{"id":6856,"date":"2016-05-14t11:29:23","date_gmt":"2016-05-14t18:29:23","guid":{"rendered":"\/\/www.catharsisit.com\/hs\/?p=6856"},"modified":"2017-09-28t09:06:56","modified_gmt":"2017-09-28t16:06:56","slug":"act-algebra-everything-need-know","status":"publish","type":"post","link":"\/\/www.catharsisit.com\/hs\/act\/act-algebra-everything-need-know\/","title":{"rendered":"act math algebra: everything you need to know"},"content":{"rendered":"
of the 60 questions on the act math section<\/a>, about 21 of them will consist of algebra problems. that’s about 1\/3 of the total math test, so you’ll need to have a firm grasp on the concepts here in order to do well.<\/p>\n first, we’ll take a quick look at the elementary algebra topics you need to know for the act, conveniently listed so that you can go through and pick out the topics you need to work on. a list of intermediate algebra topics follows. finally, we’ll go into greater depth on intermediate algebra and the top three content areas you need to know for act math<\/a>.<\/p>\n <\/p>\n photo by felixleong<\/a><\/p>\n solving expressions with substitution<\/strong><\/p>\n here you need to be very comfortable to manipulating equations, solving a problem with given expressions and numbers, and plugging in numbers. this is your basic bread and butter of algebra,.<\/p>\n simplifying algebraic expressions<\/strong><\/p>\n know how to combine like terms and how to factor. in order to do this, you’ll writing\/solving expressions and equations<\/strong><\/p>\n on some problems, you’ll need to take a word problem and convert it into multiplying binomials<\/strong><\/p>\n when you see a binomial, the first thing you should think about is foil. this inequalities<\/strong><\/p>\n you can treat these the same as equations, but remember that the sign flips and points the other way whenever you multiply or divide both sides by a negative number.<\/p>\n quadratic equations<\/strong><\/p>\n most of the time, you can factor a quadratic equation problem on the act. however, you should know the quadratic formula just in case you need it. for more on quadratic equations, see below.<\/p>\n systems of equations<\/strong><\/p>\n there are multiple ways to solve a system of equations \u2013 so make sure you relationships among variables in an equation<\/strong><\/p>\n you’ll sometimes come across some tougher questions that ask you to think functions<\/strong><\/p>\n the act likes to throw in multistep problems that involve figuring out the answer to one function and then plugging that answer into a second function. make sure you don’t confuse one function with the other!<\/p>\n logarithms<\/strong><\/p>\n you won’t see too many of these on the act, but make sure you know how matrices<\/strong><\/p>\n again, it’s a topic that is not tested very often, but make sure you familiarize three concepts—quadratics, systems of equations, and logarithms<\/strong>—are probably ones with which you\u2019re less a bit less familiar (especially logarithms), but they come up fairly regularly on the act math test. in these types of problems, you’ll use some of the exact same skills you did for elementary algebra, but in more complex ways. here are the basics you\u2019ll need to know to master these types of questions.<\/p>\n quadratic equations<\/strong> have three terms and are in the form ax\u00b2 + bx + c. an example of a quadratic is x\u00b2 \u2013 5x + 6. to find the factors of this equation, we must set up our set of two parentheses: ( )( ). the first term in both parentheses must be x, since x multiplied by x is the only way to get x\u00b2. then we look at the coefficient of the second term, -5. it\u2019s important to include the sign in front of the integer as part of the coefficient. one of the rules of quadratic equations is that the second terms in the two factors must add<\/i><\/strong> together to equal the middle term\u2019s coefficient. so we need to think of two numbers that add together to give us -5.