Quick Algebra Review PDF

A Quick Algebra Review 1. Simplifying Expressions 2. Solving Equations 3. Problem Solving 4. Inequalities 5. Absolute Values 6. Linear Equations 7. Systems of Equations 8. Laws of Exponents 9. Quadratics 10. Rationals 11. Radicals Simplifying Expressions An expr...

A Quick Algebra Review 1. Simplifying Expressions 2. Solving Equations 3. Problem Solving 4. Inequalities 5. Absolute Values 6. Linear Equations 7. Systems of Equations 8. Laws of Exponents 9. Quadratics 10. Rationals 11. Radicals Simplifying Expressions An expression is a mathematical “phrase.” Expressions contain numbers and variables, but not an equal sign. An equation has an “equal” sign. For example: Expression: Equation: 5+3 5+3=8 x+3 x+3=8 (x + 4)(x – 2) (x + 4)(x – 2) = 10 x² + 5x + 6 x² + 5x + 6 = 0 x–8 x–8>3 When we simplify an expression, we work until there are as few terms as possible. This process makes the expression easier to use, (that’s why it’s called “simplify”). The first thing we want to do when simplifying an expression is to combine like terms. For example: There are many terms to look at! Let’s start with x². There Simplify: are no other terms with x² in them, so we move on. 10x x² + 10x – 6 – 5x + 4 and 5x are like terms, so we add their coefficients = x² + 5x – 6 + 4 together. 10 + (-5) = 5, so we write 5x. -6 and 4 are also = x² + 5x – 2 like terms, so we can combine them to get -2. Isn’t the simplified expression much nicer? Now you try: x² + 5x + 3x² + x³ - 5 + 3 [You should get x³ + 4x² + 5x – 2] Order of Operations PEMDAS – Please Excuse My Dear Aunt Sally, remember that from Algebra class? It tells the order in which we can complete operations when solving an equation. First, complete any work inside PARENTHESIS, then evaluate EXPONENTS if there are any. Next MULTIPLY or DIVIDE numbers before ADDING or SUBTRACTING. For example: Inside the parenthesis, look for more order of Simplify: operation rules - PEMDAS. -2[3 - (-2)(6)] We don’t have any exponents, but we do = -2[3-(-12)] need to multiply before we subtract, = -2[3+12] then add inside the parentheses before we = -2 multiply by negative 2 on the outside. = -30  Let’s try another one… Inside the parenthesis, look for order of operation rules - Simplify: PEMDAS. We need to subtract 5 from 3 then (-4)2 + 2[12 + (3-5)] add 12 inside the parentheses. This takes care of the P in PEMDAS, = (-4)2 + 2[12 + (-2)] now for the E, Exponents. We square -4. Make sure to use (-4)2 if = (-4)2 + 2 you are relying on your calculator. If you input -42 the calculator will = 16 + 2 evaluate the expression using PEMDAS. It will do the exponent = 16 + 20 first, then multiply by -1, giving you -16, though we know the = 36 answer is 16. Now we can multiply and then add to finish up. Practice makes perfect… Since there are no like terms inside the parenthesis, we need to distribute the negative Simplify: sign and then see what we have. There is really a -1 (5a2 – 3a +1) – (2a2 – 4a + 6) there but we’re basically lazy when it comes to the number = (5a2 – 3a +1) – 1(2a2 – 4a + 6) one and don’t always write it (since 1 times anything is = (5a2 – 3a +1) – 1(2a2)–(-1)(- 4a )+(-1)( 6) itself). So we need to take -1 times EVERYTHING in the = (5a2 – 3a +1) –2a2 + 4a – 6 parenthesis, not just the first term. Once we have done = 5a2 – 3a +1 –2a2 + 4a – 6 that, we can combine like terms and rewrite the = 3a2 + a - 5 expression. Now you try: 2x + 4 [2 –(5x – 3)] [you should get -18x +20] Solving Equations An equation has an equal sign. The goal of solving equations is to get the variable by itself, to SOLVE for x =. In order to do this, we must “undo” what was done to the problem initially. Follow reverse order of operations – look for addition/subtraction first, then multiplication/division, then exponents, and parenthesis. The important rule when solving an equation is to always do to one side of the equal sign what we do to the other. For example Solve: To solve an equation we need to get our x + 9 = -6 variable by itself. To “move” the 9 to the other side, we need to subtract 9 from -9 -9 both sides of the equal sign, since 9 was added to x in the original problem. Then x = -15 we have x + 9 – 9 = -6 – 9 so x + 0 = -15 or just x = -15. Solve: 5x – 7 = 2 When the equations get more complicated, just remember to “undo” +7 +7 what was done to the problem initially using PEMDAS rules BACKWARDS 5x = 9 and move one thing at a time to leave the term with the variable until the end. 5x = 9 They subtract 7; so we add 7 (to both 5 5 sides). They multiply by 5; we divide by five. x = 9/5 Solve: When there are variables on both sides of 7(x + 4) = 6x + 24 distribute the equation, add or subtract to move them to the same side, then get the term 7x + 28 = 6x + 24 with the variable by itself. Remember, -28 - 28 we can add together terms that are alike! 