Question

solve each equation and check your solution.

1. $2k + 3 = 15$
2. $\frac{c}{6} - 7 = 1$
3. $8k + 8 = -16$
4. $\frac{n}{9} + 5 = 14$

Explanation:

Problem 5: Solve \( 2k + 3 = 15 \) and check the solution

Step 1: Subtract 3 from both sides

To isolate the term with \( k \), we subtract 3 from both sides of the equation.
\( 2k + 3 - 3 = 15 - 3 \)
\( 2k = 12 \)

Step 2: Divide both sides by 2

To solve for \( k \), we divide both sides of the equation by 2.
\( \frac{2k}{2} = \frac{12}{2} \)
\( k = 6 \)

Step 3: Check the solution

Substitute \( k = 6 \) back into the original equation:
Left - hand side: \( 2(6)+3=12 + 3=15 \)
Right - hand side: \( 15 \)
Since the left - hand side equals the right - hand side, \( k = 6 \) is a valid solution.

Step 1: Add 7 to both sides

To isolate the term with \( c \), we add 7 to both sides of the equation.
\( \frac{c}{6}-7 + 7=1 + 7 \)
\( \frac{c}{6}=8 \)

Step 2: Multiply both sides by 6

To solve for \( c \), we multiply both sides of the equation by 6.
\( 6\times\frac{c}{6}=8\times6 \)
\( c = 48 \)

Step 3: Check the solution

Substitute \( c = 48 \) back into the original equation:
Left - hand side: \( \frac{48}{6}-7 = 8-7 = 1 \)
Right - hand side: \( 1 \)
Since the left - hand side equals the right - hand side, \( c = 48 \) is a valid solution.

Step 1: Subtract 8 from both sides

To isolate the term with \( k \), we subtract 8 from both sides of the equation.
\( 8k + 8-8=-16 - 8 \)
\( 8k=-24 \)

Step 2: Divide both sides by 8

To solve for \( k \), we divide both sides of the equation by 8.
\( \frac{8k}{8}=\frac{-24}{8} \)
\( k=-3 \)

Step 3: Check the solution

Substitute \( k = - 3 \) back into the original equation:
Left - hand side: \( 8(-3)+8=-24 + 8=-16 \)
Right - hand side: \( -16 \)
Since the left - hand side equals the right - hand side, \( k=-3 \) is a valid solution.

Answer:

\( k = 6 \)

Problem 6: Solve \( \frac{c}{6}-7 = 1 \) and check the solution