: Thanks for your very thorough response!: Our system tracks students in math in 8th grade. The tracks for 8th
: grade are Accelerated Algebra, Algebra, and Pre-Algebra. This test
: was given to the middle track Algebra class. About 50% of our
: entire eighth grade population takes this class.: Much of the time this class doesn’t seem to be working out of their
: text book and according to my son, they jump around in the book
: when they do so it is very difficult for me to try to figure out
: what they have actually covered.: Supposedly the whole class did very poorly on this test. My son is
: not going to like it, but I am going to insist that he go over
: this test with me this weekend.: There are still 2 problems on this test that are gnawing at me. I’m
: almost positive that I have them worked out correctly. I think the
: reason they are still bothering me is that the school teaches a
: different system of “showing your work” than I learned.
: Theirs is visually very confusing to me. I can only solve the
: problems using my system!: Here are the problems. I would have shown my solutions but it took me
: long enough to figure out how to get this picture to work and it’s
: getting very late now…*************************************************************I looked at these problems once and winced, started to write that no way could a beginning (or even intermediate) algebra student solve them. Then I looked at them a second time and recognized that they are both trick questions; they look a lot more frightening than they really are. The teacher may be trying to teach kids to look past the fright for the trick(s); or he/she may not be sure of what level is appropriate.I personally don’t like “gotcha!” in education, ever, but some people think it amuses kids for some reason.I have no idea how to make the picture work (all this high-tech, just give me a nice quill and some oak galls to boil and a sheepskin to stetch, grumble, grumble); I can do |x| for absolute value of x, use the standard x^2 for x squared, and I’ll use instead of square root sign again.(1) 3 (x^2) + |x| - 36 = 964First trick: remember from yesterday that (x^2) = |x|(actually used for defining |x| in more advanced work)So the problem simplifies a lot to3|x| + |x| - 36 = 964Second trick: I don’t know what x is, and I don’t know what |x| is, but it is *some* number. And when I have three of something and one more of the same thing, whatever it it, I now have four of that thing, ie 3|x| + |x| = 4|x|Problem simplifies to 4|x| - 36 = 964Collect numbers by adding 36 each side, 4|x| = 1000Still don’t know x or |x|, but if four of something equals 1000, one of it must be 250: divide both sides by 4 and|x| = 250Now, two numbers have that absolute value. I could write the answer here by logic right now, but for reasons of my own (to lead to Example 3 below) I’m going to do it absolutely correctly algebraically and note that |x| = EITHER x if x is positive or zero OR -x if x is negative.So the above gives two equations:x = 250 (if x is positive) OR -x = 250 (if x is negative)Multiplying both sides of the second equation by -1 to get +x, we get the answersx = 250 OR x = -250********************Please stress the OR; it is *NOT* an ordered pair on a graph, it is *NOT* a sum (250 and -250 = 250 + -250 = 0); x is a single number.(2) 4|3t-2| - 6|3t-2| + |3t-2| = 10Looks horrible. And if it weren’t a setup, it would be almost impossible to solve. If you had different quantities in each of those absolute value signs, there would be twelve potential solutions.BUT it’s a setup! Trick: all those absolute value signs contain identical quantities. I don’t know what t is, or what 3t - 2 is, or its absolute value, and to be honest with you I don’t much care.Whatever it is, it’s a number, and if I take 4 of something, subtract six of the same, and add back one of the same, I get -1 of that something. Using u as a variable of some sort, 4u -6u + u = -u, always. So the left side of the above equation, treating |3t - 2| as one unit, simplifies down to -|3t-2|This gives -|3t-2| = 10Now, I *could* go into some algebra here and work this out and get some potential solutions and then check them in the equation. BUT — Trick #2! This is a setup.On the left is the *negative* of the absolute value of some number. I have no idea what the number might be, but it’s a real number. Its *absolute value* must be *positive or zero*. So the *negative* of that absolute value must be **negative or zero**.Can a negative or zero left side equal positive 10 on the right side, ever? No! (If this seems easier to look at, can -|u| = +10 ? NO)So the answer to this problem is **No solution************************************************************ Of course, I am assuming you entered the above without typographical error. (nasty stuff to type)What if we had a problem like this, which IS soluble? Example 3: -|2y - 5| = -7Multiply both sides by -1 to reverse signs|2y-5| = 7Now, Trick: remember |v| = v if v is positive or zero, and |v| = -v if v is negative.So we get two equations, each with a condition(2y - 5) = 7 *if* 2y - 5 >= 0or -(2y - 5) = 7 *if* 2y - 5 < 0first case, add 5 both sides, then divide both sides by 22y = 12y = 6and yes, 2y - 5 = 7 which is positivesecond case multiply both sides by -1 again (other routes also work, longer)2y - 5 = -7Add 5 both sides, divide both sides by 22y = -2y = -1and yes, 2y - 5 = -7, less than zeroAnswers y = 6 OR -1Why, why, why?? This stuff is so useless, and there are so many better and more valuable things to do in algebra.