1. solve for ( b )
( a + \frac{b}{c} = d )
2. isolate the variable ( x )
( 3 - x + k = 12 )
3. solve for ( y )
( \frac{1}{2}(2y - 6n) = 5n )
4. solve the equation for ( w )
( w + 2 = e - 4 )
5. find the expression that represents the value of ( v )
( 2(v - 8) = n + y )
6. isolate the variable ( r )
( rs - 5 = t )
7. solve the equation for ( a )
( t - ya = u )
8. solve for ( b )
( \frac{b - 5}{c} = k )

Question

Accepted Answer

### Problem 1: Solve for $ b $ in $ a + \frac{b}{c} = d $
# Explanation:
## Step 1: Subtract $ a $ from both sides
To isolate the term with $ b $, we subtract $ a $ from both sides of the equation.
$ a + \frac{b}{c} - a = d - a $
Simplifying the left side, $ a - a = 0 $, so we get $ \frac{b}{c} = d - a $.

## Step 2: Multiply both sides by $ c $
To solve for $ b $, we multiply both sides of the equation by $ c $ to cancel out the denominator.
$ \frac{b}{c} \times c = (d - a) \times c $
Simplifying, we get $ b = c(d - a) $.

# Answer:
$ b = c(d - a) $

### Problem 2: Isolate the variable $ x $ in $ 3 - x + k = 12 $
# Explanation:
## Step 1: Subtract $ 3 $ and $ k $ from both sides
First, we want to get the term with $ x $ alone. Subtract $ 3 $ and $ k $ from both sides.
$ 3 - x + k - 3 - k = 12 - 3 - k $
Simplifying the left side: $ 3 - 3 = 0 $ and $ k - k = 0 $, so we have $ -x = 9 - k $.

## Step 2: Multiply both sides by $ -1 $
To solve for $ x $, multiply both sides by $ -1 $ to make the coefficient of $ x $ positive.
$ (-1) \times (-x) = (-1) \times (9 - k) $
Simplifying, we get $ x = k - 9 $.

# Answer:
$ x = k - 9 $

### Problem 3: Solve for $ y $ in $ \frac{1}{2}(2y - 6n) = 5n $
# Explanation:
## Step 1: Distribute $ \frac{1}{2} $
First, distribute $ \frac{1}{2} $ to the terms inside the parentheses.
$ \frac{1}{2} \times 2y - \frac{1}{2} \times 6n = 5n $
Simplifying, $ \frac{1}{2} \times 2y = y $ and $ \frac{1}{2} \times 6n = 3n $, so we get $ y - 3n = 5n $.

## Step 2: Add $ 3n $ to both sides
To solve for $ y $, add $ 3n $ to both sides of the equation.
$ y - 3n + 3n = 5n + 3n $
Simplifying, $ -3n + 3n = 0 $, so we get $ y = 8n $.

# Answer:
$ y = 8n $

### Problem 4: Solve the equation for $ w $ in $ w + 2 = e - 4 $
# Explanation:
## Step 1: Subtract $ 2 $ from both sides
To isolate $ w $, subtract $ 2 $ from both sides of the equation.
$ w + 2 - 2 = e - 4 - 2 $
Simplifying the left side, $ 2 - 2 = 0 $, so we get $ w = e - 6 $.

# Answer:
$ w = e - 6 $

### Problem 5: Find the expression that represents the value of $ v $ in $ 2(v - 8) = n + y $
# Explanation:
## Step 1: Divide both sides by $ 2 $
First, divide both sides of the equation by $ 2 $ to isolate the term with $ v $.
$ \frac{2(v - 8)}{2} = \frac{n + y}{2} $
Simplifying, we get $ v - 8 = \frac{n + y}{2} $.

## Step 2: Add $ 8 $ to both sides
To solve for $ v $, add $ 8 $ to both sides of the equation.
$ v - 8 + 8 = \frac{n + y}{2} + 8 $
Simplifying, $ -8 + 8 = 0 $, so we get $ v = \frac{n + y}{2} + 8 $ (or can be written as $ v = \frac{n + y + 16}{2} $).

# Answer:
$ v = \frac{n + y}{2} + 8 $ (or $ v = \frac{n + y + 16}{2} $)

### Problem 6: Isolate the variable $ r $ in $ rs - 5 = t $
# Explanation:
## Step 1: Add $ 5 $ to both sides
First, add $ 5 $ to both sides of the equation to isolate the term with $ r $.
$ rs - 5 + 5 = t + 5 $
Simplifying, $ -5 + 5 = 0 $, so we get $ rs = t + 5 $.

## Step 2: Divide both sides by $ s $ (assuming $ s
eq 0 $)
To solve for $ r $, divide both sides of the equation by $ s $.
$ \frac{rs}{s} = \frac{t + 5}{s} $
Simplifying, we get $ r = \frac{t + 5}{s} $.

# Answer:
$ r = \frac{t + 5}{s} $ (for $ s
eq 0 $)

### Problem 7: Solve the equation for $ a $ in $ t - ya = u $
# Explanation:
## Step 1: Subtract $ t $ from both sides
First, subtract $ t $ from both sides to isolate the term with $ a $.
$ t - ya - t = u - t $
Simplifying, $ t - t = 0 $, so we get $ -ya = u - t $.

## Step 2: Divide both sides by $ -y $ (assuming $ y
eq 0 $)
To solve for $ a $, divide both sides of the equation by $ -y $.
$ \frac{-ya}{-y} = \frac{u - t}{-y} $
Simplifying, we get $ a = \frac{t - u}{y} $ (since $ \frac{u - t}{-y} = \frac{t - u}{y} $).

# Answer:
$ a = \frac{t - u}{y} $ (for $ y
eq 0 $)

### Problem 8: Solve for $ b $ in $ \frac{b - 5}{c} = k $
# Explanation:
## Step 1: Multiply both sides by $ c $
First, multiply both sides of the equation by $ c $ to cancel out the denominator.
$ \frac{b - 5}{c} \times c = k \times c $
Simplifying, we get $ b - 5 = kc $.

## Step 2: Add $ 5 $ to both sides
To solve for $ b $, add $ 5 $ to both sides of the equation.
$ b - 5 + 5 = kc + 5 $
Simplifying, $ -5 + 5 = 0 $, so we get $ b = kc + 5 $.

# Answer:
$ b = kc + 5 $

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QUESTION IMAGE

Question

Explanation:

Problem 1: Solve for \( b \) in \( a + \frac{b}{c} = d \)

Step 1: Subtract \( a \) from both sides

Step 2: Multiply both sides by \( c \)

Step 1: Subtract \( 3 \) and \( k \) from both sides

Step 2: Multiply both sides by \( -1 \)

Step 1: Distribute \( \frac{1}{2} \)

Step 2: Add \( 3n \) to both sides

Answer:

Problem 2: Isolate the variable \( x \) in \( 3 - x + k = 12 \)