Question

1. Solve for x and y:
x + y = a + b
ax - by = a2 – b2​

Answer 1

Answer

x = a

y = b

Explanation

We have,

x + y = a + b

→ x = a + b - y

Also, ax - by = a² - b²

Put the value of x as (a + b - y)

→ a(a + b - y) - by = a² - b²

→ a² + ab - ay - by = a² - b²

→ (- y)(a + b) = a² - b² - a² - ab

→ ( - y)(a + b) = -b² - ab

→ (- y)(a + b) = (- b)(b + a)

Cancel (a + b) from both sides

→ (- y) = (- b)

Cancel negative sign from both sides,

→ y = b

Put the value of y in first equation

x + y = a + b

→ x + b = a + b

→ x = a

Answer 2

Answer:

$\large\boxed{\sf{x=a\:,\:y=b}}$

Step-by-step explanation:

From the Question,

$x + y = a + b \\ \\ = > y = a + b - x \: \: \: \: \: \: \: ........(1)$

Also, from the Question,

$ax - by = {a}^{2} - {b}^{2}$

Putting the value of y from eqn (1) ,

$= > ax - b(a + b - x) = {a}^{2} - {b}^{2} \\ \\ = > ax - ab - \cancel{{b}^{2}} + bx = {a}^{2} - \cancel{ {b}^{2}} \\ \\ = > ax - ab + bx = {a}^{2} \\ \\ = > x(a + b) = {a}^{2} + ab \\ \\ = > x \cancel{(a + b) }= a \cancel{(a + b)} \\ \\ = > \sf{ x = a}$

Now, put the value of x in eqn (1),

$= > y = a + b - a \\ \\ = > \sf{y = b}$