<\/p>\n already, we can think of many combinations: -6 and 1, -2 and -3, -200 and 105. so which pair is it? now we have to look at the integer that\u2019s the third term of the quadratic. here it\u2019s + 6. another rule of quadratic equations is that the third term of the quadratic equation will equal the product<\/strong> <\/b>of the second terms in the two factors. so not only do we need the two numbers to add together to equal -5, but we need them to multiply together to equal + 6. therefore the factors must be: (x \u2013 2) (x \u2013 3). the \u201croots\u201d or the \u201csolutions\u201d for this quadratic would be 2 and 3.<\/p>\n the act math test will often present you with two or more equations<\/b> with multiple variables. remember the \u201cn equations with n variables rule<\/i>.\u201d if there are 2 variables in an equation (for example, x and y), then there must be 2 equations that each contain those variables in order to solve. the two common ways to solve are substitution and combination.<\/p>\n logarithms<\/b> are a unique way of writing exponents. we\u2019re used to seeing exponents in a format like y = xa<\/sup>. in \u201clogs\u201d that equation is equal to logx<\/sub>(y) = a. this is the most essential piece of information you\u2019ll need to solve logarithms. you can get more practice with logarithms on purple math<\/a>!<\/p>\n let\u2019s try a practice logarithm problem, just like the ones you might see on test day:<\/p>\n given that logx<\/sub>a<\/i> = 2 and logx<\/sub>b<\/i> = 3, what is the value of logx<\/sub><\/i>(ab<\/i>)3<\/sup>?<\/p>\n here, the term we are interested in, logx<\/sub><\/i>(ab<\/i>)3<\/sup>, is equivalent to 3logx<\/sub><\/i>(ab<\/i>).<\/p>\n this can also be expressed as 3logx<\/sub>a <\/i>+ 3logx<\/sub>b<\/i>, and since we know the values of logx<\/sub>a<\/i> and logx<\/sub>b<\/i>, we can substitute to find the answer. logx<\/sub><\/i>(ab<\/i>)3<\/sup> = 3logx<\/sub>a<\/i> + 3logx<\/sub>b<\/i> = 3(2) + 3(3) = 15 (choice b).<\/p>\n if you don’t know these logarithmic identities, you can still solve the problem by finding values for x<\/i>, a<\/i>, and b<\/i> that satisfy the conditions. then, simply calculate the value of log(x<\/i>)(ab<\/i>)3<\/sup>. the easiest way to do this is to work with a base of 10, which would mean that x<\/i> = 10, a <\/i>= 100, and b <\/i>= 1,000. we can then calculate the answer:<\/p>\n logx<\/sub><\/i>(ab<\/i>)3<\/sup> = log(10)(100 * 1,000)3<\/sup> = log(10)(1,000,000,000,000,000) = log(10)(1015<\/sup>) = 15. the answer is (b).<\/p>\n now it’s time for you to put your head down and study hard. if you have a mountain of studying to do, don’t despair! i strongly recommend that you break it down into manageable, small sections so that you can learn effectively while avoiding burnout.<\/p>\n good luck!<\/p>\nact math: elementary algebra topics<\/h2>\n
\nneed to know basic math properties such as the associative and distributive
\nproperty.<\/p>\n
\nsimple linear equation or expression in order to solve it. to solve equations,
\nthe most reliable way to get to the answer is by isolating the variable on one
\nside of the equation and the numbers on the other side.<\/p>\n
\nstands for first, outer, inner, and last.<\/p>\nact math: intermediate algebra topics<\/h2>\n
\nknow how to do it several ways. depending on the problem, one way will be
\nquicker than another. more info on systems of equations follows below!<\/p>\n
\nabout what happens to one variable as another variable increases or
\ndecreases. if you aren’t sure, you can try plugging in a few values and
\ncheck the answer. beware of variables to a power higher (or lower) than 1.<\/p>\n
\nthey work and the basics of logarithms. if you don’t, scroll down and take a look at intermediate algebra in-depth.<\/strong><\/p>\n
\nyourself with them so that you don’t waste too much time trying to remember
\nhow they work.<\/p>\nact math: intermediate algebra in-depth<\/h2>\n
\n
act algebra review: what to do next<\/h2>\n