( yields. Combining the terms in the numerator you arrive at the answer:


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1 Algebra Skillbuilder Solutions: 1. Starting with, you ll need to find a common denominator to add/subtract the fractions. If you choose the common denominator 15, you can multiply each fraction by one accordingly to achieve that denominator: ( yields. Combining the terms in the numerator you arrive at the answer: 2. Beginning with, factor out the 4s on the left hand side of the equation to get alone: Then divide both sides by 4 and you ll have the answer: 3. Starting with remember Order of Operations and deal with the parentheses first. Multply the exponent 2 across the parentheses to yield: Then you can combine the m terms, adding the exponents because you re multiplying their bases: 4. Beginning with:, your job is to combine like terms to solve for a. It s likely easiest to begin by combining on each side of the equals sign: 5a 5a = 3a 0 = 3a So a must equal As you start with, your immediate goal should be to get rid of denominators to have a straight linear equation to work with. In order to do that, multiply both sides of the equation by 5: Then distribute that multiplication across the parentheses: 4c + 10 = 5c 10
2 Then subtract 4c from both sides: 10 = c 10 Then add 10 to both sides: 20 = c 6. Here it s important to keep your goal in mind, which isn t to solve for any one variable but rather for the combination. You start with two equations: So your goal should first be to get rid of the denominator in the first equation by multiplying both sides by y: x + y = 3y Then, as your goal is to solve for a combination of x and z, you ll want to y in terms of x so as to eliminate the y variable. Subtract y from both sides: x = 2y Then divide both sides by 2 and you ll have solved for y: This means that in the other equation, you can plug in instead of y: Then subtract from each side to get the x term alone: z = ½ (x) And then divide both sides by x to get the fraction to match the question, and you ll find that. 7. You begin this problem with two equations: And your goal is to solve for b. Your first goal should be to reduce the number of variables, and the second equation affords you a quick opportunity to do so. Since it tells you that b = 2a, you can plug 2a in for b in the first equation: 4a 2a = 6
3 Meaning that 2a = 6 and a = 3. Since b = 2a, then b = 2(3), so b = This problem tests your familiarity with exponent rules, a very important concept on the GMAT. As you start with, you should recognize that a negative exponent like is the same thing as  essentially you can take the negative exponent, flip it over the fraction bar, and make it positive. So what you really have here is. And with that in mind, you know that if you re multiplying together the same base (in this case 5) to two different exponents, you add the exponents, giving you. 9. The answer here is 0. Remember, = 1, so even though 7 is not a factor of 31, because 1 is a factor of 31 you can still make satisfy the equation. Accordingly, the correct answer is Beginning with, look for squares within the numbers underneath the radical and you ll see that what you really have is:, which means that the answer is. 11. With root problems that involve addition or subtraction, you ll typically want to factor out the roots to eliminate squares within them. As you have, look for ways to pull out actual squares:. This allows you to take the square root of the 3 2 terms, leaving:. Combine the like terms and you ll find the answer: With this problem you should recognize that you can quickly calculate the square root portion, as you know that. That means that is really just, which you know is Remember that with inequalities, you can add and subtract values from either side the same exact way you can with equations (the only time things get tricky is when you multiply or divide by a negative, or a variable). So as you start with:, you can subtract 3a from each side to get:, then to get a alone subtract 6 from both sides, and you ll see that: 14. means that either:, or
4 Solving for each, if 2x + 3 = 5, then subtract 3 from each side to get: 2x = 2, and then divide both sides by 2 to get: x = 1 Or, if 2x + 3 = 5, then subtract three from each side to get: 2x = 8, then divide both sides by 2 to get: x = 4 Looking at the answer choices, you ll only find 4, so 4 is correct. 15. In a quadratic like, you want to factor out the numeric term (in this case 12, but don t forget that it s negative). 12 could be: 1 and 12 2 and 6 3 and 4 Then look for a combination that adds/subtracts to the middle number (4a). Since you need one factor to be negative and one to be positive, the combination that works to add to 4 is (a + 6)(a 2) = 0. But remember you still have to solve the equation. If a = 6 or if a = 2, the equation stands, so the solutions are 2 and You can split into two inequalities to help keep your algebraic manipulation clean. You know that: 3y + 3 > 6, which allows you to subtract 3 from both sides to get: 3y > 9, and then you can divide both sides by 3 to get: y > 3 So that s one part of the solution, and now you can attack the other half. Knowing that: 3y + 3 < 12 allows you to again subtract 3 from both sides to get: 3y < 9, whereby you can again divide by 3 to get: y < 3 Combining what you know, you know that 3 < y < When you re given two inequalities like x + y > 8 and x y < 4, if you can get the signs going in the same direction you can add the inequalities together to eliminate a variable, essentially using the Elimination Method for multiple equations and multiple variables. And since the question asks you about y, your goal should be to eliminate x. You can do that by taking the second inequality:
5 x y < 4, and multiplying everything by 1 to flip the sign: x + y > 4. Pair that with the other inequality: x + y > 8, and add them together 2y > 4 So y > You should recognize the opportunity to use the Difference of Squares rule here, as the given equation fits the form x 2 y 2 = 0: 4x 2 9 = 0 can be written as (2x) = 0 This allows you to transform this into: (2x + 3)(2x 3) = 0, matching answer choice B. 19. This is a great question for using the answer choices as guides, since the question asks for what could be the value of x + y and not what is the value of x + y, the algebra won t necessarily lead to a specific value. You re looking for a number that equals its own square root, and since  1 and 2 don t qualify, the answer must be Solving for c given the equation, you ll first want to create a quadratic by adding 27 to both sides to form:. Then you can divide all terms by 3: for squares:, and now you can either factor or recognize the common algebraic equation, so c must be 3.
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