Question

Solve the following system of equations by the elimination method:

`x/a + y/b = a + b, x/a^2 + y/b^2 = 2, a ≠ 0, b ≠ 0`

Answer 1

Given the system of equations:

`x/a + y/b = a + b`   ...(i)

`x/a^2 + y/b^2 = 2`   ...(ii)

Where (a ≠ 0), (b ≠ 0).

Step 1: Eliminate the denominators by multiplying appropriately

Multiply equation (i) by (ab) to clear denominators:

bx + ay = ab(a + b)   ...(iii)

Multiply equation (ii) by (a²b²) to clear denominators:

b²x + a²y = 2a²b²   ...(iv)

Step 2: Use the elimination method to solve for (x) and (y)

Multiply (iii) by (a):

abx + a²y = a²b(a + b)   ...(v)

Multiply (iii) by (b): 

b²x + aby = ab²(a + b)   ...(vi)

Note here, the original (iv) is b²x + a²y = 2a²b².

We want to eliminate one variable; let’s eliminate (y).

Multiply (iii) by (a):

abx + a²y = a²b(a + b)

Multiply (iv) by (–a):

ab²x – a³y = –2a³b²

Add the two equations:

(abx – ab²x) + (a²y – a³y) = a²b(a + b) – 2a³b² 

abx(1 – b) + a²y(1 – a) = a²b(a + b) – 2a³b²

Alternative elimination approach:

From (iii) and (iv):

bx + ay = ab(a + b)   ...(iii) 

b²x + a²y = 2a²b²   ...(iv)

Multiply (iii) by (b):

b²x + aby = ab²(a + b)   ...(v)

Multiply (iii) by (a):

abx + a²y = a²b(a + b)   ...(vi)

Subtract (vi) from (iv):

(b²x + a²y) – (abx + a²y) = 2a²b² – a²b(a + b) 

b²x – abx = 2a²b² – a²b(a + b) 

x(b² – ab) = a²b(2b – (a + b)) = a²b(b – a) 

x(b(b – a)) = a²b(b – a) 

Divide both sides by b(b – a), assuming (b ≠ 0) and (b ≠ a):

`x = (a^2b(b - a))/(b(b - a))`

x = a²

Step 3: Find (y) by substituting x = a² in (iii):

bx + ay = ab(a + b)

b(a²) + ay = ab(a + b) 

ay = ab(a + b) – ba²

ay = aba + ab² – a²b

ay = ab² 

`y = (ab^2)/a`

y = b² 

Thus, the solution to the system by elimination method is x = a² and y = b².