• M1 M4

    where did the common denominator go in the solution you have provided? after finding the common denominator, we didn't do any manipulation with denominator afterward but instead worked only with nominator - i couldn't understand this jump, i think there are some steps missing for denominators

  • ADMIN M0★ M1 M5

    @funnyorangutan-1 Thank you for asking! I'm very happy to answer your question, and I'm sorry for your confusion. It just occurred to me that a little bit of color might help with this solution.

    We started by equating the success rate after we made a number of \(x\) additional shots, and got this equation:

    $$ \frac{3}{10 + x} = \frac{1}{4} $$

    Now that you mention it, there is a simpler way to do this step, which is to "cross-multiply"! We can multiply the \(\textcolor{red}{\text{red}}\) with the \(\textcolor{red}{\text{red}}\) and \(\textcolor{blue}{\text{blue}}\) with \(\textcolor{blue}{\text{blue}}.\)
    $$ \frac{\textcolor{red}{3}}{\textcolor{blue}{10+x}} = \frac{\textcolor{blue}{1}}{\textcolor{red}{4}} $$

    The method shown in the solution was to try to make the denominators of both fractions the same so that we could equate the numerators. A fail-safe common denominator is to simply take the two different denominators, \(10 + x\) and \(4,\) and multiply them together. So let's try a common denominator of \((10 + x)(4).\) To do this, we should multiply the top and bottom of \(\frac{3}{10+x}\) by \(\textcolor{red}{4}\) and multiply the top and bottom of \(\frac{1}{4}\) by \(\textcolor{blue}{(10+x)}.\)

    $$ \frac{3}{10+x} \times \left( \frac{\textcolor{red}{4}}{\textcolor{red}{4}} \right) = \frac{1}{4} \times \left( \frac{\textcolor{blue}{10 + x}}{\textcolor{blue}{10 + x}} \right) $$

    Now the bottoms of both fractions are both the same, \( (10 + x)(4),\) so we are confident that the tops should also be the same!

    This is the step where the denominators "disappear," and that's because we know that the numerators must equal each other if the denominators are equal. It's just like how \(\frac{1}{3} = \frac{1}{3}\) or \(\frac{3}{4} = \frac{3}{4}.\)

    $$ \frac{\textcolor{green}{3 \times 4 }}{(10 + x)(4)} = \frac{\textcolor{green}{10 + x}}{(10 + x)(4)} $$

    The \(\textcolor{green}{\text{green}}\) stuff on the left is equal to the \(\textcolor{green}{\text{green}}\) stuff on the right, so we get

    $$ \textcolor{green}{12 = 10 + x} $$

    I've updated the mini-question solution to reflect this additional explanation. https://daily.poshenloh.com/courses/take/1-algebra-basics/quizzes/9834571-day-4-challenge-explanation-2-of-4

    Thanks so much for helping us to improve our explanations! Please let us know if there are any other areas that could also be explained better!

  • M1 M4

    Abundantly thankful for the explanation - everything becomes explicitly apparent after going through this explanation.

  • ADMIN M0★ M1 M5