# Algebra: Understanding Systems of Linear Equations

A system of equations is simply working with more than one equation (usually two linear equations) with different variables at the same time. This is sometimes referred to as simultaneous equations. The concept is quite simple, once we can appreciate the two ways in which we can solve the question. Once you understand the concept, it doesn’t matter how the question is structured, you will be able to work it. Again, the emphasis is not on just being able to work the question in a particular way, but on *conceptual understanding*. For the purpose of the SAT exam, you will also be required to reason through worded problems in order to generate a system of equations and then apply the concept of either method to solve for the unknown variable.

There are two methods that we use to solve a system of linear equations and or simultaneous equations: The method of elimination and the substitution method. It can also be solved graphically by graphing the two equations and examine where both graphs meet. However, we will just focus on the method of elimination and substitution. After exploring both methods, then you will decide which one best works for you. Please note that all methods should produce the *same* answers.

Example 1

1000 tickets were sold. Adult tickets cost $8.50, children’s cost $4.50, and a total of $7300 was collected. How many tickets of each kind were sold?

In answering this question, we need to find a system of linear equations that represents the following information so that we can solve for the ‘unknown’. Now we need to define the unknowns or what we want to know from the question. What we want to know is how many adult tickets were sold and how many children’s ticket were sold. Therefore we can say let *x* = number of adult tickets sold and *y* = number children’s ticket sold. Next, we must consider the information that is given in the question and how it can help us. We know that 1000 tickets were sold and $7300 was collected. Therefore, to get $7300, must add the number of adult tickets sold with the number of children’s ticket sold. To get the number of each we must multiply the amount of ticket sold (unknown) by the cost price. Therefore:

$8.50*x* + $4.50*y* = $7300.

Also the number of adult tickets plus the number of children’s ticket should be equal to 1000, therefore:

*x *+ *y* = 1000

So the system of equations is as follows:

*x *+ *y* = 1000 ……… eq. 1

8.50*x* + 4.50*y* = 7300……. Eq. 2

Using this one example we will explore both the method of elimination and substitution. Let us start with the method of elimination:

The method of elimination is when you eliminate one variable and solve for the other, then substituting the answer for the variable found to get the other variable. In order to eliminate we must ensure that the coefficients of a particular variable is the same. It doesn’t matter which variable you eliminate first. Let us eliminate the *y* from the system of equations above. However, note that the coefficients of the *y* variable aren’t the same for both equations. Therefore, what we do is to multiple equation 1 by the coefficient of the *y* of equation 2 and vice versa, therefore:

*x *+ *y* = 1000 × 4.50

8.50*x* + 4.50*y* = 7300 × 1

We do this because we do not want to change the value of the equations altogether. Therefore the new system of equation is:

4.50*x* + 4.50*y* = 4500…… eq. 3

8.50*x* + 4.50*y* = 7300……. Eq. 4

Since we have the coefficient of the *y* variable to be the same in both equations, we can eliminate it. Please note the following: you have to look at the sign before each coefficient. If both have a positive sign (+) you minus one equation from the other. If both have a negative sign (-), you add one equation to the other. In this question both have a positive sign (+), so we minus.

8.50*x *– 4.50*x* + 4.50*y* – 4.50*y* = 7300 – 4500

4*x* = 2800

*x* = 2800/4

*x* = 700

Once we find the *x* value, we can then substitute it in any one of the equations to solve for the *y* value. Let us use eq. 1, therefore:

700 + *y* = 1000

*y* = 1000 – 700

*y =* 300.

Therefore the number of adult tickets sold was 700 and the number of children’s ticket sold was 300.

Let us try the method of substitution: this method is when you make one variable the subject of the formula of any one of the equations and then substitute it in the other. Let us use eq. 1 and make *y *the subject of the formula.

*x *+ *y* = 1000

*y *= 1000 – *x *

By substituting *y *= 1000 – *x *in eq. 2 we get

8.50*x* + 4.50(1000 – *x*) = 7300

8.50*x* + 4500 – 4.50*x* = 7300

8.50x – 4.50*x* + 4500 = 7300

4*x* + 4500 = 7300

4*x* = 7300 – 4500

4*x* = 2800

*x* = 2800/4

*x* = 700

Since we find the value of *x* we substitute it in eq. 1 to solve for *y*. Therefore:

700 + *y* = 1000

*y* = 1000 – 700

*y =* 300.

Therefore, we see that both methods give us the same answer. Understanding a concept makes any question easy. We will stop here for now.