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Case Western Univ., summa cum laude
Perfect scorer on the SAT & the ACT
Devorah is the founder of Advantage Point Test Prep and the author of the book “Boost Your Score” The Unofficial Guide to the Real ACT.
Let's get started talking about the top elementary algebra concepts you'll see on the Math ACT. We're going to talk about substituting the variable, simplifying equations, solving linear equations, and inequalities. And if you're thinking, "Oh my gosh, that sounds really overwhelming," Don't worry we're going to break it down.
Let's start talking about substituting the variables. These questions give you an algebraic equation and then just the value of a variable within and they're going to plug it in. These are really straight forward. Let's look at this one, it looks complicated when you first look at it, but take a close look, when x plus y is 3, what is the value of two times parenthesis x plus y plus x plus y over three minus parenthesis x plus y to the fourth. So again it looks horrible but what do we know here, we know x plus y is equal to three and now this is the case where you'll just plug that in wherever you see x plus y. So it turns into a really straight forward problem, we have two times three, like x plus y, plus three over three or just one, minus three to the fourth, okay great. So just to simplify, this is six plus one minus 81, okay, and that is equal to six, seven minus 81 or negative 74, and we've got it. So it wasn't that bad after all, the answer choice A would be correct.
Let's keep going. Let's take a look at simplifying equations, so x squared minus four x plus nine in parenthesis and in another parenthesis, we have three x squared minus two x minus six. It looks kind of complicated you're thinking, "Oh my gosh, is there multiplication going on, what's up with the parenthesis." But if you take a close look, all we doing is subtracting, so these question types are just testing, can you combine expressions. These you remember from algebra, you know in school, combining like terms. So let's go ahead and get started with this problem. So we're going to get rid of the parenthesis, x minus four x plus nine, and now the next part, students often mess up on this so keep an eye out. You want to distribute your negative, it is so important it needs to go to every one of your terms, because your parenthesis means that, it's got to travel to every one of the parts. So minus three x squared, plus two x 'cause you've got two negatives there, so plus two x and then also plus six, great. And now this is looking pretty straight forward right? And now you know you just have to combine like terms, which you probably remember doing you know from class. So we've got x squared minus three x squared, okay just negative two x squared, so let's cross this out this becomes negative two x squared. Next you've got your negative four x plus two x, so negative four x plus two x just negative two x, and I can cross these out too. Okay last, nine plus six just 15, and we've got it. We've just simplified these expressions perfect. So which of these choices matches this? We need negative two x squared minus two x plus 15, and C is the correct answer choice here.
Let's move on, next solving linear equations. Again pretty straight forward but it's worth reviewing. If five x plus three is 23, what's 12 x minus ten? So the linear equation are just those really straight forward ones, you know we're not talking x squared, x cubed, just a plain x, often just one variable and that's what you're going to do, you're going to figure out what that x is? Five x plus three is 23 and then there is another step, okay, so what's 12 x minus ten. So you just take it into parts, if five x plus 3 is 23, first of all what's x? So five x plus three equals 23, so you know you need to isolate the variable, we're going to take three to the other side, okay so we have that five x is equal to 20. And in that case x is going to be equal to four, right? Cause you divide both sides by five. Okay, but we not done, and by the way you guys a tip for you, once in a while the ACT people are kind of mean that actually one of the answer choices would be, the answer that you'll get the first half of the problem, if it's a two part problem like this. So keep an eye out, here they're nice, we don't have a four, but sometimes there might be. Okay we know we're not done, we need to plug in our four to our x, back into 12 x minus ten, let's do that, so 12 times four minus ten, okay we know we do our multiplication before a subtraction, 48 minus ten which is going to equal 38. So B is the correct answer choice here. And you see how even though this looked you know, a couple of steps maybe a little tricky, it's really pretty straight forward if you remember the stuff you learn in school about isolating your variable.
Let's keep going and look at Inequalities. These are one of my favorite topics on algebra. An inequality states that one side of the equation is greater than the other. So just review of what the sign means, if you have A, this is A is less than B, if you've got, my teacher I think it was fifth grade, think about it you probably had this to, she says that the crocodile eats the bigger side you know, so if you had an open mouth, it would be pointing to the larger part of the equation, that's how I remember it, here because the mouth is pointing to the B, the B is larger, so A is less than B, A is greater than B, A is less than or equal to B if you've got this little line here, And A is greater than or equal to B, okay just a quick review.
Let's do a problem, now let me tell you something important, these look kind of intimidating but actually do them how you would do a straight forward algebra problem, pretending that this was actually an equal sign. The only time anything is ever going to change is if you multiply or divide by a negative number, watched out for this, they test this on the ACT and in that case, you're going to flip your inequality. Okay, so if four times x minus three minus two, is greater, okay let me write this out, four times x minus three minus two is greater than or equal to three times four x plus five. Okay, let's do it how we would do a typical equation, and again you're going to what to isolate the variables so we move some stuff around. First though, you know you need tackle out the parenthesis, so we've got to do four x minus 12 here and distribute four x minus 12 minus two is greater than or equal to 12 x plus 15. Okay this is already looking simpler, now you know this is end, this ends the thing, four x minus 14, 14 greater than or equal to 12 x plus 15, now we can move the 14 to the other sign and then we can move the 12 x to the side. Let's do that at the same time, so we have the 12 x moving over here, so you would have, negative eight x and that if you move the 14 to this side and you add it, you going to have its greater than or equal to 29. Okay we're almost finished, now all we have to do is divide by negative eight on both sides. And yes I said negative, we're dividing by a negative number, so we're going to switch the inequality and it's going to look like this, divide by negative eight, divide by negative eight. So now x is less than or equal, I flipped it, than 29 over negative eight. And actually you can't simplify that any further because 29 is prime. Great, which answer choice looks like that, and we see it's answer choice D, x is less than or equal to 29 over negative eight. And be careful because always one of the answer choices will be the same answer if you didn't flip the inequality, okay.
