How do u solve this intermediate algebra problem?

Question

-(2/w+3)=7

Thanks :)

Thanks but I'm not asking for the answer, I'm asking "How to Solve it" I know you need to get it on to one side, but I forget how. I know I need to solve it myself.

Answer 1

expand it from -(2/w+3)=7 to

-2/w -3=7 (add three to both sides)

-2/w=10 (multiply by w)

-2=10W (divide by 10)

-1/5=w

Answer 2

You need to solve this for yourself! If you let people do your homework for you, you'll never pass the tests AND you'll be recognized as a cheat because your homework was correct but you made "F's" on the tests.

The way to solve it is to rearrange the equation step-by-step until "w" is by itself on one side of the equals sign. Then simplify whatever is on the other side so that it is just a single number.

Answer 3

set up your 2 situations: C1: speed = x Distance = six hundred Time = y C1: speed = x + 10 Distance = six hundred Time = y - 2 Now form your 2 structures of equations (making use of speed = distance/time): x = six hundred/y x + 10 = (six hundred) / (y - 2) substitute (six hundred/y) for x in the 2d equation: six hundred/y + 10 = (six hundred)/(y - 2) Multiply each little thing by skill of y(y - 2) to cancel off all denominators: six hundred(y - 2) + 10y(y - 2) = 600y amplify all brackets: 600y - 1200 + 10y^2 - 20y = 600y integrate each and all of the like words: 10y^2 + 580y - 1200 = 600y circulate each little thing to a minimum of one section: 10y^2 - 20y - 1200 = 0 Divide each little thing by skill of 10 to simplify the equation: y^2 - 2y - one hundred twenty = 0 element it out: (y + 10) (y - 12) = 0 ------------------------- this supplies the potential of the time it took the 1st prepare to holiday being the two 12 hours or -10 hours. Now needless to say there is not any such element as a destructive time (or going returned in time), so the time it took the 1st prepare can only be 12 hours. Now bypass returned to a minimum of one among your situations: C1: speed = x Distance = six hundred Time = 12 resolve for the cost (making use of speed = distance/time): x = six hundred/12 x = 50 If the cost of the 2d prepare became into x + 10, this is speed became into 60. usual: the cost of the two trains are 50mph and 60mph.

Answer 4

w = - 1/5

Answer 5

2/w+3=-7
2/w+3-3=-7-3
2/w=-10
2/w *2=-10*2
w=-20

the * is a multiplication sign.

Answer 6

-(2/w+3)=7
2/w+3= -7
2/w= -7-3
2/w= -10
w/2=1/-10
w=2/-10
w=-1/5

Answer 7

2/w+3=-7
2/w=-10
w=-20