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## MATH121, Unit 1 Test Guide

by: Mallory McClurg

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6

# MATH121, Unit 1 Test Guide Math 121

Marketplace > University of Mississippi > Math > Math 121 > MATH121 Unit 1 Test Guide
Mallory McClurg
OleMiss
GPA 3.37

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These notes cover Lessons 2.1a, 2.2, 2.3, 2.4, 2.5 and 2.6!
COURSE
College Algebra
PROF.
Dirle
TYPE
Study Guide
PAGES
6
WORDS
CONCEPTS
college, Algebra
KARMA
50 ?

## Popular in Math

This 6 page Study Guide was uploaded by Mallory McClurg on Monday September 12, 2016. The Study Guide belongs to Math 121 at University of Mississippi taught by Dirle in Fall 2016. Since its upload, it has received 70 views. For similar materials see College Algebra in Math at University of Mississippi.

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Date Created: 09/12/16
Math 121 Unit 1 Test – Study Guide I took all of these directly from my practice test online. Hope it helps you! Example 1. √(3y+37) + 7 = y + 6 (First, subtract 7 from both sides.) √(3y+37) = y – 1 (Square both sides of the equation.) 3y + 37 = (y – 1)(y – 1) (FOIL the right side of the equation.) 3y + 37 = y – 2y + 1 (Now, subtract “y – 2y + 1” from both sides to get a zero on the right side of the equation.2 3y + 37 – y – 2y + 1 = 0 (Combine like terms.) -y + 5y + 36 (Now, we’ve got to use the quadratic formula, since this would be pretty hard to factor without it. Remember, it’s “x (or the variable) equals the opposite of b, plus or minus the square root of b-squared minus four time a times c, all divided by two times a”, where a, b and c are in the form “ax + bx + c”.) a = -1 b = 5 c = 36 (I’m not going to write out how I worked the quadratic formula. Sorry. Just not even gonna do it.) y = -4 y = 9 (Don’t forget to verify these solutions in the original problem, because with some radicals, one of the “solutions” isn’t actually a solution.) y ≠ -4 y = 9 (Only one solution!) Example 2. (y – 1) – 12(y – 1) + 27 = 0 (Try to look at this problem and see that it’s actually in “ax + bx + c” form. Substitute 2 “A2 for the expression “y – 1”!) A – 12A + 27 = 0 (Factor it out!) (A – 9)(A – 3) 2 A = 9 A = 3 (So now, sinc2 we know that y – 1 = A, and A = 9 and 3, then y – 1 must equal 9 and 3.) y – 1 = 9 y – 1 = 3 (Add one to both sides of each equation, then take the square root of both sides.) y = +/- √10 y = +/- 2 (There are four solutions!) Example 3. (8z – 7)/ 6 + (21/12) = (16z + 10)/12 (First we can simplify some of these fractions. “21/12” is the same as “7/4”. We can also factor out a 2 from “(16z + 10)/12” to make it look more like the first fraction in the equation.) (8z – 7)/6 + (7/4) = (8z + 5)/6 (The LCM – which is the smallest number that can be divisible by the two denominators – is 12, so multiply all of the fractions by 12. Remember, when multiplying by a fraction, only the expression in the numerator is multiplied.) (96z – 84)/6 + (84/4) = (96z + 60)/6 (Now, simplify these fractions.) 16z – 14 + 21 = 16z + 10 (Combine like terms and set the equation equal to zero to solve for z.) 0 ≠ 3 (NO SOLUTION!) Example 3. |7x + 5| - 21 = 0 (First, we should try to get the absolute value on a side by itself, so add 21 to both sides.) |7x + 5| = 21 (At this point, we can determine that what’s in the absolute value signs may be equal to a positive or a negative number for x, so take out the absolute value signs and set it equal to the positive and negative of 21.) 