If a – (b/3) = 1/3 and a + (b/4) = 3/2, then what is the value of a + b? A. 1 B. 2 C. 3 D. 4 E. 5
(C) Whenever we have fractions in algebra, it is usually a good idea to eliminate them. Let’s multiply the first equation by 3, yielding:3a – b = 1.Then let’s multiply the second equation by 4, yielding:4a + b = 6. If we add the equations, the b terms cancel out, yielding: (3a – b = 1) + (4a + b = 6) 7a = 7 a = 1.
Now, we can substitute the value of a into either of the equations to determine that b = 2.Therefore, a + b = 1 + 2 = 3.The correct answer is choice (C).
If we add the equations, the b terms cancel out, yielding: (3a – b = 1) + (4a + b = 6)
7a = 7 a = 1
My question is how do you know when you should be adding equations together (The step referenced above). I know there has to be a fundamental concept I am forgetting. I think I tried the solving for a and then substituting the answer into the other equation, therefore treating each equation as if they are mutually exclusive. Please explain the correct approach/rationale and how to make the correct determination. Thanks.
The most obvious reason to add equations together or subtract one from another is having one of the variables with the same constant multiplier (or opposite, e.g. 3 and -3). a + 3b = 7 2a – 3b = 7 or a + 3b = 7 2a + 3b = 7
If we consider this specific question then you can see that we could subtract one from another right from the beginning : a – (b/3) = 1/3 – a + (b/4) = 3/2
a – (b/3) – a – (b/4) = 1/3 – 3/2 – (b/3) – (b/4) = 1/3 – 3/2 We got rid of variable a, though we have to deal with fractions. -(7b/12) = -(1/6) b = 2
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