1. ## Tricky Word Problem

Hello everyone,

Thank you for taking the time to help me. I have never been good at math, and I'm trying to learn now a bit later in life. I've downloaded a free kindle book called CK-12 Algebra 1 - Second Edition. There is a word problem in one of the earlier chapters that is giving me no end of grief. Thanks in advance for any help you can provide. I'd really like to understand how you did it, not just the answer, if you would please:

problem: One number is 25 more than 2 times another number. If each number were multiplied by five, their sum would be 350. What are the numbers?

Here's what I have worked out:

[one number] - 25 = 2 * [another number]
x - 25 = 2y

5x + 5y = 350

Breaking that last equation down (dividing each side by 5) I get

x + y = 70

Now for the initial equation. I isolated the x by adding 25 from each side:

x - 25 + 25 = 2y + 25
x = 2y + 25

Now I'm a little lost as to what to do next. If I divide both sides by 2, I get:

x / 2 = 2y / 2 + 25
x/2 = y + 25
or 1/2x = y + 25

At this point I am just lost. I don't know where else to go. Even trying to brute force it doesn't seem to make sense to me, as I can't find two numbers that fit both equations, which makes me think that I did something wrong somewhere. Any help is REALLY appreciated while I still have hair left to pull out LOL

2. ## Re: Tricky Word Problem

Hello, Vinicide!

But you've forgotten how to solve a system of equations.

One number is 25 more than 2 times another number.
. . . $\displaystyle {\color{blue}x \:=\:2y + 25\;\;[1]}$
If each number were multiplied by five, their sum would be 350.
. . . $\displaystyle {\color{blue}5x + 5y \:=\:350\;\;[2]}$
What are the numbers?

Substitute [1] into [2]: .$\displaystyle 5(2y+25) + 5y \:=\:350 \quad\Rightarrow\quad 10y+125 + 5y \:=\:350$

. . . . . . . . . . . . . . . . . . $\displaystyle 15y \:=\:225 \quad\Rightarrow\quad \boxed{y \,=\,15}$

Substitute into [1]: .$\displaystyle x \:=\:2(15)+25 \quad\Rightarrow\quad \boxed{x \,=\,55}$

3. ## Re: Tricky Word Problem

Woah, that is great. The book hadn't gotten into combining equations like that, and it took me a little bit to understand it, but I see what you did. Thank you so much!