Question

Solve the following system of equations
1/2(x+2y)+5/3(3x-2y)=-3/2
5/4(x+2y)-3/5(3x-2y)=61/60

Answer 1

Let x + 2y = a, 3x - 2y = b;

Then, 1/2a + 5/3b = -3/2 -- (i)

and 5/4a - 3/5b = 61/60 -- (ii)

Multiply eqn. (i) by 5, then it becomes:

5/2a + 25/3b = -15/2 -- (iii)

Similarly multiply eqn. (ii) by 2, then it becomes:

10/4a - 6/5b = 61/30

= 5/2a - 6/5b = 61/30 -- (iv)

Subtracting (iv) from (iii), we get:

25/3b - ( -6/5b) = -15/2 - 61/30

25/3b + 6/5b = -286/30

(125b + 18b)/(3b.5b) = -143/15 (on taking the LCM)

143b/15b2 = -143/15

If b not equal to zero, then b and b cancel out giving:

143/15b = -143/15

1/b = -1

b = -1

We know that b cannot be zero, as if b = 0, the eqn. (i), (ii) and rest will be of indeterminate form (because anything divided by zero is not defined)

So b = -1

Substituting in (i),

1/2a + 5/(-1).3 = -3/2

1/2a - 5/3 = -3/2

1/2a = 5/3 - 3/2

1/2a = 1/6 (taking LCM)

Hence a = 3

Now we know a = x + 2y = 3 -- (v)

and b = 3x - 2y = -1 -- (vi)

Now solving by normal pair of linear eqns. in two var.s,

adding (v) and (vi) we get: 4x = 3 - 1

x = 2/4 = 1/2

Substituting in (v),

1/2 + 2y = 3

(1 + 4y)/2 = 3 (LCM)

1 + 4y = 6

So y = (6-1)/4 = 5/4

Therefore x = 1/2 and y = 5/4