
d. Each term has a base of n and an exponent of 3, so the base and exponent of your answer will be n^{3}. Subtract the coefficient of the second term from the coefficient of the first term: 20 – 17 = 3, so 20n^{3} – 17n^{3} = 3n^{3}.

d. Multiply the coefficients: (8)(12) = 96. The first term has bases of r, s, and t and the second term has bases of s and t, so your answer will have bases of r, s, and t.
The exponent of r in the answer is 3, because there is no r in the second term. Add the exponent of s in the first term to the exponent of s in the second term: 2 + 4 = 6. The exponent of s in the answer is 6.
Add the exponent of t in the first term to the exponent of t in the second term: 7 + 5 = 12. The exponent of t in the answer is 12.
(8r^{3}s^{2}t^{7})(12s^{4}t^{5}) = 96r^{3}s^{6}t^{12}.

c. Replace u with 3: 12(3)^{2}.
Exponents come before multiplication, so handle the exponent first: 3^{2} = 9.
The expression becomes 12(9).
Multiply: 12(9) = 108.

a. Use the distributive law to find 6( j – 4k). Multiply each term in parentheses by 6:
6( j – 4k) = 6j – 24k
This expression, –5j + 6j – 24k – 8k, has two j terms and one k term.
Combine the j terms: –5j + 6j = j.
Combine the k terms: –24k – 8k = –32k.
The expression is now j – 32k.
Substitute 2 for j and –1 for k: 2 – 32(–1) = 2 + 32 = 34.

b. The keyword half means multiplication by (or division by 2) and the keyword difference signals subtraction. The difference between a number and five is x – 5. Half of that quantity is .

strong>a. 1, 2, 3, 6, a^{8}, and b^{2} are factors of 6a^{8}b^{2} and 1, 3, 9, a5, b, and c5 are factors of 9a^{5}bc^{5}. The greatest common factor of 6 and 9 is 3. The variables a and b are common to both terms, but the variable c is not in the first term. The smaller exponent of a is 5 and the smaller exponent of b is 1, so 3a^{5}b can be factored out of every term. Divide both terms by 3a^{5}b: 6a^{8}b^{2} ÷ 3a^{5}b = 2a^{3}b, and 9a^{5}bc^{5} ÷ 3a^{5}b = 3c^{5.}
6a^{8}b^{2} + 9a^{5}bc^{5} factors into 3a^{5}b(2a^{3}b + 3c^{5}).

b. (e^{–2}) is raised to the negative second power, and then multiplied by 6.
Multiply –2 by –2: (–2)(–2) = 4, so (e^{–2})^{–2} = e^{4}.
Multiply e^{4} by its coefficient, 6: (6)(e^{4}) = 6e^{4}.

c. In the equation 9u = 63, u is multiplied by 9. Use the opposite operation, division, to solve the equation. Divide both sides of the equation by 9:
u = 7

d. To find f in terms of g, we must get f alone on one side of the equation, with g on the other side of the equal sign. In the equation f – 8 = g + 12, 8 is subtracted from f. Use the opposite operation, addition, to get f alone on one side of the equation. Add 8 to both sides of the equation:
f – 8 + 8 = g + 12 + 8
f = g + 20
The value of f, in terms of g, is g + 20.

a. The equation –2(3v + 5) = 14 has a constant multiplying an expression. The first step to solving this equation is to use the distributive law. Multiply 3v and 5 by –2: –2(3v + 5) = –6v – 10. The equation is now: –6v – 10 = 14. The equation shows subtraction and multiplication, so we must use addition and division to solve it. Add 10 to both sides of the equation –6v – 10 + 10 = 14 + 10
–6v = 24
Because v is multiplied by –6, divide both sides of the equation by –6:
v = –4

c. The equation shows multiplication and subtraction. We will need to use their opposites, division and addition, to find the value of c.
We add first, because addition comes before division in the inverse of the order of operations:
Divide both sides of the equation by the coefficient , which is the same as multiplying by its reciprocal , to get c alone on the left side:
c = 16

b. In the equation 3√ 12w = 18, three times the square root of 12w is equal to 18. First, divide both sides of the equation by 3:
√ 12w= 6
To remove the radical symbol from the left side of the equation, we must raise both sides of the equation to the second power.
( √ 12w)^{2} = (6)^{2}
12w = 36
Because w is multiplied by 12, divide both sides of the equation by 12:
w = 3

a. The slope of the line is , because the line is in slopeintercept form, and the coefficient of x is . Any line with a slope of , including y =x, is parallel to the line .

