The equation \(\displaystyle x+55=172\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 55\) is equal to \(\displaystyle 172\).

In order to solve this equation, we need to isolate the \(\displaystyle x\) variable on one side of the equals sign. We will do this by performing the opposite of the operations that were done on the variable,\(\displaystyle x\). We need to remember that whatever we do to one side of the equals sign, we have to do to the other. 

Since \(\displaystyle 55\) was added to the variable, we can subtract \(\displaystyle 55\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}x+55=172\\ -55\ \ -55\end{array}}{\\\\x+0=117}\)

Simplify by dropping the \(\displaystyle 0\) because it has not numerical value; therefore:

\(\displaystyle x=117\)

We can check to make sure we have the correct value for \(\displaystyle x\) by plugging \(\displaystyle 117\) into the original equation; thus:

\(\displaystyle 117+55=172\)

\(\displaystyle 172=172\)

The equation \(\displaystyle x+12=36\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 12\) is equal to \(\displaystyle 36\).

In order to solve this equation, we need to isolate the \(\displaystyle x\) variable on one side of the equals sign. We will do this by performing the opposite of the operations that were done on the variable,\(\displaystyle x\). We need to remember that whatever we do to one side of the equals sign, we have to do to the other. 

Since \(\displaystyle 12\) was added to the variable, we can subtract \(\displaystyle 12\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}x+12=36\\ -12-12\end{array}}{\\\\x+0=24}\)

Simplify by dropping the \(\displaystyle 0\) because it has not numerical value; therefore:

\(\displaystyle x=24\)

We can check to make sure we have the correct value for \(\displaystyle x\) by plugging \(\displaystyle 24\) into the original equation; thus:

\(\displaystyle 24+12=36\)

\(\displaystyle 36=36\)

The equation \(\displaystyle x+22=57\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 22\) is equal to \(\displaystyle 57\).

In order to solve this equation, we need to isolate the \(\displaystyle x\) variable on one side of the equals sign. We will do this by performing the opposite of the operations that were done on the variable,\(\displaystyle x\). We need to remember that whatever we do to one side of the equals sign, we have to do to the other. 

Since \(\displaystyle 22\) was added to the variable, we can subtract \(\displaystyle 22\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}x+22=57\\ -22-22\end{array}}{\\\\x+0=35}\)

Simplify by dropping the \(\displaystyle 0\) because it has not numerical value; therefore:

\(\displaystyle x=35\)

We can check to make sure we have the correct value for \(\displaystyle x\) by plugging \(\displaystyle 35\) into the original equation; thus:

\(\displaystyle 35+22=57\)

\(\displaystyle 57=57\)

The equation \(\displaystyle x+10=37\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 10\) is equal to \(\displaystyle 37\).

In order to solve this equation, we need to isolate the \(\displaystyle x\) variable on one side of the equals sign. We will do this by performing the opposite of the operations that were done on the variable,\(\displaystyle x\). We need to remember that whatever we do to one side of the equals sign, we have to do to the other. 

Since \(\displaystyle 10\) was added to the variable, we can subtract \(\displaystyle 10\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}x+10=37\\ -10-10\end{array}}{\\\\x+0=27}\)

Simplify by dropping the \(\displaystyle 0\) because it has not numerical value; therefore:

\(\displaystyle x=27\)

We can check to make sure we have the correct value for \(\displaystyle x\) by plugging \(\displaystyle 27\) into the original equation; thus:

\(\displaystyle 27+10=37\)

\(\displaystyle 37=37\)

The equation \(\displaystyle x+43=75\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 43\) is equal to \(\displaystyle 75\).

In order to solve this equation, we need to isolate the \(\displaystyle x\) variable on one side of the equals sign. We will do this by performing the opposite of the operations that were done on the variable,\(\displaystyle x\). We need to remember that whatever we do to one side of the equals sign, we have to do to the other. 

Since \(\displaystyle 43\) was added to the variable, we can subtract \(\displaystyle 43\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}x+43=75\\ -43-43\end{array}}{\\\\x+0=32}\)

Simplify by dropping the \(\displaystyle 0\) because it has not numerical value; therefore:

\(\displaystyle x=32\)

We can check to make sure we have the correct value for \(\displaystyle x\) by plugging \(\displaystyle 32\) into the original equation; thus:

\(\displaystyle 32+43=75\)

\(\displaystyle 75=75\)

The equation \(\displaystyle x+13=42\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 13\) is equal to \(\displaystyle 42\).

