Question

If x + y = 9 and x2 + y2 = 49, then the value of xy will be.

Answer 1

GIVEN :-

• x + y = 9 and x² + y² = 49.

TO FIND :-

• value of xy.

SOLUTION :-

$\implies \displaystyle \sf \:x + y = 9$

On squaring both the sides we get,

$\implies \displaystyle \sf \:(x + y) ^{2} = (9) ^{2}$

Now by using identity i.e (a + b)² = a² + b² + 2ab.

$\implies \displaystyle \sf \:x ^{2} + y ^{2} + 2 \times xy = 81$

$\implies \displaystyle \sf \:49 + 2 \times xy = 81$

$\implies \displaystyle \sf \:2 \times xy = 81 - 49$

$\implies \displaystyle \sf \:2 \times xy = 32$

$\implies \displaystyle \sf \:xy = \frac{32}{2}$

$\implies \underline{ \boxed{\displaystyle \sf \:xy = 16}}$

Answer 2

Answer:

CORRECT QUESTION :-

If x + y = 9 and x² + y² = 49 then find the value of xy

Given :-

• x + y = 9
• x² + y² = 49

To Find :-

Value of xy

Solution :-

According to the question

$\large \sf \bigg \lgroup \: x + y = 9 \bigg \rgroup \: ...1$

$\large \bigg \lgroup \sf {x}^{2} + {y}^{2} = 49 \bigg \rgroup \: ...2$

Now,

By using identity

(a² + b²) = a² + b² + 2ab

Now,

Rewriting

(a + b)² - (a² + b²) = 2ab

9² - 49 = 2xy

81 - 49 = 2xy

32 = 2xy

xy = 32/2

xy = 16