In order to solve for \(\displaystyle x\), we need to isolate the \(\displaystyle x\) to one side of the equation. 

For this problem, the first thing we want to do is distribute the \(\displaystyle 4\):

\(\displaystyle 32=4(x+40)\)

\(\displaystyle 32=4x+160\)

Next, we can subtract \(\displaystyle 160\) from both sides:

\(\displaystyle \frac{\begin{array}[b]{r}32=4x+160\\ -160\ \ \ \ \ \ \ \ \ \ \-160 \end{array}}{\\\\-128=4x}\)

Finally, we divide \(\displaystyle 4\) from both sides:

\(\displaystyle \frac{-128}{4}=\frac{4x}{4}\)

\(\displaystyle x=-32\)

In order to solve this problem, we need to recall our exponent rules:

When our base numbers are equal to each other, like in this problem, we can add our exponents together using the following formula:

\(\displaystyle a^m\times a^n=a^{(m+n)}\)

Let's apply this rule to our problem

\(\displaystyle 3^{-6}\times 3^{2}=3^{(-6+2)}\)

Solve for the exponents

\(\displaystyle -6+2=-4\)

\(\displaystyle 3^{-4}\)

We cannot leave this problem in this format because we cannot have a negative exponent. Instead, we can move the base and the exponent to the denominator of a fraction:

\(\displaystyle a^{(-m)}=\frac{1}{a^m}\)

Solve the problem

\(\displaystyle \frac{1}{3^{4}}=\frac{1}{81}\)

There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination. 

Substitution can be used by solving one of the equations for either \(\displaystyle x\) or \(\displaystyle y\), and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the \(\displaystyle y=\) form, and then set both equations equal to each other. 

Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable. 

For this problem, substitution makes the most sense because the first equation is already solved for a variable. We can substitute the expression that is equal to \(\displaystyle x\), into the \(\displaystyle x\) of our second equation:

\(\displaystyle 6y+3(1+3y)=-27\)

Next, we need to distribute and combine like terms:

\(\displaystyle 6y+3+9y=-27\)

\(\displaystyle 15y+3=-27\)

We are solving for the value of \(\displaystyle y\), which means we need to isolate the \(\displaystyle y\) to one side of the equation. We can subtract \(\displaystyle 3\) from both sides:

\(\displaystyle \frac{\begin{array}[b]{r}15y+3=-27\\ \ -3\ \ \ \ -3\end{array}}{\\\\15y=-30}\)

Then divide both sides by \(\displaystyle 15\) to solve for \(\displaystyle y\text:\)

\(\displaystyle \frac{15y}{15}=\frac{-30}{15}\)

\(\displaystyle y=-2\)

Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both \(\displaystyle x\) and \(\displaystyle y\) values. 

Now that we have the value of \(\displaystyle y\), we can plug that value into the \(\displaystyle y\) variable in one of our given equations and solve for \(\displaystyle x\textup:\)

\(\displaystyle x=1+3(-2)\)

\(\displaystyle x=1+(-6)\)

\(\displaystyle x=-5\)

Our point of intersection, and the solution to the two system of linear equations is \(\displaystyle (-5,-2)\)