(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) x/2 + 2y/3 = -1 and x – y/3 = 3
Solution:
(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
Therefore, 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
Therefore, 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
Therefore, 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