1. ## Solving equations

3x - 4 = x + 5

X x 3X = x2 + 4

2. Originally Posted by Shandy

3x - 4 = x + 5
$\displaystyle 3x-x = 5+4$

$\displaystyle 2x=9$

$\displaystyle x=\frac{9}{2}$

lets check it.

$\displaystyle 3 * \frac{9}{2} - 4 = \frac{9}{2} + 5$

$\displaystyle \frac{27}{2} - \frac{8}{2} = \frac{9}{2} +\frac{10}{2}$

$\displaystyle \frac{19}{2}= \frac{19}{2}$

X x 3X = x2 + 4
i assume that this is your equation. $\displaystyle x*3x = x^2 +4$

$\displaystyle 3x^2= x^2 +4$

$\displaystyle 3x^2-x^2 = 4$

$\displaystyle 2x^2=4$

$\displaystyle \sqrt{2x^2}=\sqrt{4}$

$\displaystyle x\sqrt{2}=2$

$\displaystyle x=\frac{2}{\sqrt{2}}$

$\displaystyle x=\frac{2}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}$

$\displaystyle x=\frac{2\sqrt{2}}{2}$

$\displaystyle x=\sqrt{2}$

$\displaystyle \sqrt{2}*\sqrt{2}^2= \sqrt{2}^2 + 4$

$\displaystyle 4+2= 2+4$

$\displaystyle 6=6$

3. Thank you very much, but I made a little mistake on the second one. It was supposed to be:

X * 3X = -X^2 + 4

Thanks again!

4. Originally Posted by Shandy
Thank you very much, but I made a little mistake on the second one. It was supposed to be:

X * 3X = -X^2 + 4

Thanks again!
$\displaystyle X * 3X = -X^2 + 4$

$\displaystyle 3x^2= -x^2 +4$

$\displaystyle 3x^2+x^2=4$

$\displaystyle 4x^2=4$

$\displaystyle x^2=\frac{4}{4}$

$\displaystyle \sqrt{x}^2=1$

$\displaystyle x=1$

let's check it.

$\displaystyle 1 * 3 = -1 +4$

$\displaystyle 3=3$