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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!

Note: Please provide the full solution, the more hints i get the more confused i get...

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    $\begingroup$ The first one is just one equation with two variables. You can't find both! $\endgroup$ – Dennis Gulko Sep 3 '12 at 11:30
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    $\begingroup$ What does "ax as the first word" mean? $\endgroup$ – draks ... Sep 3 '12 at 11:31
  • $\begingroup$ Well my book wants me to do it...The answer is coming at x = a/b and y = b/a $\endgroup$ – Aayush Agrawal Sep 3 '12 at 11:31
  • $\begingroup$ For equation 3, you can multiply the top equation with $\sqrt 3$ and the bottom by $\sqrt 2$, and you will see that nice things happen. That is square roots vanishing and all. $\endgroup$ – Ravi Sep 3 '12 at 11:32
  • $\begingroup$ draks, i thought the latex might have some problem rendering it, atleast on my screen it does. So the first word is "ax" just in case others get confused by it too. $\endgroup$ – Aayush Agrawal Sep 3 '12 at 11:32
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Equation 2:
$$6x+5y=7x+3y+1\Rightarrow 0=x-2y+1 \Rightarrow 2y-1=x$$ $$7x+3y+1=2(x+6y-1)\Rightarrow 5x-9y+3=0$$
Substituting the $\,x\,$ from the first equation you get:
$$5(2y-1)-9y+3=0\Rightarrow y-2=0\Rightarrow y=2\Longrightarrow x=2y-1=3$$

Equation 3:
$$\sqrt{3}x-\sqrt{8}y=0$$ Hence $$\sqrt{3}x=\sqrt{8}y$$ So $$x=\frac{\sqrt{8}}{\sqrt{3}}y$$ Substituting in the second: $$\sqrt{2}\frac{\sqrt{8}}{\sqrt{3}}y+\sqrt{3}y=0$$ Now you have $$\left(\sqrt{\frac{16}{3}}+\sqrt{3}\right)y=0\Longrightarrow y=0\Longrightarrow x=0$$

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  • $\begingroup$ YOUR A FRIGGIN GENIUS! Now lets hope someone helps me with the first one D: $\endgroup$ – Aayush Agrawal Sep 3 '12 at 11:47
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First Equation: $$ax\left({\frac {1}{a-b}-\frac {1}{a+b}}\right)+by\left({\frac {1}{b-a}-\frac {1}{b+a}}\right)=2 $$ $$ax\left({\frac {a+b-a+b}{a^2-b^2}}\right)+by\left({\frac {b+a-b+a}{b^2-a^2}}\right)=2$$ $$2abx-2aby=2(a^2-b^2)$$ $$x-y=\frac{a}{b}-\frac{b}{a}$$ From this the obvious solution is $x=a/b \text{ and } y=b/a$. But there will be infinite solutions $x=y+(a/b)-(b/a)$. Just put values for $y$ and get corresponding values for $x$.

Second Equation: $$6x+5y=7x+3y+1=2(x+6y-1)$$ Break it into $2$ equations $$6x+5y=7x+3y+1 \text{ and } 6x+5y=2(x+6y-1)$$ $$-x+2y=1 \text{ and } 4x-7y=-2$$ The value of $x$ from the first equation is $x=2y-1$. Put it in second equation. $$4(2y-1)-7y=-2$$ $$y=2$$ Hence $x=2y-1=3$.

Third equation: $$\sqrt2x+\sqrt3y=0 $$ From this $x=-\sqrt{3/2}y$. Put this in the second equation. $$\sqrt3x-\sqrt8y=0$$ $$y(3/\sqrt2 - \sqrt8)=0$$ $$y=0$$ Hence $x=0$.

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$$ax\left(\frac{1}{a-b}-\frac{1}{a+b}\right)+by\left(\frac{1}{b-a}-\frac{1}{b+a}\right)=2\Longrightarrow$$ $$\frac{ax}{a^2-b^2}(2b)+\frac{by}{b^2-a^2}(2a)=2\Longrightarrow \frac{abx-aby}{a^2-b^2}=1\Longrightarrow$$ $$x-y=\frac{a^2-b^2}{ab}\,\,,\,\,\text{assuming}\,\,ab\neq 0$$ and nothing more can be done as you're only giving one equation with two variables. Of course, the substitution method here cannot possibly work, no matter what your book requires (check this, it may well be you made a mistake while copying the question)

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  • $\begingroup$ Thats the strange thing, i triple checked it and it aint making no faults D: I will ask my teacher tomorrow, hopefully i will render him speechless, or not D: $\endgroup$ – Aayush Agrawal Sep 3 '12 at 11:55
  • $\begingroup$ Well, just take into account that if $\,b=0\,\,,\,\,a\neq 0\,$ ,then your equation becomes $\,0=2\,$, which of course has no solutions... $\endgroup$ – DonAntonio Sep 3 '12 at 11:57

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