Solve the following pair of linear equations by the elimination method and the substitution method: (i) x + y = 5 and 2x – 3y = 4 (ii) 3x + 4y = 10 and 2x – 2y = 2
Solve the following pair of linear equations by the elimination method and the substitution method:
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(i) x + y = 5 and 2x – 3y = 4
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
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