14. Linear and Quadratic Equations

Practice Question #6 (Trick Question w/ Difference of Squares)

(25 - x^9) / (5 - x^3) is equal to what? Solution: - None of the above since (25 - x^9) is NOT a difference of squares, as a difference of squares can only be when x is raised to an even power. - Trick question here.

Another way to express negative one

(x - y) / (y - x) = -1 (x - y) / (x - y) = 1

Practice Question #12 - Medium

(x^3 + x^2 + x + 1) / (x^2 + 1) Steps (our main goal is to factor out the x^2+1 on the denominator by getting that term on the numerator): 1. Factor out x^2 from the first two terms in the numerator, making it x^2 (x+1) + x + 1. 2. Now we can factor x+1 from the numerator, making it (x+1) (x^2 + 1), allowing us to factor out the bottom denominator and leaving us with x+1 as the correct answer.

Three common quadratic identities

1. (x + y)^2 = x^2 + y^2 + 2xy 2. (x - y)^2 = x^2 + y^2 - 2xy 3. (x + y) (x - y) = x^2 - y^2

Substitution Method

A method of solving a system of equations by replacing one variable with an equivalent expression containing the other variable. - Straightforward approach when it is easy to isolate a variable in one of the equations.

Practice Question #10

At a local bakery, cakes and pies are only sold for whole number of dollars. If the price of cake is more than $15, how much is the price for one pie? Two statements given: 1) 5 cakes and 30 pies cost $235. 2) Each cake costs $17. Solution (SS): 1. Set up an equation for 5c + 30p = 235, and solve for p (note that we can factor out 5 from 235 and 5c on the numerator and 30 on the denominator once they are on the right side of the equation) 2. As with most questions involving whole numbers / integers being a restriction in the problem, we now know that (47 - c) / 6 must be an integer. - Therefore, what are the values that makes the above an integer? There are multiple, so we don't know just based off the 1st statement only. 3. Once we add in the 2nd statement that the price of cake is $17, then now we know the price of pies is $5 (47-17 = 30, divided by 6 = $5). - We therefore need both statements to determine the correct answer.

Practice Question #16 - Medium (When One Equation is Sufficient to Determine Unique Values for Two Variables)

Cali purchased almonds and peanuts. Each almond cost $0.25, and each peanut cost $0.15. If the total spent on both was $4.35, what could be the number of almonds purchased? Solution (SS): 1. Set an equation and solve for "a", in this case it will be 3(29-p)/5 2. Now we have to substitute multiples of p which make the top divisible by 5 (29-p), we can't do 29 since we know Sarah purchased some almonds as well (i.e., otherwise almonds purchased would be zero if we subbed in 29 as the number of peanuts). 3. We go down the list in multiples of 5: 24, 19, 14, 9, 4, which leaves us with 4 peanuts and 15 almonds as the correct answer choice since the other # of almonds when plugging in those # of peanuts do not exist in the listed answer choices.

Practice Question #14 - Medium (FACTORING BY GROUPING)

FACTORING BY GROUPING IS KEY (try and factor x + 1 or x - 1 instead of just x or x^2): If we are given x^3 - x^2 - 4x + 4 = 0, we should look to pull out (x - 1) and group the above into two separate groups (see below), and now we can pull out the (x - 1) as a common factor. x^2(x - 1) - 4(x - 1); which makes (x-1) (x^2-4), and the resulting solutions are -2, 2, and 1, which equals 1 for the correct answer. Screenshot solution

Factoring Quadratic Equations

General quadratic equation form: ax^2 + bx + c = 0 Example: x^2 - 5x - 30 = 0; a = 1 b = -5 c = -30 Can be factored into (x + p) (x + q) - Where p and q must multiply together to c and add together to b.

Constant terms and coefficients in quadratic equations - Example Question

If -2 is a solution to the equation x^2 - 6nx = 40, in which n is a constant, which of the following could be a product of n and x? Solution (SS): 1. Plug in (-2) into the equation first for x and then solve for n. 2. Plug n back into the equation and factor the quadratic equation 3. Multiply the two results of x by n and check which of the possible answer choices fit into the result.

Practice Question #15 - Medium

If a = (b+c) / (b-c), what is the value of a^2 - 1? Solution: 1. We should square both sides and subtract it by 1 so that we can get a^2 - 1 on the left side; then it is just a matter of factoring and simplifying the right side until we get the correct answer. 2. To subtract 1 on the right side, we need to get a common denominator by multiplying the top and bottom by (b^2 - 2bc + c^2), then we can cross out the terms that cancel each other. 3. Now we get 4bc / (b^2 - 2bc + c^2), and we know that (b-c)^2 is the same as that expression, which we can substitute into the denominator and gets us to our answer.

Practice Question #3

If a and b are positive integers such that b is a multiple of 5, and 3a + 4b = 270, then "a" must be a multiple of what? Solution (SS): 1. Let b=5k for some positive integer k since b is a multiple of 5. 2. Plug the above into the 3a + 4b = 270 equation given in the question stem, and solve for a. 3. Since 10 is not divisible by 3, we see that (27 - 2k) must be some integer which is divisible by 3 instead. 4. As such, since a is the product of 10 and some integer (27 - 2k), we know that "a" must be a multiple of 10.

