How would you solve this problem? (solve this equation for x)?

(x + 12)/2x = 22/(3x - 1)
3x² - x + 36x - 12 = 44x
3x² - 9x = 12
x² - 3x = 4
x² - 3/2x = 4 + (- 3/2)²
x² - 3/2x = 16/4 + 9/4
(x - 3/2)² = 25/4
x - 3/2 = ± 5/2

x = 5/2 + 3/2, x = 8/2, x = 4
x = - 5/2 + 3/2, x = - 2/2, x = - 1

Answer: x = 4, - 1

Check (x = 4):
(4 + 12)/(2[4]) = 22/(3[4] - 1)
16/8 = 22/(12 - 1)
2 = 22/11
2 = 2

Check (x = - 1):
(- 1 + 12)/(2[- 1]) = 22/(3[- 1] - 1)
11/- 2 = 22/(- 3 - 1)
- 11/2 = 22/- 4
- 11/2 = - 11/2

your question is punctiliously ambiguous without parentheses to teach what's the best and what's the backside of the fractions. i think of you are able to advise (x + 4)/(x³ + 8) + (x+2)/(x²-2x+4) = 11/(2x+4) wherein case you need to be attentive to that x³ + 8 = (x + 2)(x² - 2x + 4) and so the liquid crystal show would be 2(x + 2)(x² - 2x + 4) we multiply all 3 fractions with the help of that and cancel out all of the denominators, leaving 2(x + 4) + 2(x + 2)(x + 2) = 11(x² - 2x + 4) 2x + 8 + 2x² + 8x + 8 = 11x² - 22x + 40 4 2x² + 10x + sixteen = 11x² - 22x + 40 4 0 = 9x² - 32x + 28 now we seem for aspects of 9•28 = 252 that upload as much as -32, so 14 and 18 9x² - 18x - 14x + 28 = 0 9x(x - 2) - 14(x - 2) = 0 (9x - 14)(x - 2) = 0 so x = 14/9 or x = 2 determine you verify each answer; you are able to't use one that would make a denominator 0.

x+12/2x =22/3x-1
(x + 12)(3x - 1) = 22(2x)
3x^2 + 15x - x - 12 = 44x
3x^2 + 14x - 12 = 44x
3x^2 - 30x - 12 = 0
3(x^2 - 10x - 4) = 0
x^2 - 10x - 4 = 0 (Since 3 cannot equal '0').
x = {--10 +/- sqrt[(-10)^2 - 4(1)(-4)]} / 2(1)
x = {10 +/- sqrt[100 + 16]}/ 2
x = { 10 +/- sqrt(116)}/2
x = {10 +/- 10.77}/2
x = 20.77/2 = 10.385
&
x = -0.77/2 = -0.385

where are the ()???????????
x+(12/2)x =(22/3x)-1
x+(12/2x) =(22/3x)-1
x+(12/2)x =(22/3)x-1
x+(12/2x) =(22/3)x-1
(x+12)/2x =(22/3x)-1

I could go on but chances are you mean
(x+12)/2x =22/(3x-1)
(x+12)(3x-1)=44x
3x²+35x-12=44x
3x²-9x=12
x²-3x-4=0
(x-4)(x+1)=0
x=4
x=-1

(x + 12)/(2x) = 22/(3x - 1)
x + 12 = 2x[22/(3x - 1)]
x + 12 = 44x/(3x - 1)
(x + 12)(3x - 1) = 44x
x(3x) + 12(3x) - x(1) - 12(1) - 44x = 0
3x^2 + 36x - x - 44x - 12 = 0
3x^2 - 9x -12 = 0
x^2 - 3x - 4 = 0
x^2 + x - 4x - 4 = 0
(x^2 + x) - (4x + 4) = 0
x(x + 1) - 4(x + 1) = 0
(x + 1)(x - 4) = 0

x + 1 = 0
x = -1

x - 4 = 0
x = 4

∴ x = -1, 4

x+12/2x =22/3x-1 is wrong, should be (x+12)/2x =22/(3x-1)
(x+12)(3x-1)=44x
3x^2-x+36x-12=44x
3x^2-9x-12=0
x^2-3x-4=0
(x-4)(x+1)=0
x=4 and -1

x + 12 = 44x/(3x - 1)

3x - 1(x+12) = 44x

3x^2 + 36x - x - 12 = 44x

3x^2 - 9x - 12 = 0

3(x^2 - 3x - 4) = 0
3((x-3/2)^2 - (3/2)^2 - 4) = 0
3((x-3/2)^2 -9/4 - 16/4) = 0
3((x-3/2)^2 - 25/4) = 0
3((x-3/2)^2 - √25/4)^2) = 0
3((x-(3-√25)/2)(x-(3+√25)/2)) = 0

x = (3±√25)/2

x = (3±5)/2
x = 4 and - 1

Yes i must agree with the first answerer, you have not specified if each term is together or not. As such i have put a completely different answer to the second person. In future put terms in brackets to make sure you have valid implied multipication or division.

x + 6x = 22/3 x - 1
7x + 1 = (7 1/3)x
1 = 1/3 x
3 = x

plzz use brackets to sepearate each term....
is it (22/3x)-1 or (22/3x-1)???