Download Section 4.1

Section 4.1 (Solving Systems of Linear Equations in Two Variables)
Hooper’s Store on Sesame Street recently decided to sell VA Tech Hokies football jerseys. The following
graph shows the estimated cost and revenue predictions.
Notice that as the number of jerseys
sold increases, the revenue grows
higher than the cost of the jerseys.
At 20 jerseys, the cost and revenue are
the same ($800). This point (20,800)
is the break-even point and signifies
where the jersey sales become profitable.
Hokie Jersey Sales
2000
Dollars
1600
Both of these lines represent a linear
equation in two variables, and both
equations form a system of linear
equations. The point where the two
are equal is called the solution of the
system ((20,800) in this case).
1200
800
400
0
0
10
20
30
40
50
Number of Jerseys
Cost
Revenue
A solution of a system of 2 equations in 2 variables is an ordered pair (x,y) that makes both true.
Example: Determine whether the given ordered pair is a solution of the system.
(4,1) solves x – y = 3 and 2x – 3y = 5
(-3,3) solves 3x – y = -12 and x – y = 0
We can estimate the solution(s) of a system by graphing each equation and estimating any intersection.
Example: Solve each system by graphing
x – y = 2 and x + 3y = 6
_______ solution
___________ system
Independent equations
y = -3x and 6x + 2y = 4
______ solution
_____________ system
Independent equations
-2x + y = 1 and 4x – 2y = -2
___________ solutions (*)
___________ system
Dependent equations
Graphing is useful in finding approximate solutions given easily graphed equations, but one method for
finding exact solutions of a system involves the substitution method. Steps are
1. Solve 1 equation for 1 of its variables
2. Substitute this into the other equation
3. Solve this substituted equation
4. Replace this value into the equation in step 1
5. Check answer in both
Example: Use the substitution method to solve the systems
6x – 4y = 10 and y = 3x – 3
3x – y = 6 and -4x + 2y = -8
Recall (from pre-algebra) that multiplying both sides of an equation does not change the equation
Example: Use the same method to solve the system 
x y 1
x y
1
  and   
2 4 2
2 2
8
Another algebraic property (addition property: if A = B and C = D then A + C = B + D) can be used to give
us the elimination method for solving systems of linear equations in two variables. Steps are
1. Rewrite each equation into standard form (Ax + By = C)
2. If needed, multiply both sides of 1 (or both) equations such that the coefficient of 1 variable in 1
equation is opposite (additive inverse) of its coefficient in the other equation
3. Add the equations
4. Using the result, find the value of the remaining variable
5. Find the value of the other variable by substituting the result into either original equation
6. Check answer in both
Example: Use the elimination method to solve the systems
5x + 2y = 1 and x – 3y = 7
2x – 5y = 6 and 3x – 4y = 9
x
+ 2y = -1 and x + 6y = 2
3
Example: Use the same method to solve the system 4x – 7y = 10 and -8x + 14y = -20
{(x,y)|4x – 7y = 10}