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Hi first question here and a newbie at maths.

I have an equation that goes: $$x = \frac {a-b}{c-d }$$

i know it is possible to isolate any individual variable here but is it possible to isolate $a$ and $c$ together in some in order to solve for them only?

for example lets say I have all other variable are known except for $a$ and $c$ only,

is it possible to solve?

A simple yes or no answer could also suffice.

by the way, how do I write a fraction here? sorry for the ultra noob question.

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  • $\begingroup$ This is a system of $1$ equation in $5$ variables. You can choose freely any $4$ of them, then isolate the $5$th (if needed) and calculate it. For fraction, encapsulate \frac{m}{n} with a dollar sign on each side. $\endgroup$ Commented Oct 19, 2016 at 20:47
  • $\begingroup$ Is your equation $x=a-\dfrac{b}{c}-d$ or $x=\dfrac{a-b}{c-d}$? $\endgroup$ Commented Oct 19, 2016 at 20:48
  • $\begingroup$ Thank you Olivier, my equation is the latter one x = $\frac{a-b}{c-d}$ $\endgroup$
    – MadPir8
    Commented Oct 19, 2016 at 20:51
  • $\begingroup$ Assuming you mean $\frac {a-b}{c-d}$, suppose you knew that $\{x,b,d\}=\{1,0,0\}$. Then your equation would read $1=\frac ac$ and all you can say is that $a=c$. The numerical value is not determined. $\endgroup$
    – lulu
    Commented Oct 19, 2016 at 20:51
  • $\begingroup$ actually my son is doing cartesian planes and i know the equation for the gradient is m = delta y / delta x. but i need to know the second point and I only know the first point x,y coordinate and the gradient. lets say i want to find the second point without using the formula y = mx + c , can this work using transposition somehow? just curious. if it wont then thats fine $\endgroup$
    – MadPir8
    Commented Oct 19, 2016 at 20:54

1 Answer 1

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And i quote "for example lets say I have all other variable are known except for $a$ and $c$ only, is it possible to solve? A simple yes or no answer could also suffice."

The only answer i could give you is, you get an equation with two unknowns: $a = f(c)$. So it is underdetermined. In other words, any couple $(a,c)$ satisfying $a = f(c)$ is a solution.

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  • $\begingroup$ Yes that makes sense now. Thank you all so much for your help. Its much appreciated $\endgroup$
    – MadPir8
    Commented Oct 19, 2016 at 22:00

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