To solve the problem [tex]\(36 + 3(4x - 9) = c(2x + 1) + 25\)[/tex] and determine the value of [tex]\(c\)[/tex] that makes the equation have no solution, we need to follow these steps:
1. Expand and simplify both sides of the equation.
[tex]\[
36 + 3(4x - 9) = c(2x + 1) + 25
\][/tex]
2. Distribute the constants inside the parentheses.
[tex]\[
36 + 3 \cdot 4x - 3 \cdot 9 = c \cdot 2x + c \cdot 1 + 25
\][/tex]
[tex]\[
36 + 12x - 27 = 2cx + c + 25
\][/tex]
3. Combine like terms on the left side.
[tex]\[
36 - 27 + 12x = 2cx + c + 25
\][/tex]
[tex]\[
9 + 12x = 2cx + c + 25
\][/tex]
4. Rearrange terms to isolate the [tex]\(x\)[/tex] terms and constant terms on each side.
[tex]\[
12x - 2cx = c + 25 - 9
\][/tex]
[tex]\[
12x - 2cx = c + 16
\][/tex]
5. Factor out [tex]\(x\)[/tex] on the left-hand side.
[tex]\[
x(12 - 2c) = c + 16
\][/tex]
For the equation to have no solution, the coefficient of [tex]\(x\)[/tex] must be zero, and the constant terms must yield a contradiction (i.e., a non-zero value). This means:
[tex]\[
12 - 2c = 0
\][/tex]
6. Solve for [tex]\(c\)[/tex].
[tex]\[
12 - 2c = 0
\][/tex]
[tex]\[
2c = 12
\][/tex]
[tex]\[
c = 6
\][/tex]
To confirm that this value of [tex]\(c\)[/tex] leads to a contradiction in the constant terms:
Substitute [tex]\(c = 6\)[/tex] back into the equation:
[tex]\[
x(12 - 2 \cdot 6) = 6 + 16
\][/tex]
[tex]\[
x(12 - 12) = 22
\][/tex]
[tex]\[
0 = 22
\][/tex]
Since [tex]\(0 = 22\)[/tex] is a contradiction, the equation has no solution when [tex]\(c = 6\)[/tex].
Thus, the value of [tex]\(c\)[/tex] is [tex]\( \boxed{6} \)[/tex].
