Question

1. 40x + 42 = 1,162
2. 2,528 - 72y = 8
3. 72y - 33 = 5,367
4. 13 + 33y = 277

Explanation:

Problem 5: \( 40x + 42 = 1162 \)

Step 1: Subtract 42 from both sides

To isolate the term with \( x \), we subtract 42 from both sides of the equation.
\( 40x + 42 - 42 = 1162 - 42 \)
\( 40x = 1120 \)

Step 2: Divide both sides by 40

To solve for \( x \), we divide both sides of the equation by 40.
\( \frac{40x}{40} = \frac{1120}{40} \)
\( x = 28 \)

Step 1: Subtract 2528 from both sides

To isolate the term with \( y \), we subtract 2528 from both sides.
\( 2528 - 72y - 2528 = 8 - 2528 \)
\( -72y = -2520 \)

Step 2: Divide both sides by -72

To solve for \( y \), we divide both sides of the equation by -72.
\( \frac{-72y}{-72} = \frac{-2520}{-72} \)
\( y = 35 \)

Step 1: Add 33 to both sides

To isolate the term with \( y \), we add 33 to both sides of the equation.
\( 72y - 33 + 33 = 5367 + 33 \)
\( 72y = 5400 \)

Step 2: Divide both sides by 72

To solve for \( y \), we divide both sides of the equation by 72.
\( \frac{72y}{72} = \frac{5400}{72} \)
\( y = 75 \)

Answer:

\( x = 28 \)

Problem 7: \( 2528 - 72y = 8 \)