<PRE><B>Question 582046</B>
1. {{{((x+2))/x}}} = {{{((2x+4))/3}}}
Since you have single fractions on each side of the =, you can cross multiply
3(x+2) = x(2x+4)
3x + 6 = 2x^2 + 4x
arrange as a quadratic equation on the right
0 = 2x^2 + 4x - 3x - 6
0 = 2x^2 + x - 6
the will factor to
(2x-3)(x+2) = 0
two solutions
2x = 3
x = 1.5
and
x = -2
You should check all our solutions in original problems
:
2. {{{4/((x+1))}}} = {{{((x-1))/((x+5))}}}
Cross multiply
4(x+5) = (x+1)(x-1)
4x + 20 = x^2 + x - x - 1
0 = x^2 - 4x - 1 - 20
0 = x^2 - 4x - 21
Factors to
(x-7)(x+3) = 0
x = +7
x = -3
:
Do 3 and 4 the same way
3. x/2x+1 = 2x/x+2
4. x/5 = -4/x-9
:
I am not sure what you mean in this one, use brackets to clarify things here
5. x^2 + 3/7x = x+1/6 </PRE>