Question

1. $73 = -6(k - 7) + 6(k + 5)$
2. $-9 + 4r = 4r - 3 - 6$
3. $4n + 5n + 15 = 5n + 7n$

Explanation:

Problem 10: \( 73 = -6(k - 7) + 6(k + 5) \)

Step 1: Distribute the terms

First, we distribute the \(-6\) and \(6\) into the parentheses:
\( 73 = -6k + 42 + 6k + 30 \)

Step 2: Combine like terms

The \(-6k\) and \(6k\) cancel each other out:
\( 73 = 42 + 30 \)
\( 73 = 72 \)
Since \(73
eq 72\), this equation has no solution.

Problem 12: \( -9 + 4r = 4r - 3 - 6 \)

Step 1: Simplify the right side

Combine the constants on the right side:
\( -9 + 4r = 4r - 9 \)

Step 2: Subtract \(4r\) from both sides

Subtract \(4r\) from both sides:
\( -9 = -9 \)
This is a true statement, so the equation has infinitely many solutions.

Problem 14: \( 4n + 5n + 15 = 5n + 7n \)

Step 1: Combine like terms on both sides

Left side: \( 9n + 15 \)
Right side: \( 12n \)
So the equation becomes:
\( 9n + 15 = 12n \)

Step 2: Subtract \(9n\) from both sides

\( 15 = 3n \)

Step 3: Solve for \(n\)

Divide both sides by \(3\):
\( n = 5 \)

Answer:

s:

• Problem 10: No solution
• Problem 12: Infinitely many solutions
• Problem 14: \( n = 5 \)