Expressing a+b in terms of x and y, is it possible?

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potato124

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Is it possible to express a+b in terms of x and y if:

2a+b = x (1)
b-a = y (2)

These were some of my failures:
b = y-a

Substitute b into 1
2a + y-a = x
a + y = x
a = x-y

and then from there I tried
a + b = x - y + b (But this definitely looks wrong)
a + b = x - y + (y - a)

then I tried
2a = x - b
a = (x - b)/2
b = y - a

a + b = (x - b)/2 + y - a
a + b = (x - b)/2 + (2y - 2a)/2
a + b = [(x - b) + (2y - 2a)]/2
a + b =[2x + x - 2a - b]/2---------> I can't seem to remove a and b from this side of the equality

and then it keeps going on and I don't even know if what I'm doing is actually right within the rules of mathematics.
One of my efforts seemed to go down the right direction towards getting an answer, but I noticed that I accidentally did something wrong...

b-a = y
a = y+b (Which is definitely wrong but I didn't notice at the time)
a = (x - b)/2

y+b = (x-b)/2
2y + 2b = (x-b)
2y - x = -3b
b = (-2y+x)/3 ------->I use this below

and then for some reason I took the equation:
a = y + b
a + b = y + 2b
a + b = y + 2[(-2y+x)/3]
a + b = y + (-4y+2x)/3
a + b = (3y + -4y + 2x)/3
a + b = (2x-y)/3

The answer is supposed to be (2x+y)/3


Could someone help nudge me in the right direction?

(I had no idea where to post this question, so I decided to post it here... if I should've posted it somewhere else, I'll move this post to that category)
 
Last edited:
potato124 said:
Is it possible to express a+b in terms of x and y if:

2a+b = x (1)
b-a = y (2)
Click to expand...
Solve the way you would solve a system of equations if x and y were just numbers. You might eliminate b by subtracting (2) from (1) and solve for a, then similarly solve for b, and add the resulting expressions together.
potato124 said:
Nevermind I solved it....
Click to expand...
If you'd like to show your work, we can check it or make further suggestions.
 
potato124 said:
Nevermind I solved it....
Click to expand...

2a+b = x (1)
b-a = y ------> b=y+a (2)

Substitute 2 into 1:
2a + (y+a) = x
3a = x-y
a = (x-y)/3

Thus a+b is:

a+b = (x-y)/3 + (y+a)
a+b = (x-y)/3 + (x-y)/3 + y
a+b = (2x-2y)/3 + y
a+b = (2x-2y)/3 +3y/3
a+b = (2x+y)/3
 
Last edited:
potato124 said:
2a+b = x (1)
b-a = y ------> b=y+a (2)

Substitute 2 into 1:
2a + (y+a) = x
3a = x-y
a = (x-y)/3

Thus a+b is:

a+b = (x-y)/3 + (y+a)
a+b = (x-y)/3 + (x-y)/3 + y
a+b = (2x-2y)/3 + y
a+b = (2x-2y)/3 +3y
a+b = (2x+y)/3
Click to expand...
Good work. Another way would be:

Subtract eqn (2) from (1):

(2a+b) - (b-a) = x - y

3a = (x -y) → a = (x-y)/3

Rest of the steps will be same as yours.
 
potato124 said:
2a+b = x (1)
b-a = y ------> b=y+a (2)

Substitute 2 into 1:
2a + (y+a) = x
3a = x-y
a = (x-y)/3

Thus a+b is:

a+b = (x-y)/3 + (y+a)
a+b = (x-y)/3 + (x-y)/3 + y
a+b = (2x-2y)/3 + y
a+b = (2x-2y)/3 +3y
a+b = (2x+y)/3
Click to expand...
Small typo above (but the answer is correct):
a+b = (2x-2y)/3 + y

a+b = (2x-2y)/3 +3y/3

a+b = (2x+y)/3
 
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