7x = 6x - 4 -6x -6x Your Turn: 2(x -1) = -3 x = -4 (you should get x = -1/2) Problem Solving Many people look at word problems and think, “I’m really bad at these!” But once we accept them, they help us solve problems in life when the equation, numbers, and variables are not given to us. They help us THINK, logically. One of the challenging parts of solving word problems is that you to take a problem given in written English and translate it into a mathematical equation. In other words, we turn words into numbers, variables, and mathematical symbols. There are three important steps to “translating” a word problem into an equation we can work with: 1. Understand the problem 2. Define the variables 3. Write an equation Let’s look at an example: The fence around my rectangular back yard is 48 feet long. My yard is 3ft longer than twice the width. What is the width of my yard? What is the length? First, we have to make sure we understand the problem. So what’s going on here? Drawing a picture often helps with this step. We know that the problem is describing a person’s rectangular yard. We also length know that one side is the width and the other side is the length. The perimeter of (distance around) the yard is width Yard 48ft. To arrive at that perimeter, we add length + length + width + width, or use the formula 2l + 2w = p (l = length, w = width, p = perimeter)  The problem also states, “The length of my yard is 3ft more Next, we have to define the variables: than twice the width.” This means that if I know the We know width, I canthemultiply perimeter is 2 it by 2w + 3 and 48ft,add but3we to determine do not know thethe length. length or width of the yard. w Yard Let’s have w = the width. We find the length by multiplying the width by 2 and adding 3, so length = 2w + 3. Let’s add these labels to the picture. Now, we write an equation. To write the equation we substitute our variables into the p = 2l + 2w equation for the perimeter. The formula already calls for w, so we can just leave that as is. Where it 48 = 2(2w + 3) + 2w calls for length, we can just plug in “2w + 3”. We already know our perimeter is 48ft, so we substitute that in for p. To solve the equation: Use the distributive property to multiply 2 by 2w and 3. 48 = 2(2w + 3) + 2w Combine like terms (4w and 2w). 48 = 4w + 6 + 2w Subtract 6 from both sides to “undo” the addition. 48 = 6w + 6 Divide by 6 on both sides to “undo” the multiplication. -6 -6 You are left with a width of 7 ft. Now we need to find 42 = 6w the length. 6 6 7ft = w=width  Check to see that it works Substitute our newly found width and simplify using order of operations  2(7) + 3 14 + 3 So now we know that the yard is 7ft wide and 17ft long. 17ft = length Your turn: I have a box. The length of the box is 12 in. The height of the box is 5 in. The box has a total volume of 360 in.2 What is the width of the box? Note: The formula for volume is V = lwh, where v = volume, l = length, w = width, and h = height. (You should get w = 6in.) Inequalities We’ve all been taught little tricks to remember the inequality sign. For example; when given x < 10, we know that x is less than ten because x has the LITTLE side of the sign and 10 has the BIG side of the sign. Solving inequalities is similar to solving equations; what you do to one side of an inequality, we must do to the other. If we are given x + 7 > 13 and asked to solve, we would undo the addition on the left side by subtracting 7 from both sides. We would then be left with x > 6, which is our answer. Suppose we were given ¼ x < 2. To undo the division, we would multiply both sides by 4. The result would be x < 8. But if we had – ¼ x < 2, we would multiply by both sided of the inequality by -4 and the rule is that when multiplying (or dividing) by a negative number, we must always flip the sign of an inequality. So we would get x > 2.