Question

pls help me with this question

Answer 1

Understanding the question:-

If we subtract a specific number from -8xy + 2x² + 3y², we will get x² + 2.

Solution :-

Let the unknown number be A.

Now, according to question,

(-8xy + 2x² + 3y²) - ( A ) = x²+2

-8xy + 2x² + 3y² - A = x² + 2

-8xy + 2x² - x² + 3y² - A = 2

-8xy + x² + 3y² - A = 2

A = -8xy + x² + 3y² - 2

So, if we subtract -8xy + x² + 3y² - 2 from -8xy + 2x² + 3y², we will get x²+2.

━━━━━━━━━━━━━━━━━

Check:-

(-8xy + 2x² + 3y²) - (-8xy + x² + 3y² - 2) = x²+2

-8xy + 2x² + 3y² + 8xy - x² - 3y² + 2 = x² + 2

-8xy + 8xy + 2x² -x² + 3y² - 3y² + 2 = x² + 2

x² + 2 = x² + 2

Hence Proved!

And so that our answer is correct.

Answer 2

$\huge\sf{\underline{ \bold{Q}uestion}}:$

What should be subtracted from (-8xy+2x²+3y²) to get (x²+2) ?⠀⠀

$\large\bf{\underline{To \: find}} \: –$

A number which when gets subtracted from -8xy+2x²+3y² gives the result as x²+2

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\huge\bold{\underline{\underline{Solution}}}:$

✯ let the number be 'p'.

Acc. to the question,

$\sf (- 8xy + {2x}^{2} + {3y}^{2}) - (p) = {x}^{2} + 2 \\ \\ \implies \sf - 8xy + {2x}^{2} + {3y}^{2} (- p) = {x}^{2} + 2 \\ \\ \implies \sf- 8xy + {2x}^{2} - {x}^{2} + {3y}^{2} (- p) = 2 \\ \\ \implies \sf- 8xy + { {x}^{2} + {3y}}^{2} (- p) = 2 \\ \\ \implies \purple{\boxed{\bf\:p = - 8xy + {x}^{2} + {3y}^{2} - 2}}$

∴ If we subtract (-8xy + x² + 3y² -2) from (-8xy + 2x² + 3y²), we get (x²+2).

_____________________________

⠀

$\huge\green{\mathbb{\underline{{V}\large{ERIFICATION}}}:}$

Here, the equation is –

(-8xy + 2x² + 3y²) - (-8xy + x² + 3y² - 2) = (x²+1)

⠀⠀⠀⠀⠀⠀⠀⠀⠀

♦ LHS :

$\sf (- 8xy + {2x}^{2} + {3y}^{2} ) - ( - 8xy + {x}^{2} + {{3y}}^{2} - 2) \\ \implies \sf - {{\large\not} 8xy} + {2x}^{2} + {{\large{\not}{3y}^{2}}} + {{ \large\not}8xy} - {x}^{2} - {{\large\not}{3y}^{2}} + 2 \\ \implies \sf {2x}^{2} - {x}^{2} + 2 \\ \implies \sf{\underline{{x}^{2} + 2}}$

♦ RHS :

$\implies \sf{ \underline{ {x}^{2} + 2}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

LHS = RHS

hence verified!

⠀⠀⠀⠀i.e, our answer is correct :)

⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀✨HOPE IT HELPS✨