I am new to the forum and was hoping someone could help with a quick question I had. It is straight from the Official Guide 12th Edition, question #148, page 233
If x, y and k are positive numbers such that (x/x+y)(10)+(y/x+y)(20)=k and if x<y, which of the following could be the value of k?
A) 10 B) 12 C) 15 D 18 E) 30
In the book they show the steps but I am stuck on the progression of one step to the next.
1) 10x+20y/x+y (I understand the first step)
2) 10(x+y)+10y/x+y (This is the next step and I do not understand how they got the numerator)
Any help would be greatly appreciated!
Simplifying Algebraic Expressions
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Yeah, they could have added a step in the middle. First, however, I'll add some brackets to make it clear to others how the OG12 solution looks:bkearney31 wrote:I am new to the forum and was hoping someone could help with a quick question I had. It is straight from the Official Guide 12th Edition, question #148, page 233
If x, y and k are positive numbers such that (x/x+y)(10)+(y/x+y)(20)=k and if x<y, which of the following could be the value of k?
A) 10 B) 12 C) 15 D 18 E) 30
In the book they show the steps but I am stuck on the progression of one step to the next.
1) 10x+20y/x+y (I understand the first step)
2) 10(x+y)+10y/x+y (This is the next step and I do not understand how they got the numerator)
Any help would be greatly appreciated!
We have:
Step 1: (10x+20y)/(x+y)
Step 2: [10(x+y)+10y]/(x+y)
Okay, in between steps 1 and 2, we take 20y and rewrite it as 10y + 10y
Here we go:
Step 1: (10x+20y)/(x+y)
Step 1.5: (10x+10y+10y)/x+y
Then factor 10x+10y to get
Step 2: [10(x+y)+10y]/(x+y)
Cheers,
Brent
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Honestly, would take this problem away from the algebra...
This is a weighted average problem. If x < y then the number has to be closer to 20 than 10. 20 cannot be the answer because x and y both have positive values. Only if x = 0 could the answer be 20.
Hope that helps.
This is a weighted average problem. If x < y then the number has to be closer to 20 than 10. 20 cannot be the answer because x and y both have positive values. Only if x = 0 could the answer be 20.
Hope that helps.
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Jim,Jim@StratusPrep wrote:Honestly, would take this problem away from the algebra...
This is a weighted average problem. If x < y then the number has to be closer to 20 than 10. 20 cannot be the answer because x and y both have positive values. Only if x = 0 could the answer be 20.
Hope that helps.
I would appreciate if you could elaborate your answer
Thanks
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