We first multiply the second equation by 4.

So our resulting equation is:

Then we subtract the first equation from the second new equation.

Left Hand Side:

Right Hand Side:

Resulting Equation:

We divide both sides by -15

Left Hand Side:

Right Hand Side:

Our result is:

When you have a set of two equations, you need to solve one for one of the variables. Then, you substitute that value into the other. For this set, we can even do something a little easier. Solve the second for :

 becomes:

Now,substitute into the first equation for :

 becomes:

Now, just simplify:

Now, use  to solve for . Substitute in  for :

Finding the common denominator, you get:

Divide both sides by :

When you have a set of two equations, you need to solve one for one of the variables. Then, you substitute that value into the other. Either equation will work rather well, so use the first. Thus, from  you get:

Now, substitute into the second equation:

Simplifying, you get:

Begin by writing out your data in equations. Based on the description, we know:

and ...

Now, to solve two equations like this, solve one equation for a variable. Then substitute into the other. The easiest one to solve is . From this, we know:

Now, substitute this into the other equation:

Simplify:

Luckily, this is just what we need!

There are many methods to finding the solution of a system of linear equations: graphing, elimination, substitution, matrix manipulations; we'll use subsitution.

Since both equations are equal to , we can use substitution to set them equal to each other and solve for x:

Now we plug the value we just found for  into either equation (it doesn't matter, check both if you're unsure)

Thus we get the ordered pair: 

The easiest way to solve this equation is to work backwards. Start with the  for the third year and divide by  to get the second year's earning which were , then subtract  to get the first year which was .

You factor the equation and get:

(x + 2)(x + 1) = 0

Then, you get x = –2 and x = –1.

There are two ways to solve this problem. One option is to use the quadratic formula:

In this problem, a = 1, b = 1, and c = –6. However, there is a lot of room for computational error using this method.

Another option is to factor the equation using the reverse FOIL method. Because we have x², we know that both factors must include x. Thus, so far we have (x   )(x   ).  We also know that one factor must be negative and one must be positive if our last value (the L in FOIL) is to be negative. Thus, we have (x–  ) (x+ ). The factors of 6 are 1, 2, 3, and 6, so our last values must multiply together to make 6. Therefore, our options are (x – 1)(x + 6), (x – 6)(x + 1), (x – 2)(x + 3), and (x – 3)(x + 2). However, because we need our middle value to be 1, the combination of the outer and inner values (O and I in FOIL), must add up to positive 1. Since only a combination of +/–2 and +/–3 can add to 1, we can eliminate our first two options. Then, because we need the positive value to be larger than the negative value, we can determine that the correct factorization is (x – 2)(x + 3). When we set each of these pieces equal to zero, we get x – 2 = 0 and x + 3 = 0. Thus, X equals 2 and –3. We can check our work using FOIL.

1. Factor the above equation:

2. Solve for :

and...

To factor this, we need to find a pair of numbers that multiplies to 6 and sums to 5. The numbers 2 and 3 work. (2 * 3 = 6 and 2 + 3 = 5)

(x + 2)(x + 3) = 0

x = –2 or x = –3