Question

x/a + y/b = 2 , ax - by = a^2 - b^2 using elimination method.
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Answer 1

Step-by-step explanation:

Given :-

(x/a) + (y/b) = 2

ax - by = a²-b²

To find :-

Solve the given equations by Elimination method ?

Solution :-

Given pair of linear equations in two variables are

(x/a) + (y/b) = 2 -----------(1)

=> (bx+ay) / (ab) = 2

=> bx+ay = 2ab

On multiplying with a both sides then

=> abx + a²y = 2a²b --------(2)

and

ax - by = a²-b² -----------(3)

On multiplying with b both sides then

=> abx - b²y = a²b-b³ -------(4)

On Subtracting (2) from (4)

abx - b²y = a²b-b³

abx + a²y = 2a²b

(-) (-) (-)

_____________________

0 -b²y-a²y = (a²b-b³)-2a²b

_____________________

=> -b²y-a²y = (a²b-b³)-2a²b

=> y(-b²-a²) = -b³-a²b

=> -y(b²+a²) = -(b³+a²b)

=> y(a²+b²) = b³+a²b

=> y = (b³+a²b)/(a²+b²)

=> y = b(b²+a²)/(a²+b²)

=> y = b

On Substituting the value of y in (3) then

ax -by = a²-b²

=> ax -b(b) = a²-b²

=> ax -b² = a²-b²

=> ax = a²-b²+b²

=> ax = a²

=> x = a²/a

=> x = a

Therefore, x = a and y = b

Answer:-

The solution for the given pair of linear equations in two variables is (a,b)

Check :-

If x = a and y = b then LHS of equation x/a + y/b =2

=> (a/a)+(b/b)

=> 1+1

=> 2

LHS = RHS

and

If x = a and y = b then LHS of equation ax - by = a²-b²

=> a(a) -b(b)

=> a²-b²

=> RHS

LHS = RHS is true for x = a and y = b

Used Method :-

→ Elimination Method