Question

If ax + by = a²-b² and bx + ay = 0 ,find the value of ( x+ y).​

Answer 1

Solution :-

Given pair of linear equations :

→ ax + by = a² + b²

→ bx + ay = 0

Finding the value of x from 2nd equation

⇒ bx + ay = 0

⇒ bx = - ay

⇒ x = - ay/b

Substituting x = - ay/b in ax + by = a² - b²

⇒ a(-ay/b) + by = a² - b²

⇒ - a²y/b + by = a² - b²

⇒ ( - a²y + b²y)/b = a² - b²

⇒ [ - y(a² - b²) ] /b = a² - b²

⇒ - y/b = (a² - b²) / (a² - b²)

⇒ - y/b = 1

⇒ - y = b

⇒ y = - b

Substituting y = - b in x = - ay/b

⇒ x = - a(-b)/b

⇒ x = ab/b

⇒ x = a

Now, x + y

x + y = a + ( - b ) = a - b

Therefore the value of (x + y) is (a - b)

Answer 2

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Given:

If ax + by = a² - b² and bx + ay = 0.

To find:

The value of (x+y).

Explanation:

We have equation,

• ax + by = a² - b²......................(1)
• bx + ay = 0..............................(2)

Using substitution Method:

From equation (2),we get;

→ bx + ay = 0

→ bx = -ay

→ x = $\frac{-ay}{b}$........................(3)

Putting the value of x in equation (1),we get;

→ $a(\frac{-ay}{b} )+by=a^{2} -b^{2}$

→ $\frac{-a^{2}y }{b} +by=a^{2} -b^{2}$

→ -a²y + b²y = a²b - b³

→ -y(a² - b²) = b(a² - b²)

→ y= -b

Putting the value of y in equation (3), we get;

→ x = $\frac{-a(-b)}{b}$

→ x = $\frac{a\cancel{b}}{\cancel{b} }$

→ x = a

Now,

→ x + y = a + (-b)

→ x + y = a - b

Thus,

The value is a-b.