Thread: a + b + c

1. a + b + c

If a+b = 11, b+c = 15, and a+c =10, what is the value of a+b+c?

I know we have 3 equations in two unknowns.
Can someone show how to find one of the variables?
I can then proceed to find the other two.

2. Re: a + b + c Originally Posted by harpazo If a+b = 11, b+c = 15, and a+c =10, what is the value of a+b+c?
TAKE NOTE: $(a+b)+(b+c)+(a+c)=2a+2b+2c=2(a+b+c)=36$ HOW?

3. Re: a + b + c

Since I am not nearly as smart as Plato, I would have done this the hard way:

Subtract b+ c= 15 from a+ b= 11 to get a- c= -4. Add that to a+ c= 10: 2a= 6 so a= 3. Then a+ c= 3+ c= 10 so c= 7. a+ b= 3+ b= 11 so b= 8. a+ b+ c= 3+ 8+ 7= 18.

4. Re: a + b + c Originally Posted by harpazo If a+b = 11, b+c = 15, and a+c =10, what is the value of a+b+c?
Since I'm not nearly as smart as Halls, I'd do it this way (which is the best way!):

a+b = 11 
b+c = 15 
a+c = 10 
Code:
 a + b =  11 
-a - c = -10 *-1
============
b - c =   1
b + c =  15 
============
2b     =  16
b = 16/2 = 8
Now go get a and c....

3 cheers fer me!!

5. Re: a + b + c Originally Posted by DenisB 3 cheers fer me!!
With you it's more like "3 beers for me!!"

-Dan

6. Re: a + b + c Originally Posted by HallsofIvy Subtract b+ c= 15 from a+ b= 11 to get a- c= -4. Add that to a+ c= 10: 2a= 6 so a= 3. Then a+ c= 3+ c= 10 so c= 7. a+ b= 3+ b= 11 so b= 8. a+ b+ c= 3+ 8+ 7= 18.
@ harpazo, In the testing business this question is known as an "eater". There are two ways to do.
One is to solve for $a,~b,~\&~c$ so to be able find the value of $a+b+c.$ That eats up time, so the name.
The other as I used really requires only simple addition & division. From other posts I know you are preparing for tests.

7. Re: a + b + c Originally Posted by Plato TAKE NOTE: $(a+b)+(b+c)+(a+c)=2a+2b+2c=2(a+b+c)=36$ HOW?

a+b = 11
b+c = 15
a+c =10

2a+2b+2c = 36

I divided both sides by 2.

a+b+c = 18

8. Re: a + b + c Originally Posted by Plato @ harpazo, In the testing business this question is known as an "eater". There are two ways to do.
One is to solve for $a,~b,~\&~c$ so to be able find the value of $a+b+c.$ That eats up time, so the name.
The other as I used really requires only simple addition & division. From other posts I know you are preparing for tests.
Do not step beyond the question. I am not preparing for a test. I am 53 years old. Born in 1965. Do the math. Do not assume. I simply use test prep books to practice word problems. It's that simple, bro.

9. Re: a + b + c

Plato, you need to kiss his feet...

10. Re: a + b + c

Since I would do it Halls of Ivy's way, and Denis said that he is not as smart as Halls of Ivy, does it mean that I am smarter than Denis?

11. Re: a + b + c Originally Posted by Monoxdifly Since I would do it Halls of Ivy's way, and Denis said that he is not as smart as Halls of Ivy, does it mean that I am smarter than Denis?
On set of all people the relation $\mathscr{S}$ defined as $\mathscr{S}(A,B)\iff A\text{ is smarter than }B$ is
• irreflexive
• asymmetric
• transitive

I can absolutely assure that $(plato,HallsofIvey)\notin\mathscr{S}$ that said the jury is out on if $(HallsofIvey,plato)\in\mathscr{S}~.$

12. Re: a + b + c Originally Posted by Monoxdifly Since I would do it Halls of Ivy's way, and Denis said that he is not as smart as Halls of Ivy, does it mean that I am smarter than Denis?
Mr.Fly, I ain't smart enough to figure that out!

13. Re: a + b + c Originally Posted by DenisB Mr.Fly, I ain't smart enough to figure that out!
Me, too. My brain short-circuited.

14. Re: a + b + c Originally Posted by HallsofIvy Since I am not nearly as smart as Plato, ...
Why are people shortchanging themselves, even in jest!?
In imparting lessons, having arrogance/being full of yourself is a debit in
a ledger. You have a credit in the ledger if you are wise, and you
do not discredit yourself (or others).

15. Re: a + b + c

a=11-b
b+c=15
c+a=10

b+c=15
11-b+c=10

b+c=15
-b+c= -1

2c=14
c=7
b+7=15
b=8
a+8=11
a=3

Here are unknowns
a=3
b=8
c=7
And
a+b+c=3+8+7=18