More formally, these are called “rational expressions” in mathematics — “rational” in the sense of “having do with a ratio”, as the word is used in the phrase “rational numbers“. These rational expression appear in some of the most challenging Quant problems on the GMAT. Here are a few practice questions.
(D) a finite number greater than two
(E) infinitely many
5. As y increases from y = 247 to y = 248, which of the following decreases?
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
Solutions will follow this article.
Thoughts on rational algebraic expression
First of all, to understand this stuff, you should be clear on the basic rules of fractions: how to add, subtract, multiply, and divide them. If you can’t do this basic arithmetic with numerical fractions, it will be very hard to do it with algebraic rational expressions! Some further tips:
1. Suppose you have an equation involving one or more algebraic rational expressions. Suppose you are asked to solve for values of the variable. It’s important to note that any value of the variable that makes any individual denominator equal to zero cannot possibly be a solution of the equation. This can be a powerful tool in “which of the following could be a solution” question, because usually you can immediately eliminate a few answers right away, which sets you up very well for backsolving or solution behavior.
2. When adding or subtracting rational expressions, as when adding or subtracting ordinary fractions, we must find a common denominator to combine. We must do precisely the same thing with rational expressions. Here are a couple examples of this process.
3. Whenever you have just one algebraic fraction on one side of the equation equal to just one algebraic fraction on the other side of the equation, then you can cross-multiply. If either side has more than one fraction, added or subtracted, you would have to combine them, via the previous hint, before you are ready to cross-multiply.
4. If the whole equation has only one or two denominators, you also can simply multiply every term on both sides by the denominators. That can be a very efficient way to get rid of all the fractions in one fell swoop. For example, the equation:
can be simplified by multiplying each term by (x – 2) —- with the fraction, it cancels the denominator, simply leaving the numerator.
x(x – 2) = 2(x – 2) + 1
If we were to multiply this all out, we would get a quadratic that we could solve.
5. One could always use a direct algebraic solution: that may be efficient or that may take several steps and be time-consuming, even if you know you are doing. Remember that backsolving may be quicker. For a compound fraction (a big fraction with a little fraction in the numerator or denominator), it may well be quicker to step back and perform a more holistic solution, looking at what must be true about each piece: I demonstrate this in the solution for practice problem #2 below.
If this article gave you any insights, you may want to give the practice problems another look before jumping into the explanations below. Here’s another practice question from inside Magoosh.
If you would like to express anything or ask for clarification, please let us know in the comments section below.
Explanations to the practice problems
1) We could multiply by the three denominators, (x – 2), then (x – 1), then (x + 1), get three quadratics, and simplify. That’s a lot of work.
We could find a common denominator on the right, combine them into a single ugly fraction, then when we had a single fraction on each side, we could cross multiply. That’s also a lot of work.
The simple solution for which this problem is crying out is backsolving. Notice that three of the answer choices — (A) & (D) & (E) — are illegal because each makes one of the denominators zero. We can immediately eliminate all three of those. Now, it’s just a matter of plugging in the other two answer choices.
The two sides are not equal, so (B) can’t be the right answer. At this point, we pretty much know that (C) must be the answer, but it’s always good to verify that it works.
Both sides are equal, so x = 0 satisfies this equation. Therefore, answer = (C).
2) Rather than do a ton of algebraic re-arranging, let’s think about this. We have 3 divided by (something) equals 1/2. This means, the “something” must equal 6. That immediately produces the much simpler equation:
Answer = (A)
3) Multiply all three terms by x and we get
This equation is unfactorable. It is not a perfect square. Think about its graph, which is a parabola:
When x = 0, y is negative, and when x = 2, y is positive. Therefore, the parabola intersects the x-axis twice, which means the equation has two real solutions.
Answer = (C).
BTW, this is a special mathematical equation. One solution is the Golden Ratio, and the other solution is the negative reciprocal of the Golden Ratio.
4) This is an easy one to solve. Subtract 2 from both sides:
Now, add 3/y to both sides. Because the two fractions have the same denominator, y, we can just add the numerators:
Answer = (E)
5) As y gets larger, what happens to each one of these?
For statement I, as y gets larger, the 2y gets larger. Since the subtracting 100 stays the same as the value of y changes, that makes no difference. This one increases as y increases, so it is not a correct choice.
For statement II, as the denominator of a fraction increases, the value of the fraction overall decreases. When y increases, 50/y has to decrease. Again, adding 80 remains the same as y changes, so this doesn’t make any difference. This is a correct choice.
For statement III, as long as y > 3, then y^2 – 3y will increase as y increase. That means the entire fraction decrease. We are subtracting 100 minus the fraction, and if the fraction gets smaller, then we are subtracting something smaller and therefore are left with more. This means the entire expression, the difference, gets bigger as y increase. This one increases as y increases, so it is not a correct choice.
Answer = (B)