- c.We rewrite the given equation as an equiva-
lent one solved for |x|:
–3|x| + 2 = 5|x| – 14
–3|x| + 16 = 5|x|
16 = 8|x|
2 = |x|
Using the fact that |a| = bif and only ifa=
b, it follows that the two solutions of the
equation 2 = |x| are x=2. Thus, there are
two distinct values ofxthat satisfy the given
equation.
- a.Using the fact that |a| = bif and only ifa=b,
we see that solving the equation |3x– 32 | – ^19 = 0,
or equivalently |3x–^23 | = 91 , is equivalent to solving
3 x– ^23 =^19 . We solve these two equations
separately:
3 x– ^23 = – ^19 3 x– ^23 = ^19
3 x= ^23 – ^19 = ^59 3 x = ^23 + ^19 = ^79
x= 257 x= 277
So, both 257 and 277 are solutions to this
equation.
- b.Using the fact that |a| = bif and only ifa=b,
we see that solving the equation |3x+ 5 | = 8 is
equivalent to solving 3x+ 5 =8. We solve these
two equations separately, as follows:
3 x+ 5 = –8 3 x+ 5 = 8
3 x= –13 3 x= 3
x= –^133 x= 1
Thus, the solutions to the equation are x= –^133
and x= 1.
- b.First, we rewrite the equation in an equiva-
lent form:
–6(4 – |2x+ 3|) = –24
–24 + 6|2x+ 3| = – 24
6|2x+ 3| = 0
|2x+ 3| = 0
Now, using the fact that |a| = bif and only if
a=b, we see that solving the equation |2x+ 3|
= 0 is equivalent to solving 2x+ 3 = 0. The
solution of this equation is x= –^32 .So,we
conclude that there is only one value ofx
that satisfies this equation.
- a.First, we rewrite the equation in an equiva-
lent form:
1 – (1 –(2 –|1 – 3x|))= 5
1 – (1 – 2 + |1 – 3x|) = 5
1 – (–1 + |1 – 3x|) = 5
1 + 1 – |1 – 3x| = 5
2 – |1 – 3x| = 5
–|1 – 3x| = 3
1 –3x| = – 3
Since the left side is non-negative (being the
absolute value of a quantity) and the right side
is negative, there can be no value ofxthat sat-
isfies this equation.
- c.Note that |a| =|b| if and only if a =b.
Using this fact, we see that solving the equa-
tion |2x+ 1| = |4x– 5| is equivalent to solving
2 x+ 1 =(4x– 5). We solve these two equa-
tions separately:
2 x+ 1 = (4x– 5) 2 x+ 1 = –(4x– 5)
6 = 2x 2 x+ 1 = –4x+ 5
^13 = x 6 x= 4
x= ^23
Thus, there are two solutions to the original
equation.
ANSWERS & EXPLANATIONS–
