Question

a²x+b²y = c²
b²x + a²y = d²​

Answer 1

Answer:

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✤ Required Answer:

✒️ GiveN:

Equation 1) a²x + b²y = c²

Equation 2) b²x + a²y = d²

✒️ To FinD:

Solving for x and y

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✤ How to solve?

Here, we have two simultaneous linear equations in two variables i.e x and y. We can see that there coefficients are a² and b² alternatively, and hence we can use the elimination method to solve this equation.

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✤ Solution:

We have,

a²x + b²y = c² ---------(1)

b²x + a²y = d² ----------(2)

Multiplying eq.(1) with b²

➝ b²(a²x + b²y) = b²c²

➝ a²b²x + b⁴y = b²c²

Now, multiplying eq.(2) with a²

➝ a²(b²x + a²y) = a²d²

➝ a²b²x + a⁴y = a²d²

Subtracting the equations will got above,

a²b²x + b⁴y = b²c²

a²b²x + a⁴y = a²d²

➝ a²b²x + b⁴y - (a²b²x + a⁴y) = b²c² - a²d²

➝ a²b²x + b⁴y - a²b²x - a⁴y = b²c² - a²d²

[ +a²b²x and -a²b²x cancels out]

➝ b⁴y - a⁴y = b²c² - a²d²

➝ y(b⁴ - a⁴) = b²c² - a²d²

➝ y = b²c² - a²d² / b⁴ - a⁴

➝ y = a²d² - b²c² / a⁴ - b⁴

Putting the value of y in eq.(1),

➝ a²x + b²(b²c² - a²d² / b⁴ - a⁴) = c²

➝ a²x + b⁴c² - a²b²d² / b⁴ - a⁴ = c²

➝ a²x = c² - (b⁴c² - a²b²d² / b⁴ - a⁴)

[ Taking LCM in RHS ]

➝ a²x = c²(b⁴ - a⁴) - b⁴c² + a²b²d / b⁴ - a⁴

➝ a²x = b⁴c² - a⁴c² - b⁴c² + a²b²d² / b⁴ - a⁴

➝ a²x = -a⁴c² + a²b²d² / b⁴ - a⁴

[ Taking a² common from both sides ]

➝ a²x = a² ( -a²c² + b²d² / b⁴ - a⁴ )

➝ x = -a²c²+ b²d² / b⁴ - a⁴

➝ x = a²c² - b²d² / a⁴ - b⁴

❒ So, Values of x and y:

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Answer 2

Answer:

answer for the given problem is given