# Algebra help please, substitution method for these equations

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!

• Your first problem has infinitely many solutions, are you sure you typed it correctly? Commented Sep 3, 2012 at 4:41
• 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 Commented Sep 3, 2012 at 4:43
• Oh my i forgot two more pieces of information, a is not equivalent to b and a is not equivalent to -b Commented Sep 3, 2012 at 4:45
• In your edit, is $A=a?$ They are usually distinct. Commented Sep 3, 2012 at 4:52
• Yes, A is a in my edit Commented Sep 3, 2012 at 4:56

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 $y$.
• @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? Commented Sep 3, 2012 at 5:19
• @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. Commented Sep 3, 2012 at 5:45