Question

How do I do this 7/8c-3(1/8c-7)=-6

Answer 1

Answer:

$f(c)=-384c^2+480c-133$

$c=\frac{179.31}{768}$ and $c=-\frac{1139.31}{768}$

Step-by-step explanation:

$(\frac{7}{8c-3})(\frac{1}{8c-7})=-6$

• First, let's multiply these two fractions together. I'm trying to follow the order of operations as listed below:

1. Parentheses
2. Exponents
3. Multiply
4. Divide
5. Add
6. Subtract

• Because we don't have any math that would make things easier to simply within the parentheses, and no exponents to deal with either, our first operation is multiplication of our two fractions.

$(\frac{7}{8c-3})(\frac{1}{8c-7})=-6\\\frac{7(1)}{(8c-3)(8x-7)}=-6\\\frac{7}{(8c-3)(8x-7)}=-6$

• Multiplying fractions is much easier to do than adding or subtracting them. You don't need a common denominator, but instead you just multiply the two fractions' numerators and denominators together, as seen above.
• Next, I'm going to simplify my denominator; although this step isn't necessary it will save time in the long run.

$\frac{7}{(8c-3)(8x-7)}=-6\\\frac{7}{64c^2-56c-24c+21}=-6\\\frac{7}{64c^2-80c+21}=-6$

• I used distribution to solve. Remember that every value is being multiplied by each other. Not just one.
• Now, I'm going to multiply what's in the denominator onto either side of the equation, thus canceling out what's below.

$\frac{7}{64c^2-80c+21}=-6\\(64c^2-80c+21)\frac{7}{64c^2-80c+21}=-6(64c^2-80c+21)\\7=-6(64c^2-80c+21)$

• Again, I'm going to distribute. In this case the $6$ is being distributed into the values within the parentheses. Every single value in the parentheses is affected by this, not just one.

$7=-6(64c^2-80c+21)\\7=(-6)64c^2-(-6)80c+(-6)21\\7=-384c^2+480c-126\\0=-384c^2+480c-133$

• Please let me know if I went wrong somewhere along the way, but according to my math so far, this is our equation when put into standard form. In this form, we have a couple options:

1. Factor
2. Quadratic Formula

• Factoring, once you are used to it, is easier than the Quadratic Formula, so you should use factoring in most cases. There are many sites with great content on factoring if you find it confusing. Sadly, this equation doesn't seem to be obviously factorable, so I'm going to use the Quadratic Formula:

• $x=\frac{-b+_-\sqrt{b^2-4(a)(c)} }{2a}$

• Treat $+_-$, as seen in the numerator of the fraction on the right side, as ± (plus and/or minus). This is based on the parent standard form of a quadratic:
• $f(x)=ax^2+bx+c$
• $a$, $b$, and $c$ can be seen as set integers like $4$ or $5$.

$0=-384c^2+480c-133\\c=\frac{-480+_-\sqrt{480^2-4(-384)(-133)} }{2(-384)} \\c=\frac{-480+_-659.31}{768}$

• This is a long, tedious, and maybe incorrect answer, but I hope you get the general idea of how to go about doing a problem using the Quadratic Formula in the future.
• What we were just trying to do is find where in the graph $f(c)=-384c^2+480c-133$ hits the $x$-axis on a number line. I did this by setting our $f(c)$, or our $y$ output for our graph to $0$ and solving for $c$ in this case. Now that we know that, it may make sense that there will be 2 answers for our question, as this is a quadratic line, which is usually shaped like a U. Now I'll continue to solve for $c$.

$c=\frac{-480+_-659.31}{768}\\c=\frac{-480+659.31}{768} , c=\frac{-480-659.31}{768}\\c=\frac{179.31}{768} , \frac{1139.31}{768}$

• These are our final values for $c$, but I'm not quite sure if I did my math correctly, or made some type of error in my calculator, or just solved too far into the question.