Practice Question #9 (Trick Question -- Part of an Equation can be substituted into another equation)

If pq = 2, what is the value of pq (p - q) 1) pq^2 = 4 2) q = 2 Solution (SS): - Each statement alone is sufficient. - We can substitute p = 2/q from the question stem into the first equation, making it 2q^2/q = 4, or q = 2, which is the same information as #2. And in this case p = 1.

Practice Question #11 - Hard

Is (x+8) / (y+8) = (-1)? Two statements: 1) x + y = (16x - 16y) / (y - x) 2) x = 3y Statement #1: 1. Try and isolate y so that we can substitute it back into the equation. - Main key here is to factor out (-16) from the numerator so that we can cross out (-x + y) from both the numerator and the denominator and set the equation for x + y = -16. 2. Plug it back into the question stem's equation and confirm whether the initial question stem's equation is accurate. Statement #2: 1. Plug in example numbers into the initial question stem (e.g., 1 for x, 3 for y), and check whether it equals -1. - Be careful not to just plug in only 1 or 2 sets of numbers. In the above example 1 for x and 3 for y doesn't work, but -12 for x and -4 for y would make that equation correct. Therefore this equation on its own is not sufficient for us to determine the answer.

Practice Question #2: Multiply Entire Expression by (b-3)/(b-3) to cancel out terms in numerator and denominator

Or we can simplify the expression by getting common denominators and combining terms - But a quicker way is to multiply the top and bottom by (b-3) to remove that from the denominator for both sides Question: What is equivalent to [2 - (b-2/b-3)] / [b-(4/b-3)] Solution (SS):

Practice Question #1

Question: (x/y - y/x) * [(x-y)^2 - (x+y)^2] / (x=y) = 8 What does (x-y) equal? Solution (SS): Tips: 1. Remember the rules of expanding quadratics (e.g., x^2-y^2 is the same as (x+y)(x-y). 2. We can subtract fractions with different denominators by having them share a common base (e.g., in this case we make the fraction on the left "xy" for both fractions before subtracting).

Practice Question #13 - Medium (Try and remove nested fractions in an equation by multiplying by the LCM of the two fractions)

Solution (SS): - Main key is to multiply the top and bottom by 32 to remove the nested fractions. - Then we can factor using the difference of squares, and we're left with (x+8) / 2, meaning that we can plug in the answer choices until we get an even number on top to make sure it is an integer.

Practice Question #8 (FOIL)

Solution: - Remember PEMDAS and FOIL it out - Remember that we may have to flip around our signs in the answer to fit the available choices. - For example, -2x^4 + x^2 + 4x is the answer, but that is equivalent to if we flipped all of the signs (2x^4-x^2-4x) as well.

Practice Question #17 - Medium (Use rule that x-y/y-x = -1)

Solution: 1. Factor out x^2 from top and bottom and then factor out the remaining (x^2-1) into (x+1)(x-1) / (1-x) 2. Using the rule (x-y)/(y-x) = -1, we can simplify the above into (x+1) * -1, which gives us the answer of -x - 1.

Combination Method

Subtract or add an entire equation from another equation when the values are the same to make the process quicker. Addition screenshot example (SS) 1) 2x + 2y = 100 2) 2x + y = 50 We can subtract equation #2 from equation #1 and directly get a result of y = 50. As an

Solving System of Equations for Two Variables

Two primary methods to solve systems of linear equations for two variables: 1. Substitution Method 2. Combination Method

Practice Question #5

What is the solution to (a-b+1) / [ab-b+(a^0-a^2)] Solution (SS) 1. We know that anything to the power of 0 is equal to 1, so we can set a^0 to 1. 2. Then we should factor out b from (ab - b) since it looks similar to the numerator 3. Expand out (1-a^2) using difference of squares; so it will equal (1+a)(1-a). 4. Try and create common terms in the denominator by re-expressing b (a-1) as -b (1-a) since these are equivalent expressions. - This is the key step as part of the equation to try and get like terms in the denominator that we can then factor out. 5. Now we can factor out (1-a) from the denominator, and we can divide the numerator and denominator by (a - b + 1) to get the final answer of 1 / (1-a).

Practice Question #4 - Reciprocals

Which of the following can be used to determine the value of a/b? Solution (SS) 1. We can reciprocate both sides to get a/b - In this case if b/a = 3, then we know a/b = 1/3 (the reciprocal) 2. We can cancel out the third variable c and then solve for a/b. 3. This is not sufficient to tell us the answer since we do not know the value of c and it does not further cancel out.

Practice Question #7 (Factoring out Common Factors)

b = (a+b-bc) / (a+b-c), what does "a" equal? Solution (SS): - Cross multiply and cancel out like terms ("bc" in this case) - Combine all terms with "a" on one side and the other terms on the other side. - Look to factor out common terms such that we are just left with "a" on the side w/ "a"', and this will be our answer. In this case we should factor out (b-1) from each side, and then we can divide both sides by that factor. General GMAT Note: - We can't divide by (b-1) on each side unless we know that b is NOT equal to 1, which is given in the question stem. Because otherwise we would be dividing by zero which is undefined.