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:

\(\displaystyle 32+28=60\)

\(\displaystyle 60=60\)

The equation \(\displaystyle x+7=34\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 7\) is equal to \(\displaystyle 34\).

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 \(\displaystyle 7\) was added to the variable, we can subtract \(\displaystyle 7\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}x+7=34\\ -7\ \ -7\end{array}}{\\\\x+0=27}\)

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:

\(\displaystyle 27+7=34\)

\(\displaystyle 34=34\)

The equation \(\displaystyle x+17=82\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 17\) is equal to \(\displaystyle 82\).

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 \(\displaystyle 17\) was added to the variable, we can subtract \(\displaystyle 17\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}x+17=82\\ -17-17\end{array}}{\\\\x+0=65}\)

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

\(\displaystyle x=65\)

We can check to make sure we have the correct value for \(\displaystyle x\) by plugging \(\displaystyle 65\) into the original equation; thus:

\(\displaystyle 65+17=82\)

\(\displaystyle 82=82\)

The equation \(\displaystyle x+55=173\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 55\) is equal to \(\displaystyle 173\).

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. 

\(\displaystyle \frac{\begin{array}[b]{r}x+55=173\\ -55\ \ -55\end{array}}{\\\\x+0=118}\)

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

\(\displaystyle x=118\)

We can check to make sure we have the correct value for \(\displaystyle x\) by plugging \(\displaystyle 118\) into the original equation; thus:

\(\displaystyle 118+55=173\)

\(\displaystyle 173=173\)

The equation \(\displaystyle x+12=37\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 12\) 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 \(\displaystyle 12\) was added to the variable, we can subtract \(\displaystyle 12\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}x+12=37\\ -12-12\end{array}}{\\\\x+0=25}\)

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:

\(\displaystyle 25+12=37\)

\(\displaystyle 37=37\)

The equation \(\displaystyle x+22=58\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 22\) is equal to \(\displaystyle 58\).

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. 

\(\displaystyle \frac{\begin{array}[b]{r}x+22=58\\ -22-22\end{array}}{\\\\x+0=36}\)

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

\(\displaystyle x=36\)

We can check to make sure we have the correct value for \(\displaystyle x\) by plugging \(\displaystyle 36\) into the original equation; thus:

\(\displaystyle 36+22=58\)

\(\displaystyle 58=58\)

The equation \(\displaystyle x+10=38\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 10\) is equal to \(\displaystyle 38\).

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. 

\(\displaystyle \frac{\begin{array}[b]{r}x+10=38\\ -10-10\end{array}}{\\\\x+0=28}\)

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

\(\displaystyle x=28\)

We can check to make sure we have the correct value for \(\displaystyle x\) by plugging \(\displaystyle 28\) into the original equation; thus:

\(\displaystyle 28+10=38\)

\(\displaystyle 38=38\)

The equation \(\displaystyle x+13=43\) asks us to identify which number, \(\displaystyle x\), plus \(\displaystyle 13\) is equal to \(\displaystyle 43\).

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. 

\(\displaystyle \frac{\begin{array}[b]{r}x+13=43\\ -13-13\end{array}}{\\\\x+0=30}\)

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

\(\displaystyle x=30\)

We can check to make sure we have the correct value for \(\displaystyle x\) by plugging \(\displaystyle 30\) into the original equation; thus:

\(\displaystyle 30+13=43\)

\(\displaystyle 43=43\)