- b.
4– 3 ^2 x– 1 = ^12 – y
6 4– 3 ^2 x– 61 = 6^12 – y
2(4 – 2x) – 6 = 3(1 – y)
8 – 4x– 6 = 3 – 3y
2 – 4x= 3 – 3y
3 y= 1 + 4x
y= - e.Let xbe the smallest of five consecutive odd
integers. The next four consecutive odd integers
are given by x+2,x+ 4,x+ 6, and x+ 8. The
average of these five integers is equal to their
sum divided by 5, which is expressed symboli-
cally by. Since
this quantity is given as –21, we must solve the
equation = –21,
as follows:
= –21
= –21
x+ 4 = –21
x= –25
Thus, the least of the five integers is –25.
- a.First, we solve ab+ 6 = 4 for a:
ab+ 6 = 4
ab= –2
a= –2b
Next, substitute this expression for ainto the
equation –6b+ 2a– 25 = 5 and solve for b:
–6b+ 2a– 25 = 5
–6b+ 2(–2b) – 25 = 5
–6b–4b– 25 = 5
–10b– 25 = 5
–10b= 30
b= –3
Plugging this in for bin the expression a= –2b
yields a= –2(–3) = 6. Finally, we substitute
these numerical values for aand binto ab
2
to
obtain – 63
2
= –^12
2
= ^14 .
- d.Let x be the unknown number. The sentence
“Three more than one-fourth of a number is
three less than the number” can be expressed
as the equation ^14 x+ 3 = x– 3. We must solve
this equation for xas follows:
^14 x+ 3 = x– 3
4 (^14 x+ 3) = 4(x– 3)
x+ 12 = 4x– 12
x+ 24 = 4x
24 = 3x
x= 8
- c.
(2 – x)^52 x––x^2 = (2 – x)y
5 x– 2 = y(2 – x)
5 x– 2 = 2y– xy
5 x+xy= 2 + 2y
x(5 + y)= 2 + 2y
x= - b.Solve this problem by determining the
weight of each portion. The sum of the
weights of the initial corn is equal to the
weight of the final mixture. Therefore,
(20 bushels)+
(xbushels)=
[(20 + x) bushels]
Suppressing units yields the equation 20 56
+ 50x = (x+ 20)54.
54 pounds
bushel
50 pounds
bushel
56 pounds
bushel
2 + 2y
5 + y
^5 x+ 20
5
x+ (x+ 2) + (x+ 4) + (x+ 6) + (x+ 8)
5
x+ (x+ 2) + (x+ 4) + (x+ 6) + (x+ 8)
5
x+ (x+ 2) + (x+ 4) + (x+ 6) + (x+ 8)
5
1 + 4x
3
ANSWERS & EXPLANATIONS–