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I have $2$ equations :

$$-\frac{a}{2}x-\frac{b}{2}+cx+d=x+2$$
$$-2ax-2b+cx+d=2x+1$$
with $a, b, c, x \neq 0$

We have to find all possible solutions of $a, b, c, d$ that make the equations true for all $x$.

I found one solution
$a=-\frac{2}{3}$, $b=\frac{2}{3}$, $c=\frac{2}{3}$, $d=\frac{7}{3}$

But, I wonder if there are any other solutions to this system of equations?

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  • $\begingroup$ You have two equations and 5 unknowns. Your system is not determined, which means there is an infinite number of solutions. $\endgroup$ – Anna SdTC Dec 8 '17 at 23:11
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    $\begingroup$ The phrase "for any $x$" is ambiguous. It might mean you want to find $a,b,c,d$ as functions of $x$, or it might mean you want to find $a,b,c,d$ that make the equations true for all $x$. $\endgroup$ – Robert Israel Dec 9 '17 at 0:00
  • $\begingroup$ @RobertIsrael You are right. I edited the question. $\endgroup$ – Snip3r Dec 9 '17 at 0:24
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Write the equations as:

$$ \begin{cases} \begin{align} (a-2c+2)x+b-2d+4 = 0 \\ (2a-c+2)x+2b-d+1 = 0 \end{align} \end{cases} $$

A polynomial is identical $0$ (i.e. for all $x$) iff all its coefficients are $0\,$, which gives the system to solve:

$$ \begin{cases} \begin{align} a - 2c + 2 = 0 \\ 2a - c + 2 = 0 \\[7px] b - 2d + 4 = 0 \\ 2b -d + 1 = 0 \end{align} \end{cases} $$

The system has the unique solution as posted $\;a=-\frac{2}{3}$, $b=\frac{2}{3}$, $c=\frac{2}{3}$, $d=\frac{7}{3}\,$.

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Your $2\times 2$ system is consisted of $5$ variables (unknowns) to which you're being asked to find the solutions. Since you have only two equations for $5$ unknowns, this means that there is an infinite number of solutions (infinite number of numbers $a,b,c,d,x$ that satisfy the system given).

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  • $\begingroup$ I little edited the question, I want to find all solutions $a, b, c, d$ for any $x$. Are there still be infinite number of solutions? Could you add one example of solution? $\endgroup$ – Snip3r Dec 8 '17 at 23:23
  • $\begingroup$ @Snip3r Yes. Try solving one with respect to $x$ and plugging it down. You'll notice it ! $\endgroup$ – Rebellos Dec 8 '17 at 23:36

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