have to find x but not sure how to on this problem?

Question

x+2/x+3=3/4

Answer 1

x + 2/x + 3 = 3/4
Multiply everything by 4x to get rid of fractions
4x(x) + 4x(2/x) + 4x(3) = 4x(3/4)
4x^2 + 8 + 12x = 3x
Move everything to one side
4x^2 + 9x + 8 = 0
Now use the quadratic formula.
x = [-9 +/- sqrt(9^2 -4(4)(8))] / (2*4)
x = [-9 +/- sqrt(-47)] / 8
x = [-9 +/- i*sqrt(47)] / 8


If you meant x + 2/(x+3) = 3/4
or (x+2) / (x+3) = 3/4, then you need to use parentheses.

(x+2) / (x+3) = 3/4
Cross multiply
4(x+2) = 3(x+3)
Distribute
4x + 8 = 3x + 9
Subtrct 3x from each side and subtract 8 from each side
x = 1

Answer 2

Assume that question is meant to read as:-
(x + 2) / (x + 3) = 3 / 4
4(x + 2) = 3 (x + 3)
4x + 8 = 3x + 9
x = 1
PS
The question can have a number of different meanings if brackets are not used.
As a result, incorrect answers may be supplied by those responding.
Suppose you learn nothing else from this question, remember to use brackets.

Answer 3

hi!

if i'm reading the problem correctly, it's:

(x+2)/(x+3) = 3/4

multiply both sides by x+3

(x+3)(x+2)/(x+3) = 3(x+3)/4

x+2 = (3x+9)/4

multiply both sides by 4

4(x+2) = 4(3x+9)/4

4x+8 = 3x + 9

4x + 8 - 3x - 8 = 3x + 9 -3x - 8

x = 1

and to verify, plug in that value into all x of the original equation:

(1+2)/(1+3)
=3/4

which is correct

good luck!

Answer 4

multiply both sides by (x+3)

x+2 = 3/4 (x+3)

then multiply the 3/4 through
x+2=3/4 * x +9/4

then subtract 3/4 * x from both sides, and subtract 2 from both sides
x-3/4*x=9/4-2
x/4=1/4

multiply both sides by 4

x=1

Answer 5

Don't you just love fractions!
Well, we're going to get rid of them.
Multiply EVERY term by x
x^2+2 +3x =3x/4
Multiply again by 4
4x^2 +8 +12x =3x
I could have done this multiplication in one step by using 4x. Just wanted you to see each move.
4x^2 +12x-3x+8=0
4x^2+9x+8=0
Now this is bad news. There is no solution in the Real Number system. I know this because when I try to use the quadratic formula to solve for x, I get this:
The formula is x={-b+ or -rt(b^2-4ac)}/2a. where
a=coefficient of the x^2 term, in this case +4
b=coefficient of the x term, in this case +9
c= the constant, in this case +8

x={-9 + or - rt(81-128)}/8
x={-9 + or - rt(-47)}/8
Now there's no way I can take the square root of-47.
It cannot be done in the Real Number System.
I could if I used the Complex Number System, but that's a whole different set of rules.
Your answer to this question is "NO SOLUTION".
Tell them that and rest easy.
Good luck with your math.