Introduction to Systems of Equations
A system of equations is a collection of two or more equations whose graphs might or might not intersect (share a common point or points). If the graphs do intersect, then we say that the solution to the system is the point or points where the graphs intersect. For example, the solution to the system
is (1, 3) because the point (1, 3) is on both graphs. See Figure 10.1.
Fig. 10.1
We say that (1,3) satisfies the system because if we let x = 1 and y = 3 in each equation, they will both be true.
1 + 3 = 4 
This is a true statement. 
3(1) − 3 = 0 
This is a true statement. 
Solving Systems of Equations Using the Substitution and Elimination Methods
There are several methods for solving systems of equations. One of them is by sketching the graphs and seeing where, if anywhere, the graphs intersect. Even with a graphing calculator, though, these solutions might only be approximations. When the equations are lines, matrices can be used. Graphing calculators are useful for these, too. We will use two algebraic methods in this study guide and two matrix methods in the next. One of the algebraic methods is substitution and the other is elimination by addition. Both methods will work with many kinds of systems of equations, but we will start out with systems of linear equations.
Substitution Method
Substitution works by solving for one variable in one equation and making a substitution in the other equation. Usually, it does not matter which variable we use or which equation we begin with, but some choices are easier than others.
Examples
Solve the systems of equations. Put your solutions in the form of a point, (x, y) .
We have four places to start.
 Solve for x in the first equation: x = 5 − y
 Solve for y in the first equation: y = 5 − x
 Solve for x in the second equation:
 Solve for y in the second equation: y = 2 x − 1
The third option looks like it would be the most trouble, so we will use one of the others. We will use the first option. Because x = 5 − y came from the first equation, we will substitute 5 − y for x in the second equation. Then −2 x + y = −1 becomes −2(5 − y ) + y = −1. Now we have one equation with one variable.
−2(5 − y ) + y = −1
−10 + 2 y + y = −1
3 y = 9
y = 3
Now that we know y = 3, we could use any of the equations above to find x . We know that x = 5 − y , so we will use this.
x = 5 − 3 = 2
The solution is x = 2 and y = 3 or the point (2, 3). It is a good idea to check the solution.
2 + 3 = 5 This is true.
−2(2) + 3 = −1 This is true.

We will solve for y in equation B: y = 2 − 3 x . Next we will substitute 2 − 3 x for y in equation A and solve for x .
4 x − y = 12
4 x − (2 − 3 x ) = 12
4 x − 2 + 3 x = 12
7 x = 14
x = 2
Now that we know x = 2, we will put x = 2 in one of the above equations. We will use y = 2 − 3 x : y = 2 − 3(2) = −4. The solution is x = 2, y = −4, or (2, −4). The graphs in Figure 10.2 verify that the solution (2, −4) is on both lines.

Both equations are already solved for y , so all we need to do is to set them equal to each other.
A = B
4 x + 1 = 3 x + 2
x = 1
Use either equation A or equation B to find y when x = 1. We will use A: y = 4 x + 1 = 4(1) + 1 = 5. The solution is x = 1 and y = 5, or (1, 5). We can see from the graphs in Figure 10.3 that (1, 5) is the solution to the system.

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 PreCalculus: Systems of Equations and Inequalities
 PreCalculus: The Slope and Equation of a Line
 PreCalculus: Introduction to Functions
 PreCalculus: Functions and their Graphs
 PreCalculus: Combinations of Functions and Inverse Functions
 PreCalculus: Quadratic Functions
 PreCalculus: Polynomial Functions
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