SOLUTION: I need help with the equation 3(4x-9)=x+100. So far I have done two steps and then i am completely lost? My first step I got 12x-27=x+100, then the second step I got 12x=x+127, n

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I need help with the equation 3(4x-9)=x+100. So far I have done two steps and then i am completely lost? My first step I got 12x-27=x+100, then the second step I got 12x=x+127, n      Log On


   



Question 262205: I need help with the equation 3(4x-9)=x+100. So far I have done two steps and then i am completely lost? My first step I got 12x-27=x+100, then the second step I got 12x=x+127, now I cant figure out how do I divided the 12 with the x or the 127?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
your equation is:

3*(4*x-9) = x+100

simplify to get:

12*x - 27 = x + 100

so far you did good.

add 27 to both sides of this equation to get:

12*x = x + 100 + 27

you simplified this to get:

12*x = x + 127

so far you still did good.

you needed to get all the x's on one side of the equation.

you do that by subtracting x from both sides of the equation to get:

12*x - x = 127

simplify that to get:

11*x = 127

now you divide both sides of the equation by 11 to get:

x = 127 / 11 = 11.54545454

that should be your answer.

you confirm by substituting for x in the original equation.

original equation is:

3*(4*x-9) = x+100

substitute 11.54545454 for x to get:

3*(4*11.54545454 - 9) = 11.54545454 + 100

simplify to get:

3 * (37.18181818) = 111.54545454

simplify further to get:

111.54545454 = 111.54545454 confirming that the value calculated for x is good.

the numbers displayed on the calculator are usually rounded to the nearest 8 decimal places or thereabouts.

I used the internally stored numbers which are carried out to more digits than can be displayed.

If you just use the 8 or so decimal places displayed, your answer will be very close.

If you then round to the nearest 1 or 2 decimal places, your answer should be right on.