i need help despratly

[kylee]

if you have a problem that says
4\b-1\-7=17
first you would divide both sides by 4
and you get
\b-1\-7=4.25
then you take the problem out of the absolute value lines and you get:
b-1-7=4.25 b-1-7=-4.25
then you add 1 to each sides and you get:
b-6=5.25 and b-6=-3.25
and then you add 6 to both sides and then you get
b=11.25 b=2.75
then thats your answer right?
please just tell me if im right...
or if im wrong then will you tell me wat i did wrong and not tell me the answer... that would be greatly appreciated... thanx

 

[garf]

the correct answers are. 7 and -5
isolate the absolute value |b-1| then try to solve by cases, suppose b-1<0, and b-1>0, this will laed you to 2 different solutions.

 

[soroban]

Hello, kylee!

if you have a problem that says : .\(\displaystyle 4|b-1|-7\:=\:17\)

First you would divide both sides by 4

and you get: .\(\displaystyle |b-1|-7\:=\:4.25\) . . no!

\(\displaystyle \text{Divide }everything\text{ by 4: }\;\frac{4|b-1|}{4} - \frac{7}{4} \:=\:\frac{17}{4} \quad\Rightarrow\quad |b-1| - 1.75 \:=\:4.25 \quad\Rightarrow\quad |b-1| \:=\:6\)


\(\displaystyle \text{Then: }\;\begin{Bmatrix}b-1 \;=\; 6 & \Rightarrow & b \;=\; 7 \\ \\[-3mm] b-1 \;=\: \text{-}6 & \Rightarrow & b \:=\: \text{-}5 \end{Bmatrix}\)