Question

x+y=6 and x-y=4 find x^2+y^2

Answer 1

$\huge \boxed{ \mathbb{ \ulcorner ANSWER: \urcorner}}$



$\huge \bf \sf{It \: is \: given \: that:-)}$


→ x + y = 6..............(1).

→ x - y = 4................(2).


➡ Add in equation (1) and (2).


x + y = 6.
x - y = 4.
(+)..(-)....(+)
________
=> 2x = 10.

$=> x = \frac{10}{2} .$


$\huge \boxed{=> x = 5.}$



➡ Put the value of ‘x’ in equation (1).


=> 5 + y = 6.

=> y = 6 - 5.

$\huge \boxed{=> y = 1.}$


▶ Now, A/Q.


$\huge \bf = {x}^{2} + {y}^{2} .$



$\bf = {(5)}^{2} + {(1)}^{2} .$


$\bf = 25 + 1.$


$\huge \boxed{ = 26.}$


✔✔ Hence, it is founded ✅✅.

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Answer 2

here is your answer OK


Given,

x + y= 6 (equation 1)

x - y = 4 (equation 2)

Squaring and adding both the equations,

(x + y)² + (x - y)² = 6² + 4²

=> x² + y² + 2xy + x² + y² - 2xy = 52

=> 2(x²+y²) = 52

=> x²+y²=26