c. First, find the equation that was used to build the table. Use the first two rows of the table to find the slope: . Use the second row of the table and the equation y = mx + b to find the yintercept of the equation:
11 = –2(–2) + b
11 = 4 + b b = 7 The equation for this table is y = –2x + 7. To find the value of y when x = 2, substitute 2 for x in the equation and solve for y:
y = –2(2) + 7
y = –4 + 7
y = 3
The missing value in the table is 3.

d. The graph of y = –3 is a horizontal line along which every point has a y value of –3. Only the graphs in choices c and d are horizontal lines, and only along the graph in choice d does every point have a y value of –3.

b. To find the equation of the line, begin by finding the slope using any two points on the line. When x = 3, y = 7, and when x = 0, y = 5. Use these points to find the slope: . The yintercept can be found right on the graph. The line crosses the yaxis where y = 5, which means that 5 is the yintercept. This is the graph of the equation .

d. Use the distance formula to find the distance between (–5,4) and (4,–7). Because (–5,4) is the first point, x_{1} will be –5 and y_{1} will be 4. (4,–7) is the second point, so x 2 will be 4 and y_{2} will be –7:
D=√ (x_{2}x_{1})^{2} + (y_{2}y_{1})^{2}
D=√4(5))2 + (74)2
D=√ (9) 2 + (11) 2
D=√81 + 121
D=√202units

c. The denominator of a fraction cannot be equal to zero, so we can substitute any real number for x in the equation except –8, because –8 would make the fraction undefined. The domain of the equation is all real numbers except –8.

b. This system could be solved using either substitution or elimination. Because adding the two equations would eliminate the x terms, use elimination to solve.
Add the equations:
Divide by 12 to solve for y:
12y = –12
y = –1
Substitute –1 for y in either equation and solve for x:
5(–1) – 2x = 1
–5 – 2x = 1
–2x = 6
x = –3
The solution to this system of equations is x = –3, y = –1.

a. To solve the inequality, we must get x alone on one side of the inequality. Subtract 11 from both sides of the inequality and divide by –8. Because we are dividing by a negative number, change the inequality symbol from less than to greater than:
–8x + 11 < 83
–8x < 72
x > –9

a. The graphed line, y = 3x + 1, is solid, which means that the points on the line are part of the solution set. The answer must be either choice a or choice d. The point (0,4) is in the solution set, so it can be used to test each inequality. It is true that 4 ≥ 3(0) + 1, because 4 ≥ 1, so this is the graph of the inequality y ≥ 3x + 1. The same point, (0,4), fails when inserted into the inequality y ≤ 3x + 1.

a. To find the product of two binomials, use FOIL and combine like terms:
First: (10x)(5x) = 50x^{2}
Outside: (10x)(–2) = –20x
Inside: (4)(5x) = 20x
Last: (4)(–2) = –8
50x^{2} – 20x + 20x – 8 = 50x^{2} – 8

b. The binomial 25x^{16} – 144 is the difference between two perfect squares. The coefficient of x, 25, is a perfect square, and the exponent of x is even. The constant, 144, is also a perfect square. The square root of 25x^{16} is 5x^{8}, and the square root of 144 is 12. The square root of the first term, 5x^{8}, plus the square root of the second term, 12, is the first factor: 5x^{8} + 12. The square root of the first term minus the square root of the second term is the second factor: 5x^{8} – 12. Therefore, 25x^{16} – 144 factors to (5x^{8} + 12)(5x^{8} – 12).

d. The equation x2 + 7x + 5 = 0 is in the form ax2 + bx + c = 0, but this trinomial cannot be factored easily because its roots are not integers. Use the quadratic formula. The coefficient of x^{2} is 1, so a = 1. The coefficient of x is 7, so b = 7, and the constant is 5, so c = 5. Substitute these values into the quadratic formula:
Next, (7)2 = 49 and (4)(1)(5) = 20:
Under the square root symbol, 49 – 20 = 29, and in the denominator, 2(1) = 2:
The square root of 29 cannot be simplified. One root is equal to , and the other root is equal to .

