In order to solve this problem, we need to recall the formula for perimeter of a rectangle:

\(\displaystyle P=2l+2w\)

We can substitute in our known values and solve for our unknown variable (i.e. length):

\(\displaystyle 16=2l+2(3)\)

\(\displaystyle 16=2l+6\)

We want to isolate the \(\displaystyle l\) to one side of the equation. In order to do this, we will first subtract \(\displaystyle 6\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}16=2l+6\\ -6\ \ \ \ \ \ -6\end{array}}{\\\\10=2l}\)

Next, we can divide each side by \(\displaystyle 2\)

\(\displaystyle \frac{\begin{array}[b]{r}\frac{10}{2}=\frac{2l}{2}\\\end{array}}{5=l}\)

The length of the rectangle is \(\displaystyle 5\textup{ inches}\)

In order to solve this problem, we need to recall the formula for perimeter of a rectangle:

\(\displaystyle P=2l+2w\)

We can substitute in our known values and solve for our unknown variable (i.e. length):

\(\displaystyle 60=2l+2(20)\)

\(\displaystyle 60=2l+40\)

We want to isolate the \(\displaystyle l\) to one side of the equation. In order to do this, we will first subtract \(\displaystyle 40\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}60=2l+40\\ -40\ \ \ \ \ \ -40\end{array}}{\\\\20=2l}\)

Next, we can divide each side by \(\displaystyle 2\)

\(\displaystyle \frac{\begin{array}[b]{r}\frac{20}{2}=\frac{2l}{2}\\\end{array}}{10=l}\)

The length of the rectangle is \(\displaystyle 10\textup{ inches}\)

In order to solve this problem, we need to recall the formula for perimeter of a rectangle:

\(\displaystyle P=2l+2w\)

We can substitute in our known values and solve for our unknown variable (i.e. length):

\(\displaystyle 42=2l+2(13)\)

\(\displaystyle 42=2l+26\)

We want to isolate the \(\displaystyle l\) to one side of the equation. In order to do this, we will first subtract \(\displaystyle 26\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}42=2l+26\\ -26\ \ \ \ \ \ -26\end{array}}{\\\\16=2l}\)

Next, we can divide each side by \(\displaystyle 2\)

\(\displaystyle \frac{\begin{array}[b]{r}\frac{16}{2}=\frac{2l}{2}\\\end{array}}{8=l}\)

The length of the rectangle is \(\displaystyle 8\textup{ inches}\)

In order to solve this problem, we need to recall the formula for perimeter of a rectangle:

\(\displaystyle P=2l+2w\)

We can substitute in our known values and solve for our unknown variable (i.e. length):

\(\displaystyle 56=2l+2(16)\)

\(\displaystyle 56=2l+32\)

We want to isolate the \(\displaystyle l\) to one side of the equation. In order to do this, we will first subtract \(\displaystyle 32\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}56=2l+32\\ -32\ \ \ \ \ \ -32\end{array}}{\\\\24=2l}\)

Next, we can divide each side by \(\displaystyle 2\)

\(\displaystyle \frac{\begin{array}[b]{r}\frac{24}{2}=\frac{2l}{2}\\\end{array}}{12=l}\)

The length of the rectangle is \(\displaystyle 12\textup{ inches}\)

In order to solve this problem, we need to recall the formula for perimeter of a rectangle:

\(\displaystyle P=2l+2w\)

We can substitute in our known values and solve for our unknown variable (i.e. length):

\(\displaystyle 68=2l+2(14)\)

\(\displaystyle 68=2l+28\)

We want to isolate the \(\displaystyle l\) to one side of the equation. In order to do this, we will first subtract \(\displaystyle 28\) from both sides of the equation. 

\(\displaystyle \frac{\begin{array}[b]{r}68=2l+28\\ -28\ \ \ \ \ \ -28\end{array}}{\\\\40=2l}\)

Next, we can divide each side by \(\displaystyle 2\)

\(\displaystyle \frac{\begin{array}[b]{r}\frac{40}{2}=\frac{2l}{2}\\\end{array}}{20=l}\)

The length of the rectangle is \(\displaystyle 20\textup{ inches}\)

"The product of seven and a number " is \(\displaystyle 7x\). "Twenty subtracted from the product of seven and a number" is \(\displaystyle 7x - 20\) . "Exceeds one hundred" means that this is greater than one hundred, so the correct inequality is

\(\displaystyle 7x - 20 > 100\)

"The sum of a number and sixteen" is translates to \(\displaystyle x + 16\); twice that sum is \(\displaystyle 2(x+16)\). " Is no less than sixty" means that this is greater than or equal to sixty, so the desired inequality is

 \(\displaystyle 2(x+16) \geq 60\).

"The sum of a number and sixteen" translates to \(\displaystyle x + 16\); twice that sum is \(\displaystyle 2 (x+16)\). "Does not exceed eighty" means that it is less than or equal to eighty, so the desired inequality is

\(\displaystyle 2 (x+16) \leq 80\)

The way the sentence is phrased suggests that the person can spend up to \(\displaystyle 20\) dollars but not a penny more. This suggests that \(\displaystyle s\), the amount spend can be \(\displaystyle 20\) but not exceed it. 

So your answer is: \(\displaystyle s\leq20\)

Seven less than two times a number is greater than fourteen.

Let's look at the problem step by step.

If we do not know the value of a number, we give it a variable name.  Let's say x.  So, we see in the problem

Seven less than two times a number is greater than fourteen.

So, we will replace a number with x.

Seven less than two times x is greater than fourteen.

Now, we see that is says "two times" x, so we will write it like

Seven less than 2x is greater than fourteen.

The problem says "seven less" than 2x.  This simply means we are taking 2x and subtracting seven.  So we get

2x - 7 is greater than fourteen

We know the symbol for "is greater than".  We can write

2x - 7 > fourteen

Finally, we write out the number fourteen.  

2x - 7 > 14