Exercise Answers
Discrete Quantitative Questions
- B To solve for (x − 5 ), subtract 12 from both sides of the equation: x − 5 = 2.
Substitute 2 for (x − 5): 2^2 = 4. - 30 To get rid of the denominators, multiply across the equation by 30 (which
is a multiple of the values in all the denominators):
30(^12 a +^13 a + 101 a = 28)
15 a + 10a + 3a = 840
Combine like terms:
28 a = 840
a = 30
- A To express y in terms of x, you need to isolate y. Subtract 2y and 2x from
both sides of the equation and arrive at y = 3 − 2x. - B Notice that 4a + 6b = 2(2a + 3b). Therefore, to combine the equations,
multiply the first equation by 2 and arrive at 4a + 6b = 8. Substitute 4a + 6b
into the second equation and arrive at x = 8. - D To express a 2 in terms of x, you need to isolate a 2. To do so, first subtract x
from both sides:^92 a = 3 − x. To go from^92 a to a 2 , divide both sides by 9 and arrive
at a 2 = 3 – 9 x. - E Since you have two linear equations, you have to decide between
combination and substitution. Since the coefficient in front of p in the second
equation is 1, substitution is the preferred method. Write p in terms of q:
p = 15 − 2q. Substitute (15 − 2q) for p in the first equation and arrive at
3(15 − 2q) + 5q = 27. Distribute: 45 − 6q + 5q = 27. Solve for q. q = 18. - 203 Since none of the variables has a coefficient of 1, use combination. You want
to solve for b, so multiply across the first equation by −4:
− 4(1. 5a + 3b = 20)
−6a − 12b = −80
Now the equations are:
−6a − 12b = −80
6 a + 9b = 60
Add the equations and arrive at −3b = −20. Solve for b: b = 20/3.
- A What is the relationship between x^2 + 5x + 6 and x^2 + 5x − 12? x^2 + 5x + 6 is
18 more than x^2 + 5 x − 12. Since x^2 + 5x + 6 = 32, x^2 + 5x − 12 = 32 − 18 = 14. - 361 Use PEMDAS: First evaluate the expression within the parentheses:
3 – 5 = –2:
–2(–2) ÷ 12^2
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