ID: e6cb2402
(SAT Suite Question Bank > Find Questions > Assessment: SAT + Test: Math + Domain: Algebra)
Comment: Make sure you know what no solution can mean on the SAT.
Method 1: If the released official practice questions are anything to go by, then an equation or a system of equations is the most popular way to test the no solution concept. The easiest way to solve these is to make sure the algebraic part matches perfectly, sign and all, but that any detached numbers do not. In other words, if you see a line y = mx + b and a second line y = mx + b, you want everything to match except for b, because the lines would then be parallel, and no intersection equates to no solution. In the question at hand, there is only x to consider as a variable, so we just need to make sure that
3kx = (48/17)x
There is no need to divide each side by 3, unless that helps you reduce the fraction, as in
(48/17)/3 --> 48/17 * 1/3 --> 16/17
Most people, myself included, find multiplication easier than division, so you might just ask yourself 3 times what equals 48/17? Well, 3 * 16 = 48, leaving only the denominator. Thus, k must be 16/17. Whenever possible, it is a good idea to input a fraction rather than a decimal equivalent, and to reduce that fraction as much as possible, just in case you end up with something unwieldy, such as an equivalent 480/510, which could not be entered properly using only four characters or might take a bit of time to reduce. Also, one of my students once rounded when the question did not explicitly ask the test-taker to do so, and that question would have been counted incorrect on the exam. You want to make these kinds of mistakes in practice if you make them at all.
Method 2: Although you could easily split the equation into two and graph, using a slider for k, you would probably decide that 0.94 was close enough, and that answer would be considered incorrect. Since the problem does not tell you to round, you would need to find the decimal equivalent of 16/17 to the ten-thousandths (4th) place, which is not going to be apparent to the naked eye, even if you have set the proper parameters for the slider. A purely graphic solution is probably not an efficient way to approach this question.
Method 3: You could work out an algebraic solution using the entire equation if dealing with the algebraic part only, as shown above in method 1, makes you uncomfortable.
3(kx + 13) = (48/17)x + 36
3kx + 39 = (48/17)x + 36
3kx - (48/17)x = -3
Now, the only way to guarantee that no combination of k and x could make the left-hand side equal -3 (since there are no restrictions on k, which the problem only identifies as a constant) is to annihilate the left-hand side. In other words, if 3kx = (48/17)x, then all bets are off, since the left-hand side would have to be 0. Method 1 picks up solving from this step in either of two ways. Whichever way you approach the problem, k must be 16/17.
Comments
Post a Comment