Question

1. $x - 14 = 42$, $x = 56$

56 ______ a solution.

1. $k + 66 = 98$, $k = 28$

28 ______ a solution.
for problems 9–10, solve the equation.

1. $55 + s = 110$, $s = $
2. $n - 52 = 12$, $n = $

module 11 • lesson 2

Explanation:

Problem 7: Determine if 56 is a solution for \( x - 14 = 42 \)

Step 1: Substitute \( x = 56 \) into the equation

Substitute \( x = 56 \) into the left - hand side of the equation \( x-14 \). We get \( 56 - 14 \).

Step 2: Calculate the result

\( 56-14=42 \), which is equal to the right - hand side of the equation \( x - 14 = 42 \).

Step 1: Substitute \( k = 28 \) into the equation

Substitute \( k = 28 \) into the left - hand side of the equation \( k + 66 \). We get \( 28+66 \).

Step 2: Calculate the result

\( 28 + 66=94
eq98 \)

Step 1: Isolate the variable \( s \)

To solve for \( s \), we use the subtraction property of equality. Subtract 55 from both sides of the equation \( 55 + s=110 \). The equation becomes \( s=110 - 55 \).

Step 2: Calculate the value of \( s \)

\( 110-55 = 55 \)

Answer:

is

Problem 8: Determine if 28 is a solution for \( k + 66 = 98 \)