Question

puzzle #2
1: solve for b:
$a = bc + d$
2: solve for b:
$ab = c + d$
3: solve for b:
$dfrac{cd}{b} = a$
4: solve for b:
$bc + a = d$
answer choices
a: $dfrac{ac}{d}$
b: $dfrac{cd}{a}$
c: $dfrac{c - a}{d}$
d: $dfrac{c + d}{a}$
e: $c + d - a$
f: $dfrac{a - d}{c}$
g: $ab - c$
h: $dfrac{d - a}{c}$
i: $da - c$
type the 4-letter code into the answer box. all caps, no spaces.

Explanation:

Sub - Question 1: Solve for \( b \) in \( a = bc + d \)

Step 1: Isolate the term with \( b \)

Subtract \( d \) from both sides of the equation \( a = bc + d \) to get \( a - d=bc \).

Step 2: Solve for \( b \)

Divide both sides of the equation \( a - d = bc \) by \( c \) (assuming \( c
eq0 \)) to get \( b=\frac{a - d}{c} \), which corresponds to option F.

Sub - Question 2: Solve for \( b \) in \( ab=c + d \)

Step 1: Solve for \( b \)

Divide both sides of the equation \( ab=c + d \) by \( a \) (assuming \( a
eq0 \)) to get \( b=\frac{c + d}{a} \), which corresponds to option D.

Sub - Question 3: Solve for \( b \) in \( \frac{cd}{b}=a \)

Step 1: Cross - multiply

Cross - multiply the equation \( \frac{cd}{b}=a \) to get \( cd = ab \).

Step 2: Solve for \( b \)

Divide both sides of the equation \( cd = ab \) by \( a \) (assuming \( a
eq0 \)) to get \( b=\frac{cd}{a} \), which corresponds to option B.

Sub - Question 4: Solve for \( b \) in \( bc + a=d \)

Answer:

F
D
B
H
The 4 - letter code is FDBH.