In order to solve this equation, we need to isolate the \(\displaystyle x\) variable on one side of the equals sign. We will do this by performing the opposite of the operations that were done on the variable, \(\displaystyle x\). We need to remember that whatever we do to one side of the equals sign, we have to do to the other. 

Since \(\displaystyle 13\) was added to the variable, we can subtract \(\displaystyle 13\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}x+13=42\\ -13-13\end{array}}{\\\\x+0=29}\)

Simplify by dropping the \(\displaystyle 0\) because it has not numerical value; therefore:

\(\displaystyle x=29\)

We can check to make sure we have the correct value for \(\displaystyle x\) by plugging \(\displaystyle 29\) into the original equation; thus:

\(\displaystyle 29+13=42\)

\(\displaystyle 42=42\)

The equation \(\displaystyle x+26=51\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 26\) is equal to \(\displaystyle 51\).

In order to solve this equation, we need to isolate the \(\displaystyle x\) variable on one side of the equals sign. We will do this by performing the opposite of the operations that were done on the variable,\(\displaystyle x\). We need to remember that whatever we do to one side of the equals sign, we have to do to the other. 

Since \(\displaystyle 26\) was added to the variable, we can subtract \(\displaystyle 26\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}x+26=51\\ -26-26\end{array}}{\\\\x+0=25}\)

Simplify by dropping the \(\displaystyle 0\) because it has not numerical value; therefore:

\(\displaystyle x=25\)

We can check to make sure we have the correct value for \(\displaystyle x\) by plugging \(\displaystyle 25\) into the original equation; thus:

\(\displaystyle 25+26=51\)

\(\displaystyle 51=51\)

The equation \(\displaystyle x+19=24\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 19\) is equal to \(\displaystyle 24\).

In order to solve this equation, we need to isolate the \(\displaystyle x\) variable on one side of the equals sign. We will do this by performing the opposite of the operations that were done on the variable,\(\displaystyle x\). We need to remember that whatever we do to one side of the equals sign, we have to do to the other. 

Since \(\displaystyle 19\) was added to the variable, we can subtract \(\displaystyle 19\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}x+19=24\\ -19-19\end{array}}{\\\\x+0=5}\)

Simplify by dropping the \(\displaystyle 0\) because it has not numerical value; therefore:

\(\displaystyle x=5\)

We can check to make sure we have the correct value for \(\displaystyle x\) by plugging \(\displaystyle 5\) into the original equation; thus:

\(\displaystyle 5+19=24\)

\(\displaystyle 24=24\)

The equation \(\displaystyle x+5=30\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 5\) is equal to \(\displaystyle 30\).

In order to solve this equation, we need to isolate the \(\displaystyle x\) variable on one side of the equals sign. We will do this by performing the opposite of the operations that were done on the variable,\(\displaystyle x\). We need to remember that whatever we do to one side of the equals sign, we have to do to the other. 

Since \(\displaystyle 5\) was added to the variable, we can subtract \(\displaystyle 5\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}x+5=30\\ -5\ \ -5\end{array}}{\\\\x+0=25}\)

Simplify by dropping the \(\displaystyle 0\) because it has not numerical value; therefore:

\(\displaystyle x=25\)

We can check to make sure we have the correct value for \(\displaystyle x\) by plugging \(\displaystyle 25\) into the original equation; thus:

\(\displaystyle 25+5=30\)

\(\displaystyle 30=30\)

The equation \(\displaystyle x+28=60\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 28\) is equal to \(\displaystyle 60\).

In order to solve this equation, we need to isolate the \(\displaystyle x\) variable on one side of the equals sign. We will do this by performing the opposite of the operations that were done on the variable,\(\displaystyle x\). We need to remember that whatever we do to one side of the equals sign, we have to do to the other. 

Since \(\displaystyle 28\) was added to the variable, we can subtract \(\displaystyle 28\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}x+28=60\\ -28-28\end{array}}{\\\\x+0=32}\)

Simplify by dropping the \(\displaystyle 0\) because it has not numerical value; therefore:

\(\displaystyle x=32\)

We can check to make sure we have the correct value for \(\displaystyle x\) by plugging \(\displaystyle 32\) into the original equation; thus:

\(\displaystyle 32+28=60\)

\(\displaystyle 60=60\)