((x+2)/x )+((12)/3x-5)? dont understand but i know the answer is -5 or 2/3, how?
Answers:
((x+2)/x )+((12)/3x-5)=0
The LCM of x and (3x -5) is x(3x-5)
Make equal denominators
{(x+2)(3x-5)/x(3x-5) }+{(12)(x)/x(3x-5)}=0
Simplify or open all brackets
{3x^2-5x+6x-10}/{x (3x-5) }+{(12x)/{x(3x-5)}=0
Collect all like terms and simplify them
{3x^2+13x-10}/x(3x-5)}=0
It means denominator is zero
3x^2+13x-10=0
Factor it
3x^2 +15x -2x +10=0
3x(x+5) -2(x+5)=0
(x+5)(3x-2) =0
Either x+5=0 or x+-5
or 3x-2=0 or x=2/3
thast just means x is - 5 or 2/3 as thats the question whats x
This is not an equation so I guess you have not typed it correctly or you have misread the question. If you know the answer is -5 or 2/3 then I assume you mean that x = -5 or 2/3. Since you have two values for x then it looks like the correct equation is a quadratic. You can solve this by factorising, completing the square, graphically or using the standard formula. Can't help you any more without the corrections.
ı could't find it out
((x+2)(3x-5)+12x)/(3x^2-5x);
(3x^2-5x+6x-10+12x)/(3x^2-5x);
1+((12)/(3x-5))+(2/5) (or,the right answer is this)
Err. let's just sit down and enjoy some cookies.
Is this the whole question? You cannot solve for a variable unless you have an equation or inequality.
x plus 2 over x =(x over x) plus 2=(x over x)=1 then add 2 so the first bit is 3.next bit (12) over 3x-5 is typed wrong and should read (12 over 3x-5) or something..i dont know how to work that bit out anyway..good luck
((x+2)/x )+((12)/3x-5)
= [(x+2)(3x-5) + 12x] / x(3x-5)
= [3x^2 - 5x + 6x - 10 + 12x] / x(3x-5)
= [3x^2 + 13x - 10] / x(3x-5)
= (3x-2)(x+5) / x(3x-5)
there can only be an answer if an equation is given.
Assuming that you're asking for ((x+2)/x )+((12)/3x-5) = 0,
(3x-2)(x+5) / x(3x-5) = 0
(3x-2)(x+5) = 0
x = 2/3 or -5
Both answers satisfy the equation given.
Do you mean:
(x+2)/x+12/(3x)-5=0 ?
If not, you need to state the problem more clearly.
multiply through by x:
x+2+12/3-5x=0
2+4-4x=0
6-4x=0
4x=6
x=3/2
try (x+2)/x +12/(3x-5)=0
mult through by (3x-5)x
(3x-5)(x+2) + 12x =0
3x^2-5x+6x-10 +12x =0
3x^2+13x -10 =0
in any quadratic, ax^2+bx+c =0
x= (-b + or-(b^(2)-4ac)^(1/2))/2a
x= (-13 +/- (13^2-4*3(-10))^(1/2))/2*3
= (-13+/- (169 -(4*3(-10))^(1/2))/6
= (-13 +/-(169+120)^(1/2))/6
= (-13 +/- (289)^(1/2))/6 = (-13+/- (289)^(1/2))/6
= ( -13 +/- 17)/6 = -30/6 or 4/6
therefore, x = -5 or 2/3 as required
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