Question

Find the missing number so that the equation has no solutions.
2x+13=
x–19

Answer 1

To determine a value for the missing number in the equation \( 2x + 13 = x - 19 \) such that it has no solutions, we first look to set up the equation with the missing number represented as \( c \):

\[
2x + 13 = x + c
\]

To make the equation have no solutions, we need the coefficients of \( x \) on both sides to be the same, yet the constants to be different.

1. Start by rearranging the equation:
\[
2x - x = c - 13
\]
This simplifies to:
\[
x = c - 13
\]

2. For the equation to have no solutions, we need to choose a value for \( c \) such that the equation leads to a contradiction. This means we must set the constant terms so that \( c - 13 \) cannot yield a valid value for \( x \).

If we set \( c = 13 \), we get:
\[
x = 13 - 13 = 0
\]
This does yield a solution (\( x = 0 \)), so it won't work.

Instead, we can try:
- Let’s choose a value for \( c \) such that \( c \) is not equal to \( 13 \). If we set \( c = 12 \), we have:
\[
x = 12 - 13 = -1
\]
This gives a solution.

Now, let's choose a value for \( c \) more distant from \( 13 \).

Assume that \( c = 14 \):
\[
2x + 13 = x + 14
\]
Rearranging gives:
\[
2x - x = 14 - 13
\]
Simplify to:
\[
x = 1
\]
Again, this has a solution.

Let’s try the contrary idea by setting constants, which would lead us to \( 13 \):
If \( c \) is anything but \( 13 \), you'll eventually have it resolving to:
Assuming \( c = -20 \):
\[
2x + 13 = x - 20
\]
gives,
\[
2x - x = -20 - 13
\]
\[
x = -33
\]

At this point, rather we observe if \( x \) gets independent of any number c leading to just a case where
choose \( c = 19 \)
\[
2x + 13 = x - 19
\]
gives,
\[
x = -32
\]
so in contradiction,

The value \( c \) takes places if its distance lies where you find \( 19\).

Thus, the missing number \( c \) for the equation \( 2x + 13 = x + c \) so that it has no solutions can effectively be any value except for \( 13\).

With that reasoning, the general conclusion leads us to these observations can guide us to:
In simplest terms if we set \( c \neq –6\) then \( x \neq c + 13 =-19\) leads no solution paths.

Thus \( c = -32 \) to discover “no values”. Any value consistent with \( c + 6=19\) arrive at constants yielding beyond \( x\) would land it to show no valid options leading insights through constants on similar equations.

Final answer can show values such as \(-6\).