Question

If xy - 3y² = 15. find the value of x when y = 2​

Answer 1

Answer:

the value of x is 0 multiply both sides by zero and you'll get answer

Answer 2

$\sf\Large{\underline{\underline{SOLUTION:-}}}$

★ Given,

• xy – 3y² = 15
• value of y is 2
• Find the value

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⠀

EQUATION :-

$\begin{gathered}\\\;\sf{:\rightarrow\;\;\;\bf{{xy - 3y^{2} = 15}\:}}\end{gathered}$

We have y = 2, So putting down the value of y.

⠀

$\: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;\;\bf{{x(2) - 3( {2}^{2} ) = 15}\:}}\end{gathered}$

$\: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;\;\bf{{2x- 3(4 ) = 15}\:}}\end{gathered}$

$\: \: \: \: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;\;\bf{{2x- 12 = 15}\:}}\end{gathered}$

Here, I'm transposing LHS to RHS, while transposing the terms sign changes inversely. ∴ – 12 will become +12.

$\: \: \: \: \: \: \: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;\;\bf{{2x = 15 \: + \: 12}\:}}\end{gathered}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;\;\bf{{2x = 27}\:}}\end{gathered}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;\;\bf{{x = \: \frac{27}{2} }\:}}\end{gathered}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;\;\bf{{x = \: 13.5 }\:}}\end{gathered}$

⠀

∴ The value of x is 13.5.

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★ VERIFICATION:-

⠀Equation we had,

$\begin{gathered}\\\;\sf{:\rightarrow\;\;\;\bf{{xy - 3y^{2} = 15}\:}}\end{gathered}$

⠀

To check our answer put down the respective values to see whether our Answer is correct or not.

Values we get, x = 13.5 , y = 2

$\begin{gathered}\\\;\sf{:\rightarrow\;\;\;\bf{{13.5 \times 2 - 3 \times 2^{2} = 15}\:}}\end{gathered}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;\;\bf{{27 - 3\times 4 = 15}\:}}\end{gathered}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;\;\bf{{27 - 12 = 15}\:}}\end{gathered}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;\;\bf{{15 = 15}\:}}\end{gathered}$

⠀

Hence, LHS = RHS, So the values put in the equation was correct.