I dont understand this

[brady s]

Solve: r=0.9r+30

 

[Deleted member 4993]

Solve: r=0.9r + 30

Let me do a similar but different problem.

Solve: x = 9x + 30 The word "solve" here and your problem means:

Calculate the value of the unknown parameter, 'x' in this case and 'r' in your problem, such that the given equation is true.

x = 9x + 30 ........................... 'unknown' is 'x' and we will move numbers around such that all the numbers containing 'x' will be on side of the equation,

x - 9x = 30

-8x = 30

x = 30/(-8) = - 15/4

If you are still unsure, please refer to:

Literal equations? As opposed to metaphorical ones?

In this context, "literal" refers to the letters or variables in the equation. Generally, you solve for one variable in terms of the others.

www.purplemath.com

 

[HallsofIvy]

I truly dislike the words "move numbers around"! Too many students get confused about whether you should "move" the numbers by adding or multiplying.

I would word it like this:

The equation is r= 0.9 r+ 30. To "solve" or "solve for r" means to get an equation of the form "r= number". So the first thing we would like to do is get "r" on only one side of the equation. We can get rid of the "r" in "0.9r" on the right side of the equation by subtracting 0.9r. Of course, any thing we do to one side of the equation we must do to both sides: r- 0.9r= (1- 0.9)r= 0.1r= 30. Now we need to clear that "0.1" from r on the right. Since that multiplies r, we can get rid of it by dividing both sides by 0.1:
0.1r/0.1= r= 30/0.1= 300.

Check: If r=300 then 0.9r= 0.9(300)= 27. The equation becomes 300= 270+ 30 which is true.

 

[Steven G]

1r = .9r + .1r, so .1r = 30. Then r=....?