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Note: This isn't homework, I'm skipping ahead of class. Please answer all these equations, I'm deathly stuck on them.

Use the substitution method only please. (Find $x$ and $y$.)

\begin{align} ax\left({\frac {1}{a-b}-\frac {1}{a+b}}\right)+by\left({\frac {1}{b-a}-\frac {1}{b+a}}\right)=2 \end{align}

$a$ is not equal to $b$ or $-b$

Note - It's $ax$ as the first word, I am worried the latex might mess up there.

Equation 2:

\begin{align} 6x+5y=7x+3y+1=2\left({x+6y-1}\right) \end{align}

Equation 3:

\begin{align} \sqrt{2}x+\sqrt{3}y=0 \end{align} \begin{align} \sqrt{3}x-\sqrt{8}y=0 \end{align}

Thank you for the help!

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  • $\begingroup$ Your first problem has infinitely many solutions, are you sure you typed it correctly? $\endgroup$
    – mboratko
    Commented Sep 3, 2012 at 4:41
  • $\begingroup$ Yes, and the answer doesnt have to be in numbers. What is required is just the value of x and y, in my book they are given as x = a/b and y = b/a $\endgroup$
    – Aayush Agrawal
    Commented Sep 3, 2012 at 4:43
  • $\begingroup$ Oh my i forgot two more pieces of information, a is not equivalent to b and a is not equivalent to -b $\endgroup$
    – Aayush Agrawal
    Commented Sep 3, 2012 at 4:45
  • $\begingroup$ In your edit, is $A=a?$ They are usually distinct. $\endgroup$
    – Ross Millikan
    Commented Sep 3, 2012 at 4:52
  • $\begingroup$ Yes, A is a in my edit $\endgroup$
    – Aayush Agrawal
    Commented Sep 3, 2012 at 4:56

1 Answer 1

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For the first note ${\frac {1}{a-b}-\frac {1}{a+b}}=\frac {2b}{a^2-b^2}$ so you have $2abx-2aby=2(a^2-b^2)$

For equation 2 (it is really two equations) you should split them apart into $6x+5y=7x+3y+1$, which means $2y=x+1, y=\frac {x+1}2$ and $6x+5y=2x+12y-2, 4x=7y-2$

For Equation 3 (though you have two equations) the basic substitution method works. From the first $x=-\sqrt \frac 32 y$, plug that into the second and you have an equation in $y$.

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  • $\begingroup$ Im sorry but can you please solve the entire equations? Im majorly confused about them D: $\endgroup$
    – Aayush Agrawal
    Commented Sep 3, 2012 at 4:58
  • $\begingroup$ @aayush: please be more explicit in describing what you don't understand. Set 3 is the most basic. Given $x=-\sqrt{\frac 32} y$ you have from the second $\sqrt 3 \sqrt{\frac 32} y-\sqrt 8 y =0$ Can you solve this? $\endgroup$
    – Ross Millikan
    Commented Sep 3, 2012 at 5:19
  • $\begingroup$ I got the third one, but the other two are out of my bounds. Can you please post the full solution to them? $\endgroup$
    – Aayush Agrawal
    Commented Sep 3, 2012 at 5:24
  • $\begingroup$ @aayush: For 2, I gave you the substitution. Can you plug it into the other equation? For the first, you have only one equation in two unknowns. You should expect a one parameter solution family. The best you will do (unless you have not copied the problem entirely) is one step away from the last I gave. $\endgroup$
    – Ross Millikan
    Commented Sep 3, 2012 at 5:42
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    $\begingroup$ @aayush: You should be confused about the first one, as it's one equation, which you're supposed to solve for two variables--impossible to solve using basic substitution only. For the second and third, Ross has pretty much set you up completely. All you've got to do is substitute. For the second, I'd change the first conclusion equation to $x=2y-1$, substitute $x$ into the second conclusion equation. Then back-substitute to get your $x$-value. The third is even more straightforward. $\endgroup$
    – Cameron Buie
    Commented Sep 3, 2012 at 5:45

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