d. We're looking for the original size of Nancy's farm, so we can use x to represent that number. Sherry buys oneeighth of Nancy's farm, which means that she buys acres. Sherry already had 6 acres, which means that the size of her farm is now . We know that her farm is 10 acres, so we can set equal to 10 and solve for x:
+ 10
x = 32
Nancy's farm was 32 acres.

b. The number of students who wear glasses is unknown, so represent that number with x.
The number of students who wear contact lenses is 36 less than four times the number of students who wear glasses, which means that the number of students who wear contact lenses is equal to 4x – 36. The ratio of students who wear contact lenses to students who wear glasses is 7:2, or . The ratio of actual students who wear contact lenses to actual students who wear glasses is (4x – 36): x, or . These ratios are equal. Write a proportion and solve for .
Multiply both sides by 2x: 7x = 8x – 72.
Subtract 8x from both sides: –x = –72, so x = 72.
Therefore, 72 students wear glasses.

a. The probability of selecting a can of mushroom soup is equal to the number of cans of mushroom soup divided by the total number of cans of soup. Let x represent the number of cans of mushroom soup Kara buys. There are now 4 + x cans of mushroom soup, and 4 + x + 10 + 8 = 22 + x total cans. The probability of selecting a can of mushroom soup is equal to .
To find the value of x that will make this probability equal to , set the two fractions equal to each other, cross multiply, and solve for x:
4(4 + x) = 1(22 + x)
16 + 4x = 22 + x
16 + 3x = 22
3x = 6
x = 2
For the probability of selecting a can of mushroom soup to become , Kara must buy 2 cans of mushroom soup.

d. The original value is unknown, so let x represent the original value. The new value is 93. Percent decrease is equal to the original value minus the new value, all divided by the original value. Subtract 93 from x and divide by x. Set that fraction equal to 22.5%, which is 0.225:
Multiply both sides of the equation by x, and then subtract x from both sides:
0.225x = x – 93
–0.775x = –93
Divide by –0.775: x = 120.
Therefore, 120, after a 22.5% decrease, is 93.

c. The formula t_{n} = t_{1}(r^{n – 1}) gives us the value of any term in a geometric sequence. The first term, t_{1}, is x – 8.
The ratio, r, can be found by dividing any term by the previous term. Divide the fourth term by the third term, because this will cause the variable x to be canceled: 9x ÷ –3x = –3, so r = –3. Because we are looking for the sixth term, n = 6:
t_{6} = (x – 8)(–3^{6} – 1)
t_{6} = (x – 8)(–3^{5})
t_{6} = (x – 8)(–243)
t_{6} = –243x + 1,944
The ratio of any term in the sequence to the previous term is –3. That means that the second term, 2x – 6, divided by the first term, x – 8, is equal to –3. Solve this equation for x:
–3x + 24 = 2x – 6
–5x + 24 = –6
–5x = –30
x = 6
The value of x in this sequence is 6, which means that the sixth term, –243x + 1,944, is equal to –243(6) + 1,944= –1,458 + 1,944 = 486.

c. The formula for area of a triangle is . Let x represent the base of the triangle. Because the height is two less than four times the base, the height of the triangle is 4x – 2. Substitute x for b and 4x – 2 for h. Substitute 45 for A:
45 = 2x^{2} – x
2x2 – x – 45 = 0
Factor this quadratic equation and find the solutions for x.
The factors of 2x^{2} are –2, –1, 1, 2, –x, and x. The first terms of the binomials may be 2x and x.
The factors of –45 are –45, –15, –9, –5, –3, –1, 1, 3, 5, 9, 15, and 45.
We need two factors that multiply to –45, which means that we need one positive number and one negative number:
(2x – 5)(x + 9) = 2x^{2} + 13x – 45. The middle term is too large.
(2x – 9)(x + 5) = 2x^{2} + x – 45. The middle term is the wrong sign.
(2x + 9)(x – 5) = 2x^{2} – x – 45.
Set each factor equal to zero and solve for x: ;
2x + 9 = 0
2x = –9
x = –4.5
A base cannot be a negative number, so this cannot be the answer.
x – 5 = 0
x = 5
The base of the triangle is 5 inches. Since the height is two inches less than four times the base, the height of the triangle is 4(5) – 2 = 20 – 2 = 18 inches.