The document discusses methods for solving systems of linear equations in two variables, including graphing, elimination, and substitution. It provides examples of using each method to solve systems and determine if they have one solution, no solution, or infinitely many solutions. Key points covered are the three possible outcomes when graphing systems, the steps for the elimination and substitution methods, and how to determine the solution set.

1. 1
Systems of Linear Equations
• 4-1 Systems of Linear Equations in Two Variables

2. 2
4-1 Systems of Linear Equations in Two Variables
Deciding whether an ordered pair is a
solution of a linear system.
The solution set of a linear system of equations contains all
ordered pairs that satisfy all the equations at the same time.
• Example 1: Is the ordered pair a solution of the given system?
2x + y = -6 Substitute the ordered pair into each equation.
x + 3y = 2 Both equations must be satisfied.
A) (-4, 2) B) (3, -12)
2(-4) + 2 = -6 2(3) + (-12) = -6
(-4) + 3(2) = 2 (3) + 3(-12) = 2
-6 = -6 -6 = -6
2 = 2 -33 ≠ -6
∴ Yes ∴ No

3. 3
4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems by Graphing.
One way to find the solution set of a linear system of equations is to
graph each equation and find the point where the graphs intersect.
• Example 1: Solve the system of equations by graphing.
A) x + y = 5 B) 2x + y = -5
2x - y = 4 -x + 3y = 6
Solution: {(3,2)} Solution: {(-3,1)}

4. 4
4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems by Graphing.
There are three possible solutions to a system of linear equations in two
variables that have been graphed:
• 1) The two graphs intersect at a single point. The coordinates give the
solution of the system. In this case, the solution is “consistent” and the
equations are “independent”.
• 2) The graphs are parallel lines. (Slopes are equal) In this case the system
is “inconsistent” and the solution set is 0 or null.
• 3) The graphs are the same line. (Slopes and y-intercepts are the same) In
this case, the equations are “dependent” and the solution set is an infinite
set of ordered pairs.

5. 5
4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by
Method of Elimination.
Remember: If a=b and c=d, then a + c = b + d.
Step 1: Write both equations in standard form
Step 2: Make the coefficients of one pair of variable terms opposite
(Multiply one or both equations by appropriate numbers so that the
sum of the coefficients of either x or y will be zero.)
Step 3: Add the new equations to eliminate a variable
Step 4: Solve the equation formed in step 3
Step 5: Substitute the result of Step 4 into either of the original
equations and solve for the other value.
Step 6: Check the solution and write the solution set.

6. 6
4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by
Method of Elimination.
• Example 2: Solve the system : 2x + 3y = 19
Step 1: Both equations are in standard form 3x - 7y = -6
Step 2: Choose the variable x to eliminate: Multiply the top equation by 3, the bottom
equation by -2
3[2x + 3y = 19] 6x + 9y = 57
-2[3x - 7y = -6] -6x +14y = 12
Step 3: Add the new equations to eliminate a variable
0x + 23y = 69
Step 4: Solve the equation formed in step y = 3
Step 5: Substitute the result of Step 4 into either of the original equations and solve for the
other value. 2x + 3(3) = 19
2x = 10
x = 5 Solution Set: {(5,3)}
Step 6: Check the solution and write the solution set.

7. 7
4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by
Method of Elimination.
• Example 3:
Solve the system :
2[2x - 3y = 1] 4x - 6y = 2
-3[3x - 2y = 9] -9x + 6y = -27
-5x + 0y = -25
x = 5 3(5) - 2y = 9
-2y = -6
Solution Set: {(5,3)} y = 3
1 1 1
rewrite as 6[ ] 2 3 1
3 2 6
: 2 3 1
1 1 1
3 2 6
3 2 9
3 2 9
x y x y
Solv
x y
x y
e x y
x y
⇒ − = ⇒ − =
− =
−
− =
=
=
−

8. 8
4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by
Method of Elimination.
• Example 4:
Solve the system : 2x + y = 6
-8x - 4y = -24
4[2x + y = 6] 8x + 4y = 24
-8x -4y = -24 -8x - 4y = -24
0 = 0 True
Solution Set: {(x,y)| 2x + y = 6}
Note: When a system has dependent equations and an infinite number
of solutions, either equation can be used to produce the solution set.
Answer is given in set-builder notation.

9. 9
4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables
by Method of Elimination.
• Example 5:
Solve the system : 4x - 3y = 8
8x - 6y = 14
-2[4x - 3y = 8] -8x + 6y = -16
8x - 6y = 14 8x - 6y = 24
0 = 8 False
Solution Set: 0 or null
Note: There are no ordered pairs that satisfy both
equations. The lines are parallel. There is no solution.

10. 10
4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables
by Method of Substitution.
Step 1: Solve one of the equations for either variable
Step 2: Substitute for that variable in the other equation
(The result should be an equation with just one variable)
Step 3: Solve the equation from step 2
Step 4: Substitute the result of Step 3 into either of the
original equations and solve for the other value.
Step 6: Check the solution and write the solution set.

11. 11
4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by
Method of Substitution.
• Example 6: Solve the system : 4x + y = 5
2x - 3y =13
Step 1: Choose the variable y to solve for in the top equation:
y = -4x + 5
Step 2: Substitute this variable into the bottom equation
2x - 3(-4x + 5) = 13 2x + 12x - 15 = 13
Step 3: Solve the equation formed in step 2
14x = 28 x = 2
Step 4: Substitute the result of Step 3 into either of the original equations and solve for the
other value. 4(2) + y = 5
y = -3
Solution Set: {(2,-3)}
Step 5: Check the solution and write the solution set.

12. 12
4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by
Method of Substitution.
• Example 7:
Solve the system :
y = -2x + 2
-2x + 5(-2x + 2) = 22 -2x - 10x + 10 = 22
-12x = 12
x = -1 2(-1) + y = 2
y = 4
Solution Set: {(-1,4)}
1 1 1
2 4 2
2
1 1 1
rewrite as 4[ ] 2 2
2 4 2
: 2 2
-2 5 2
5 2
2
2
x y
x y
x y x y
Solve x y
x y
⇒ + = ⇒ + =
+
+
=
+
=
=
=
+
−