- b.First, clear the fractions from all terms in
the equation by multiplying both sides by the
least common denominator. Then, solve the
resulting equation using factoring techniques:
= 4 + 2 x^3 –1
(2x– 1)^2 = 4(2x– 1)^2 +
(2x– 1)^2
10 = 4(2x– 1)^2 + 3(2x– 1)
10 = 16x^2 – 16x+ 4 + 6x– 3
10 = 16x^2 – 10x+ 1
16 x^2 – 10x– 9 = 0
(2x+ 1)(8x– 9) = 0
x= – 21 ,^98
Since neither of these values makes any of the
expressions in the original equation, or any
subsequent step of the solution, undefined,
both of them are solutions to the original
equation.
- b.
^1 f= (k– 1)p^1 q+ ^1 q
=+ ^1 q
f(k^1 –1)f(k– 1)pq =[p^1 q+ ^1 q]f(k– 1)pq
pq=f(k– 1) + f)k– 1)p
q =
q=
- d.First, clear the fractions from all terms in
the equation by multiplying both sides by the
least common denominator. Then, solve the
resulting equation using factoring techniques,
as follows:
= x–^4 5
(x– 5) xx–– 51 = (x– 5)x–^4 5
x– 1 = 4
x= 5
Because this value ofxmakes the expressions
in the original equation undefined, we con-
clude that the equation has no solution.
- d.First, clear the fractions from all terms in
the equation by multiplying both sides by the
least common denominator. Then, solve the
resulting equation using factoring techniques:
(2p+1)(p– 5) –
(2p+ 1)(p– 5) 2 p^3 +1= (2p+ 1)(p– 5)p^2 –5
22 – 3(p– 5) = 2(2p+ 1)
22 – 3p+ 15 = 4p+ 2
–3p+ 37 = 4p+ 2
35 = 7p
p= 5
This value ofpmakes the expressions in the
original equation undefined, so the equation
has no solution.
^22
(2p+ 1)(p– 5)
^2
p– 5
^3
2 p+ 1
^22
(2p+ 1)(p– 5)
^22
2 p^2 – 9p– 5
x– 1
x– 5
f(k–1)(1 + p)
p
f(k–1) + f(k – 1)p
p
^1
pq
^1
f(k– 1)
^3
2 x– 1
^10
(2x– 1)^2
^10
(2x– 1)^2
ANSWERS & EXPLANATIONS–
