When the Equation (i) and (iii) are added, we get:
7x = 14
x = 2 ……………………………….(iv)
Substituting equation (iv) in (i) we get,
6 + 4y = 10
4y = 4
y = 1
Hence, x = 2 and y = 1
By the method of Substitution
From equation (ii) we get,
x = 1 + y……………………………… (v)
Substituting equation (v) in equation (i) we get,
3(1 + y) + 4y = 10
7y = 7
y = 1
When y = 1 is substituted in equation (v) we get,
A = 1 + 1 = 2
Therefore, A = 2 and B = 1
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
By the method of elimination:
3x – 5y – 4 = 0 ………………………………… (i)
9x = 2y + 7
9x – 2y – 7 = 0 …………………………………(ii)
When the equation (i) and (iii) is multiplied we get,
9x – 15y – 12 = 0 ………………………………(iii)
When the equation (iii) is subtracted from equation (ii) we get,
13y = –5
y = –5/13 ………………………………………….(iv)
When equation (iv) is substituted in equation (i) we get,
3x + 25/13 – 4=0
3x = 27/13
x = 9/13
∴ x = 9/13 and y = – 5/13
By the method of Substitution:
From the equation (i) we get,
x = (5y + 4) / 3 ……………………………… (v)
Putting the value (v) in equation (ii) we get,
9 (5y + 4) / 3 – 2y –7= 0
13y = – 5
y = – 5/13
Substituting this value in equation (v) we get,
x = {5 ( –5 / 13 ) + 4} / 3
x = 9 /13
∴ x = 9/13, y = – 5/13
(iv) x / 2 + 2y / 3 = –1 and x – y / 3 = 3
By the method of Elimination.
3x + 4y = – 6 …………………………. (i)
x – y / 3 = 3
3x – y = 9 ……………………………. (ii)
When the equation (ii) is subtracted from equation (i) we get,
–5y = –15
y = 3 ………………………………….(iii)
When the equation (iii) is substituted in (i) we get,
3x – 12 = –6
3x = 6
x = 2
Hence, x = 2 , y = – 3
By the method of Substitution:
From the equation (ii) we get,
x = (y + 9) / 3…………………………………(v)
Putting the value obtained from equation (v) in equation (i) we get,
3 (y + 9) / 3 + 4y = – 6
5y = –15
y = –3
When y = – 3 is substituted in equation (v) we get,
x = (–3 + 9) / 3 = 2
Therefore, x = 2 and y = – 3
Question 2. Form the pair of linear equations for the following problems and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 12, if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw Rs2000. She asked the cashier to give her Rs50 and Rs100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
