- d.
ax + b = cx + d
ax– cx=d– b
(a – c)x = (d – b)
= ((da––bc))
x= da––bc
- a.Let x be the smallest of the four whole num-
bers. The next three consecutive odd whole
numbers are then x+2,x+ 4, and x+ 6. The
sentence “The sum of four consecutive, odd
whole numbers is 48” can be expressed as the
equation x+ (x+ 2) + (x+ 4) + (x+ 6) = 48.
We solve this equation for xas follows:
x+ (x+2) + (x+ 4) + (x+ 6) = 48
4 x+ 12 = 48
4 x= 36
x= 9
Thus, the smallest of the four whole numbers
is 9.
- a.In order to solve for T, we must simply
divide both sides of the equation by nR. This
results in the equation T= PV\nR.
- a.
B =
B=(D –A) = C+ A
BD– BA= C+ A
BD– C= A+ BA
BD– C= A(1 + B)
A=
183. b.30% ofris represented symbolically as 0.30r,
and 75% ofsis represented symbolically as
0.75s. The fact that these two quantities are
equal is represented by the equation 0.30r=
0.75s. We are interested in 50% ofswhen r=
30. So, we substitute r= 30 into this equation,
solve for s, and then multiply the result by 0.50:
0.30(30) = 0.75s
9 = 0.75s
s= 0.^975 = 12
So, 50% ofsis equal to 0.50(12) = 6.
- e.We must solve the given equation for g:
fg+ 2f– g= 2 – (f+ g)
fg+ 2f– g= 2 – ( f+ g)
fg+ 2f– g= 2 – f– g
fg= 2 – f– g– 2f+ g
fg = 2 – 3f
g= 2–f3f
- b.Let xbe the width of the room. Then,
the length of the room is equal to 2x+ 3.
The perimeter of the room is given by 2x+
2(2x+ 3). Since this quantity is known to be
66, we must solve the equation 2x+ 2(2x+ 3)
= 66 as follows:
2 x+ 2(2x+ 3) = 66
2 x+ 4x+ 6 = 66
6 x+ 6 + 66
6 x= 60
x= 10
Thus, the length of the room is 2(10) + 3 = 23
feet.
BD– C
1 + B
C +A
D –A
(a – c)x
(a – c)
ANSWERS & EXPLANATIONS–
