Lesson 3: Algebraic Solution of Systems

New Concept

This lesson introduces two algebraic methods—substitution and elimination—for solving systems of linear equations. You will learn to find the precise intersection point $(x, y)$ without graphing, and identify inconsistent or dependent systems.

What’s next

Next, you’ll master these methods through interactive examples and practice cards. Then, test your skills on challenge problems involving real-world applications.

Property

To Solve a System by Substitution.

1. Choose one of the variables in one of the equations. (It is best to choose a variable whose coefficient is 1 or -1.) Solve the equation for that variable.

2. Substitute the result of Step 1 into the other equation. This gives an equation in one variable.

Property

The elimination method, also known as the method of linear combinations, is used to solve a system by eliminating one of the variables. First, put both equations into the general linear form $Ax + By = C$. Then, add the two equations. If the coefficients of one variable are not opposites, multiply one or both equations by suitable constants to make them opposites before adding.

Examples

• To solve the system $2x + 3y = 11$ and $4x - 3y = 7$, simply add the two equations. The $y$-terms are eliminated, leaving $6x = 18$, so $x=3$. Substitute $x=3$ into the first equation: $2(3) + 3y = 11$, which gives $6 + 3y = 11$, so $3y=5$ and $y = \frac{5}{3}$. The solution is $(3, \frac{5}{3})$.

• For the system $x + 4y = 10$ and $3x + 2y = 10$, multiply the first equation by $-3$ to get $-3x - 12y = -30$. Add this to the second equation: $(3x + 2y) + (-3x - 12y) = 10 - 30$. This simplifies to $-10y = -20$, so $y=2$. Then $x + 4(2) = 10$, so $x=2$. The solution is $(2, 2)$.

Property

When using elimination to solve a system:

1. If combining the two equations results in an equation of the form

then the system is inconsistent.

2. If combining the two equations results in an equation of the form

then the system is dependent.