Solve 2a + b = 15 algebraically

[Hoodleehoo]

I'm not really a beginner per se, but I'm sure this is a beginner level question lol

2a + b = 15

How do you solve this equation algebraically? I can intuit that if I divide 15 by 3 I can figure out the answer, but how do you solve it with algebra? Seems like it should be simple, but I'm stumped!

 

[pka]

How do you solve this equation algebraicly? I can intuit that if I divide 15 by 3 I can figure out the answer, but how do you solve it with algebra? Seems like it should be simple, but I'm stumped!

It cannot be solved. There are two variables but only one equation, making a solution impossible.
Check and see if you have listed all of the actual question.

 

[HallsofIvy]

What do you mean by "solve it"? This is one equation in two variables, a and b. There exist an infinite number of (a, b) pairs that satisfy that equation. You could solve for b "in terms of a": b= 15- 2a. Choose any value for a you wish and find the corresponding value of b. Or you can solve for a "in terms of b": a= (15- b)/2. Choose any value for b you wish and find the corresponding value of a.

Or do you mean this to be a "Diophantine equation"- that a and b must both be integers. We can find all pairs of integers, a and b, such that 2a+ b= 15. One obvious solution is a= 7, b= 1. But it is easy to see that if k is any integer, a= 7- k, b= 1+ 2k is also a solution: 2(7- k)+ (1+ 2k)= 14- 2k+ 1+ 2k= 15 for every k.

 

[Hoodleehoo]

Okay I see what you guys are saying. I came to this equation from another one. It started from 2a = b and then the second equation is 2a + 6 + b + 6 = 27.

A and B are ages of people. As in (6 years ago a was half the age of B. Now the sum of their ages is 27).

How would you plug in the 2a = b into the second equation to make it solvable using algebra?

 

[Hoodleehoo]

Oh wait, it shouldn't be 2a in the second problem that's where I went wrong.

If I change that I can subtract the other equation and get rid of B and end up with 3a = 15 which will get me to her correct age of 5.

Thanks guys!

 

[ksdhart2]

Okay I see what you guys are saying. I came to this equation from another one. It started from 2a = b and then the second equation is 2a + 6 + b + 6 = 27.

A and B are ages of people. As in (6 years ago a was half the age of B. Now the sum of their ages is 27).

How would you plug in the 2a = b into the second equation to make it solvable using algebra?

Well, this seems like a case where understanding what it really means for two expressions to be equal would be helpful. It means the expressions always have the same value. In this case, it means that, for any a you pick, b will always be twice that value. It also means that in any other expression, you can replace any b terms with 2a and not change the value of the expression. If 3b = 7 is a true statement, then 3(2a)=7 is also a true statement.

 

[pka]

Okay I see what you guys are saying. I came to this equation from another one. It started from 2a = b and then the second equation is 2a + 6 + b + 6 = 27.
A and B are ages of people. As in (6 years ago a was half the age of B. Now the sum of their ages is 27).
How would you plug in the 2a = b into the second equation to make it solvable using algebra?

Now you could have avoided all confusion if only you had given us the whole complete problem to begin with. But no you caused a great waste of time.
Please give us the problem.

 

[Hoodleehoo]

Well I wasn't trying to solve the problem, I solved it another way. I was trying to figure out why I couldn't solve the original equation I posted. It wasn't a waste of time because you guys answered my question and that really helped! So thank you!