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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE => VCE Mathematical Methods CAS => Topic started by: Jsy443 on February 03, 2019, 07:10:57 pm

Title: Linear literal equation
Post by: Jsy443 on February 03, 2019, 07:10:57 pm
1/x+a + 1/x+2a = 2/x+3a

How do I solve this?
Title: Re: Linear literal equation
Post by: aspiringantelope on February 03, 2019, 10:21:38 pm
1/x+a + 1/x+2a = 2/x+3a

How do I solve this?
Hey sorry for my late reply!
\(\frac{1}{x+a}+\frac{1}{x+2a}=\frac{2}{x+3a}\)
Try multiplying  \(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\) to the entire top to cancel out the bottom by simplifying and then expand. Sorry it's late I have to go to school tomorrow so I can only do this right now.
Please reply if you still have problems!
Title: Re: Linear literal equation
Post by: vox nihili on February 03, 2019, 10:46:18 pm
Hey sorry for my late reply!
\(\frac{1}{x+a}+\frac{1}{x+2a}=\frac{2}{x+3a}\)
Try multiplying  \(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\) to the entire top to cancel out the bottom by simplifying and then expand. Sorry it's late I have to go to school tomorrow so I can only do this right now.
Please reply if you still have problems!


This is a good answer! It asks OP to actually think through the question still :D They might like to put the completed solution in themselves.
Title: Re: Linear literal equation
Post by: aspiringantelope on February 04, 2019, 07:25:20 am
I may as well follow up with the solution
\(\frac{1}{x+a}+\frac{1}{x+2a}=\frac{2}{x+3a}\)
= \(\left(x+2a\right)\left(x+3a\right)+\left(x+a\right)\left(x+3a\right)=2\left(x+a\right)\left(x+2a\right)\)
expanded \(x^2+3ax+2ax+6a^2+x^2+3ax+ax+3a^2=2x^2+4ax+2ax+4a^2\)
Now we look subtract stuff or add stuff (only like terms(
= \(9ax+9a^2+=6ax+4a^2\)
which I can simplify to
= \(9ax-6ax=4a^2-9a^2\)
= \(3ax=-5a^2\)
\(x=-\frac{5a^2}{3a}\)
(You can simplify this btw) try doing that
I hope I didn't write any mistakes because I'm leaving for school soon aha but Good luck anyways and post more questions if needed
Title: Re: Linear literal equation
Post by: lzxnl on February 04, 2019, 10:20:46 am
I may as well follow up with the solution
\(\frac{1}{x+a}+\frac{1}{x+2a}=\frac{2}{x+3a}\)
= \(\left(x+2a\right)\left(x+3a\right)+\left(x+a\right)\left(x+3a\right)=2\left(x+a\right)\left(x+2a\right)\)
expanded \(x^2+3ax+2ax+6a^2+x^2+3ax+ax+3a^2=2x^2+4ax+2ax+4a^2\)
Now we look subtract stuff or add stuff (only like terms(
= \(9ax+9a^2+=6ax+4a^2\)
which I can simplify to
= \(9ax-6ax=4a^2-9a^2\)
= \(3ax=-5a^2\)
\(x=-\frac{5a^2}{3a}\)
(You can simplify this btw) try doing that
I hope I didn't write any mistakes because I'm leaving for school soon aha but Good luck anyways and post more questions if needed

Very small point, but the following is poor notation and working.

While the second line follows from the first, you can't write an equality there because the lines aren't equal. You're saying one equation follows from another. Just write it on another line.