That's it for the Top Elementary Algebra Topics you'll see on the ACT. And again there are ten of these that you will see in the Math section. Now if you still feel like you need a little bit of work that's no problem, head to the bonus material and there you will find a lot of resources with extra practice problems.
On to intermediate algebra. These are the two concepts you'll see showing up most frequently for intermediate algebra on the ACT, and remember there are nine intermediate algebra questions. First solving quadratic equations and then solving systems of equations we'll go over this. Let's start with the first one quadratic equations. Hopefully you guys remember what these are from Math class. This an equation where something is squared and they're usually in the form AX squared plus BX plus C and this when you factor them with your two sets of parenthesis and you find the solutions. So let me tell you something really neat actually your calculator has a really good program that will calculate this for you. Students don't know that programs on your TI calculator are okay on the test but they are. So check your bonus material and I actually have a program there that will do this equation for you all you have to do is put in the number that shows up here in each of them. So here there's no number you put one you'd put four, you'd put four and you would do tup tup turup, take a second calculate your roots no brain power required on your end.
But just in case let's look at an example that you would actually do by hand. We've got a real problem here X squared minus three equals two X find the solutions. So okay we're looking for the possible values of X. If you look at this and you're thinking huh it doesn't quite look right remember you can't factor unless it says equal to zero so we need to move this around a little bit. Let's do that first so we've got, X squared minus three we'll subtract the two X to the other side equal zero. Now still doesn't look great because you want it in the form of you know AX squared plus or minus BX plus or minus C where the number appears third and that number attached to the X appears second. So let's switch this around minus three, minus two X is the same thing as minus two X minus three so let's write that out. X squared minus two X minus three equals zero great. And now we can factor we'll put out our two sets of parenthesis okay we got an X on each side. So as you remember from school if you have a minus and a minus in your actual equation you know something about what's going to happen in your parenthesis? It's not going to be a plus here and a plus here 'cause that would give you all addition right? And it's not going to be a minus and a minus 'cause when you multiply negatives you also just get pluses. You're going to need one of these to be adding and one of these to be subtracting. So one of these is going to be a plus, one of these is going to be a minus. Remember when you're filling this out you're looking for two numbers that will multiply together to give you negative three, but add to give you negative two. Let's think about it for a second what about one and three? Let's say if you had one and three they combine to give you three and they would add to give you or subtract to give you two. The question is what's going to go where? Let's try to have you've got your three here and your one here and let's see how that would work. So we'd have X squared minus X plus three X that would give us a positive two X so that's not the right way to do it. We want to flip these and have the one appearing here and the three appearing here. Let's try it again, so X plus one and X minus three and now when you think about it that works right? You'd have if you've worked it back you would have X squared minus three plus one X that's still negative two X minus three and that's perfect.
Still the question is we're not done what's the solutions here? I have a lot of students that would look at this and they'll say oh you know one negative three perfect. Don't make that mistake because you know what there will always be an answer choice and that's wrong. Remember now you know that X plus one and X minus three each of those is equal to zero. So if you have X plus one is equal to zero X could be negative one right? So one of your roots is X equals negative one. Also we've got X minus three is equal to zero so X is going to be three. So our two roots are X equals negative one and X equals three which of these has those numbers? C we want negative one and we want positive three so C will be the correct answer choice here.
Okay next topic solving systems of equations. So this is a case where you're going to have two different questions with two variables that they both share. Here Y equals five X minus one and we have two Y equals six X plus 24 so what you do here? Hopefully you remember from school there's actually a couple of ways to solve a problem like this. You could use substitution and actually plug in we know since Y is five X minus one everywhere we see Y we can just plug in over here five X minus one. There's also something called combination you probably remember this you would line up your equations and either add or subtract to see if you can cross off a variable and solve it that way. So what I'm going to do here is just substitution. Within your bonus materials there will be a really good example of combination 'cause you know depending on the situation one thing might be easier than the other but you can always do either strategy. So here let's substitute it's easier because it's just really straight forward that Y is equal to five X minus one so for Y we can just plug that in. So here two times five X minus one is equal to six X plus 24. And now we can just simplify so we know ten X minus two is equal to six X plus twenty four. So if you move the six X over here and the two over here let's see you would have that four X 'cause we're subtracting, the six X, four X is equal adding the two over here to 26 and therefore X is equal to 26 over four which simplifies right? It simplifies to 13 halves. And there we go D is the correct answer choice here.
Let's recap in this episode we talked about elementary algebra and we talked about intermediate algebra. And there was a separate episode that dealt with pre-algebra. So you see algebra is a really big topic on the ACT and I want you to spend a lot of time practicing and head to your bonus materials if they're any topics here that you think you need more practice with.
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