Question

\frac{ 8+2x }{ 4x }=\frac{ 3x }{ 4x }

Answer 1

I did this:

Answer 2

ok... so start with a little factoring

\[\frac{2(4 + x)}{4x} = \frac{3x}{4x}\]

if you remove the common factors you get

\[\frac{4 + x}{2x} = \frac{3}{4}\]

now cross multiply

\[4(4 + x) = 2x(3)\]

now distribute and solve for x

Answer 3

\[(4x)(8+2x)=(4x)(3x)\]\[32x+8x^2=12x^2\]\[32x+8x^2-8x^2=12x^2-8x^2\]\[32x-32x=4x^2-32x\]\[0=4x^2-32x\] I've done that so far.

Answer 4

here is my problem... looking at your work... x = 0 is a solution...

which means you have on the left side

\[\frac{8 + 2 \times 0}{4 \times 0} = \frac{3 \times 0}{4 \times 0}\]

I think this is an issue

Answer 5

\[4x=0\]\[\frac{ 4x }{ 4 }=\frac{ 0 }{ 4 }\]\[x=0\]

Answer 6

the problem I see is that dividing by zero is undefined...? when you use the original equation.

Answer 7

\[\frac{ 8+2x }{ 4x }=\frac{ 3x }{ 4x }\]

Answer 8

That's the original.

Answer 9

;^)

Answer 10

dot.