Question

If x-y =3 and x^2+y^2 = 10 , find xy is
(With steps)

Answer 1

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Answer 2

Given:-

• $x - y = 3$
• ${x}^{2} + {y}^{2} = 10$

To find:-

• The value of xy.

Answer:-

Given that,

$x - y = 3$

Squaring both sides,

$\longrightarrow {(x - y)}^{2} = {3}^{2}$

Expanding L.H.S. using (a - b)² = a² + b² - 2ab with a = x and b = y,

$\longrightarrow {x}^{2} + {y}^{2} - 2xy = 9$

$\longrightarrow ({x}^{2} + {y}^{2}) - 2xy = 9$

Putting the value fo $x^2+y^2=10$ as given in the question,

$\longrightarrow 10 - 2xy = 9$

$\longrightarrow 2xy = 10 - 9$

$\longrightarrow 2xy = 1$

$\longrightarrow \underline{ \underline{xy = \dfrac{1}{2} }}$

Extra Knowledge:-

• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab
• a² - b² = (a + b)(a - b)
• (a + b)³ = a³ + b³ + 3ab(a + b)
• (a - b)³ = a³ - b³ - 3ab(a - b)
• a³ + b³ = (a + b)(a² + b² - ab)
• a³ - b³ = (a - b)(a² + b² + ab)
• (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
• a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)