Solve the equation by the method of your choice. Click to see problem
Where's the problem?
\[\frac{ 1 }{ x+13 }+\frac{ 1 }{ x }=\frac{ 1}{ 14 }\]
ok so i got x^2-5x+182=0
Whenever you have an equation containing fractions, first factor each numerator and denominator completely and then simplify each fraction by dividing like factors. Next multiply both sides by LCD (least common multiple of the denominators). Are any numerators or denominators factorable in your problem?
wait thats 15x not 5
yes x+13 and 14x
14(x+13)+14x=x(x+13)
No! How are they factorable?
well you factor them out on top
No! You have the wrong idea about what I mean by factorable. Observe each numerator and denominator, individually. I never said multiply them. That's not factoring.
The answer is no, no numerators or denominators can be factored. OK so then the next question is, what is the LCD of the denominators? Can you tell me?
LCD is 14x(x + 13). Do you understand?
thats what i had
Right but that is LCD. That's not factoring each numerator and denominator. You must have a proper understanding of the vocabulary in mathematics. OK so the next logical step is to multiply both sides by 14x(x + 13). So what do you get?
14x^2+182x
No. Multiply each term individually on both sides by 14x(x + 13). What do you get?
i dont understand
OK let's do this in baby steps. What do you get when you multiply the first term (the first fraction) on the left side of the equation by 14x(x + 13)?
14x+182+14x=x^2+13x
or 14(x+13)+14x
\[14x(x +13)(\frac{ 1 }{ x +13 })\]=?
14x?
im sorry its been a long day and i am just not getting it tonight
Right! Now what do you get when you multiply the second fraction on the left side by 14x(x + 13). In other words,\[14x(x +13) \frac{ 1 }{ x }\]= ?
14(x+13)
so is the -x^2+15x+182=0
Correct! Therefore the expression on the left side is\[14x +14(x +3)\]RIGHT?!
yeah so we do -b+-\[-15+-\sqrt{15^2-4(-1)(182)}/2(-1)\]
Not yet! Baby steps, remember? Now what do you get when you multiply the fraction on the right side by 14x(x + 13). In other words,\[14x(x +13)(\frac{ 1 }{ 14 })\]= ?
sorry x(x+13)
Correct! Thus we have the simplified equation\[14x +14(x +13)=x(x +13)\]Agree or disagree?
14x(x+13)+14x=x(x+13)
oh ok
Which is correct, mine or yours? What's the difference and do you spot the error?
yours you14x(x+13) not 14x just 14
Excellent! OK so now simplify the left side and the right side individually and right the simplified equation for me. Can you do that please?
*write not right. Sorry for the grammatical error.
14x+182=x^2+13x
-x^2+x+182=0
No, there seems to be a term missing on the left side when you simplified. Can you spot the error?
thats ok i didnt even notice way to sleepy to read right
i know hold on
28x+182=x^2+13x
It's not good, not having enough sleep. You should get at least seven or eight hours of sleep per night.
i try but midterms are coming up and i have to study
I love and appreciate your dedication to your education, however writing an exam tired won't help. Yes! Alright now always try to keep the leading term positive. Since the quadratic term is positive on the right side of the equation, manipulate all terms to the right side. Thus the left side will be zero. Can you do that?
0=x^2+13x-28x+182 im used to putting them on the left side0= x^2-15x+182
Are you sure that it is positive 182?
oh see told ya lol
So could you please write the correct equation for me? Thank you.
15\[15+-\sqrt{953}\div2\]
omg just one little negative threw my whole answer off ugh
No, that is not what I asked.
well i got my answer with your wonderful help
Yes, you did get the answer but you didn't answer my question. lol. Please do not skip steps on the test. Never be brave and skip steps unless you know your stuff well.
Good work! GOOD LUCK on all your exams!
i didnt skip I had done that part really easy
Thank you again for you help probably will need your help again soon. lol
I have to log out now, but I will probably be back in a few hours if you need me. Take care and please try to get some sleep. lol
ok thank you and you to.
Join our real-time social learning platform and learn together with your friends!