7x + 5 = 21 7x + 5 = -21 (Now, all we have to do is solve both for x, by subtracting 5 from both sides and then dividing by the coefficient of x.) x = (16/7) x = (-26/7) (Two solutions!) Example 4. (9x/5) – (5/9x) / (9 – (5/x)) (First, look at the two fractions in the numerator and figure out how we can give them a common denominator. Multiply both fractions in the numerator by 45x.) (9x(45x)/5) – (5(45x)/9x) / (9 – (5/x)) (Simplify the fractions in the numerator.) (405x /5) – (225x/9x) / (9 – (5/x)) (Simplify even more, then divide your new numerator by 45x! We can also simplify the (9 – (5/x)) by making it ((9x – 5)/x).) 2 ((81x – 25)/45x) / (9x – 5)/x (We should notice that the numerator of our fraction on top is the difference of two squares, so we can factor it out easily. We also know that dividing two fractions is the same as multiplying fractions with one of the fractions flipped, so do that here!) ((9x + 5)(9x – 5)/45x) * x / (9x – 5) (Cross out the common expressions in the numerators and denominators.) (9x + 5)/45 (This is your answer!) Example 5. y + 3y – 36y – 108 = 0 (We need to look at this problem in two pairs. Find the greatest common factor in y and 2 2 3y , which is y , and factor it out of the first pair. Then, find the GCF in -36y and -108, which is -36, and factor it out of the second pair.) 2 y (y + 3) – 36(y + 3) = 0 (We can clearly see that (y + 3), or y = -3 is one of the final solutions to the problem. Now, we need to take the GCFs that we found in both pairs and subtract them.) (y + 3) (y – 36) (Recognize that the second expression is the difference of two squares, so factor it out.) (y + 3) (y – 6) (y + 6) (Set all of these equal to zero to find the solutions to y.) y = -3 y = 6 y = -6 (Three solutions!) Example 6. √(96+ 25y) – y = 10 (First, get the square root expression on a side by itself by adding a y to both sides.) √(96+ 25y) = 10 + y (Square both sides.) 96 + 25y = (10 + y)(10 + y) 2 (Foil out the right side of the equation.) 96 + 25y = 100 + 20y + y (Now, subtract the right side of the equation from the left side to set it equal to zero.) -4 + 5y – y = 0 2 -y + 5y – 4 = 0 (Solve for y with the quadratic formula.) y = 1 y = 4 Example 7. √(8z + 4) = √(6z + 1) +1 (First, we need to square both sides of the equation.) 8z + 4 = (√(6z + 1) + 1)(√(6z + 1)+ 1) (FOIL out the right side of the equation.) 8z + 4 = 6z + 1 + 2√(6z + 1) + 1 (Combine like terms.) 8z + 4 = 2√(6z + 1) + 6z + 2 (Subtract “6z + 2” from both sides to get the square root on a side by itself.) 2z + 2 = 2√(6z + 1) (Square both sides of the equation.) 2 (2z + 2)(2x + 2) = 2 * (6z + 1) (FOIL out the left side of the equation, and simplify the right side of the equation.) 4z + 8z + 4 = 24z + 4 (Subtract “24z + 4” from both sides to get the equation equal to zero.) 4z – 16z = 0 (Factor out a z.) z (4z – 16) = 0 (Set “4z – 16” to zero.) z = 0 z = 4 (Two solutions!) Example 8. 2 ((y + 6)/(y – 4)) – ((y – 1)/(y + 6)) – (5/(y + 2y – 24)) (First, FOIL out the denominator of the last fraction.) ((y + 6)/(y – 4)) – ((y – 1)/(y + 6)) – (5/(y – 4)(y + 6)) (Now we need to try and get a common denominator across all of the fractions. Here it will be (y – 4)(y + 6), so multiply the numerator and denominator of the first fraction by (y + 6), and then multiply the numerator and denominator of the second fraction by (y – 4).) ((y + 6)(y + 6)/(y – 4)(y + 6)) – ((y – 1)(y – 4)/(y + 6)(y – 4)) – (5/(y – 4)(y + 6)) (FOIL the numerators of the first two fractions, then subtract all of the numerators of the three fractions.) 2 2 (y + 12x + 36 – y + 5y – 4 – 5)/(y – 4)(y + 6) (Combine like terms in the numerator to simplify.) 