I am a beginner still practicing a lot. I think math is fun but a bit hard.

I would like to know how this is solved:

$$\frac{x+3}2 = \frac{x-3}4$$


$x = -9$

How? What are the exact steps here, I cannot seem to figure this one out. As I said, I am still practicing a lot. Thanks in advance.

  • $\begingroup$ First you can replace the solution and check if it works. Then, if you dont know the solution, you can multiply both sides of the equality to get something simpler (for example with $4$) and then you can add or remove something to each side to have $x$ only on one side. $\endgroup$ – Jérémy Blanc Apr 2 '14 at 15:55
  • $\begingroup$ $$\frac{x+3}2=\frac{x-3}4\overset{\rm BS\times 4}\implies 2(x+3)=x-3\implies 2x+6=x-3\implies x=-9$$ $\endgroup$ – Hakim Apr 3 '14 at 11:52

$$\frac{x+3}2=\frac{x-3}4\iff 4(x+3)=2(x-3)\iff4x-2x=-6-12$$

Alternatively, $$\frac{x+3}2=\frac{x-3}4=\frac{x+3-(x-3)}{2-4}=-3\text{( Subtrahendo )}$$

$$\iff x+3=2\cdot(-3)\text{ or } x-3=4\cdot(-3)$$


Multiply both sides with 4. Do you know how to multiply with brackerts? Can you now rearrange this equation to $x= \text{something}$

  • $\begingroup$ Although I know considering OP's level this won't help, I have a huge urge to use componendo dividendo. The simulataneous presence of $x+3$ and $x-3$ is too delicious to pass up. :D $\endgroup$ – Guy Apr 2 '14 at 15:55
  • $\begingroup$ @Sabyasachi Componendo and dividendo was the first thing I wanted to do! $\endgroup$ – evil999man Apr 2 '14 at 15:56

$$\frac{x+3}{2} = $$$$\frac{2(x+3)}{4} = \frac{x-3}{4}$$

Because the $4$ in the denominator is common on both sides, set the numerators equal to each other:

$$2(x+3) = x-3$$

Distribute the left side:

$$2x + 6 = x - 3 $$

After subtracting an $x$ and subtracting a $6$ to each side:

$$ x + 6 = -3 $$

$$ x = -9 $$


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