# Math Help - Solving equations

1. ## Solving equations

3x - 4 = x + 5

X x 3X = x2 + 4

2. Originally Posted by Shandy

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

$2x=9$

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

lets check it.

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

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

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

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

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

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

$2x^2=4$

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

$x\sqrt{2}=2$

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

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

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

$x=\sqrt{2}$

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

$4+2= 2+4$

$6=6$

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!
$X * 3X = -X^2 + 4$

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

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

$4x^2=4$

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

$\sqrt{x}^2=1$

$
x=1
$

let's check it.

$1 * 3 = -1 +4$

$3=3$