17y + 27/(y – 4)(y + 6) (This is as simplified as it gets!) Example 9. y1/2 – 6y1/4+ 8 = 0 (We’re asked to solve this equation. First, we 2eed to recognize that this is in the proper “x + bx + c” form, so we can substitute “A” for “y ” since that is the common variable.) A – 6A + 8 = 0 (Factor this out to find what solutions we can find for A.) (A – 2)(A – 4) = 0 (Set each group equal to zero and solve for A.) A = 2 A = 4 (Now, since A equals “y ” , we can set that equal to 2 and 4, to find the solutions for “y”.) y = 16 y = 256 (Two solutions!) Example 10. (4/13)(y – 3) – (62/13) = -5 (First, distribute the “4/13” to the “y” and the “-3”.) (4y/13) – (12/13) – (62/13) = -5 (Combine the last two fractions that are on the left side of the equation.) (4y/13) – (74/13) = -5 (Now, multiply both sides of the equation by 13 to get rid of the fractions.) 4y – 74 = -65 (Get “y” on a side by itself.) y = (9/4) Example 11. |3p – 5| > p + 13 (We know that an expression in an absolute value can be a positive or a negative number. So, written in two different inequalities, replace the absolute value with “3p – 5” and “-3p + 5”.) 3p – 5 > p + 13 -3p + 5 > p + 13 (Get the variable “p” by itself on the left side of the equation by adding/subtracting 5 and then subtracting “p” from both sides.) p > 9 p < -2 (Now put this in interval notation.) (-∞, -2) U (9, ∞) (This is your answer!) Example 12. (2/(x + 2)) + (5/(x + 8)) = 1 (First we want to get the least common multiplier (LCM), which is (x + 2)(x + 8), so we should multiply the numerators of both fractions and the “1” on the right side of the equation by (x + 2)(x + 8).) (2(x + 2)(x + 8) / (x + 2)) + (5(x + 2)(x + 8) / (x + 8)) = 1 (x + 2)(x + 8) (Now, cross out similar groups in the numerator and denominator of each fraction to simplify.) 2(x + 8) + 5(x + 2) = (x + 2)(x + 8) (Distribute the 2 and 5 to their respective groups, then FOIL out the right side of the equation.) 2 2x + 16 + 5x + 10 = x + 10x + 16 (Subtract the right side of the equation from the left and then combine like terms.) -x – 3x + 10 = 0 (Now, since this is hard to factor out, we need to use the quadratic formula – see Example 1 if you don’t remember the formula.) x = -5 x = 2 (These are your two solutions!) Example 13. -6(y + 7) = -294 (First, divide both sides by -6 to get the expression with the exponent on a side by itself.)2 (y + 7) = 49 (Take the square root of both sides. Don’t forget! Taking the square root of a constant will result in a positive and a negative number.) y + 7 = 7 y + 7 = -7 (Get the variable on a side by itself to solve for y.) y = 0 y = -14 (Two solutions!) Example 14.2 2 33z + 7z = 10 (First, we need to put this in proper “ax + bx + c” form.) 7z + 33z – 10 = 0 (Factor it out.) (7z – 2)(z + 5) = 0 (Set each group equal to zero.) 7z – 2 = 0 z + 5 = 0 z = 2/7 z = -5 (These are the two solutions!) Example 15. |5y – 4| = |4y + 8| (Since we know the expression in an absolute value sign may be equal to the negative or positive of that expression, we can split this into four equations…although two of them will result in the same answer as the other two. We can set it up like (+A = +A) and (+A = -A), or (-A = -A) and (-A = +A).) 5y – 4 = 4y + 8 5y – 4 = -4y – 8 (Combine like terms on each side of the equal sign.) y = 12 y = -4/9 Example 16. -2 ≤ -2(n – 2) < 2 (Separate this out into two inequality equations.) -2 ≤ -2(n – 2) -2(n – 2) < 2 (Divide both sides of both equations by -2. Don’t forget to flip the signs!) 1 ≥ n – 2 n – 2 > -1 (Add 2 to both sides to get “n” on a side by itself.) 3 ≥ n n > 1 (Set this up in the combined form like it was in the original problem.) 1 < n ≤ 3 (Now, put this in interval notation.)

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