1. WARM UP
Find each equation, determine whether the indicated pair (x, y) is a solution of the equation.
2x + y = 5; (1, 3)
4x – 3y = 14; (5, 2)

2. SYSTEMS OF EQUATIONS

3. OBJECTIVES Solve real world problems using algebraic reasoning skills.
Visualize the solution of a system
Solve real world problems using algebraic reasoning skills.
Solve a system of equations in two variables by the substitution method
Solve a system of equations in two variables by linear combinations.

4. VOCABULARY Unique solution Linear combinations Method of elimination
Cramer’s Rule
Linear combinations
Method of elimination
Substitution method
System of equations
Unique solution

5. WHAT IS A SYSTEM OF EQUATIONS?
A set of two or more equations with the same variables is called a system of equations.
The solution set of a system of equations in two variables consists of all ordered pairs that make all of the equations in the system true.
If a system has only one solution it is the unique solution.

6. SOLVING A SYSTEM GRAPHICALLY
One way to find solutions of a system is to graph the equations and look for points of intersection.
Solve graphically:
y – x = 1
y + x = 3
The graph shows the solution set. The intersection appears to be the single ordered pair (1, 2).
We check by substituting.
y – x = 1
2 –
1 1
y + x = 3
3 3
Since both are true the solution is (1, 2)

7. TRY THIS… Solve graphically x + y = 11 3x – y = 5 2x – y = 7

8. SOLVING SYSTEMS OF EQUATIONS
Graphing may not be an efficient method of solving a system of equations in two variables.
We will now consider a more efficient method.

9. THE SUBSTITUTION METHOD
The substitution method is a useful technique for solving systems in which a variable has a coefficient of 1.
Example:
Use the substitution method to solve this system.
2x + y = 6
3x + 4y = 4

10. EXAMPLE 1
We solve the first equation for y because its y-term has a coefficient of 1.
y = 6 – 2x
Thus y and 6 – 2x are equivalent. We can substitute 6 – 2x for y in the second equation.
3x + 4y = 4
3x + 4(6 – 2x) = 4
Substituting 6 – 2x for y
This gives us an equation in one variable. We can then solve for x.
3x + 24 – 8x = 4
Using the distributive property
-5x = -20
x = 4

11. EXAMPLE CONT…
Now we can substitute 4 for x in either equations and solve for y.
2x + y = 6
Choosing the first equation.
2(4) + y = 6
8 + y = 6
y = -2
We obtain (4, -2). This checks, so it is the solution of the system.

12. TRY THIS… Use the substitution method to solve these equations.
y + x = 1
3y – 2x = 12
2y - 3 = 1
2y = -4
3y = 2x + 12
y = 2
7x = -21
The solution for the system is (-3, 2)
x = - 3

13. TRY THIS… Use the substitution method to solve these equations.
x + 3y = 6
x – y = -1
-y = -x – 1
y = x + 1
The solution for the system is
5x + 3(x + 1) = 6
8x = 3

14. LINEAR COMBINATIONS
If all variables have coefficients other than 1, we can use the multiplication and addition properties to find a combination of the linear equations that will eliminate a variable.
This is called the method of linear combinations

15. EXAMPLE 2 Use linear combinations to solve this system: 3x – 4y = -1
The first equation tells us that 3x – 4y and -1 are equivalent expression. We can use the addition property to add the same quantity to both sides of the second equation.
-3x – 4y = -1 + (3x – 4y) = 0 + 1(-1)
Using the addition property
When we do this, we have actually added one multiple of the first equation to one multiple of the second equation.

16. EXAMPLE CONT… It is usually easier to add equations in column form:
3x – 4y = -1
-3x + 2y = 0
-2y = -1
Adding
We can solve this equation easily, finding y = Next we substitute for y in either of the original equations.
-3x + 2y = 0
Solving for x, we find x = Substitution will show that checks. Thus, this is the solution
-3x + 2( ) = 0
-3x + 1 = 0
x =

17. THE METHOD OF ELIMINATION
Using this method we try to get an equation with only one variable, or to eliminate a variable.
This method is thus also known as the Method of Elimination.
Note in Example 2 that a term in one equation and a term in the other were additive inverses of each other, thus their sum was 0. That enable us to eliminate a variable.
We may need to multiply an equation by a constant in order to make two terms additive inverses of each other.

18. EXAMPLE 3 Use linear combinations to solve this system: 3x + 3y = 15
If we add the equations, no variables will be eliminated. We could eliminate the y-variable, however, if the 3y in the first equation were -6y.
-2(3x + 3y) = -2(15)
-6x – 6y = -30
2x + 6y = 22
2x + 6y = 22
-4x = -8
Adding
x = 2
Solving for x

19. EXAMPLE CONT… Substitute 2 for x in either of the original equations:
2x + 6y = 22
2(2) + 6y = 22
Substituting 2 for x in the second equation.
4 + 6y = 22
6y = 18
y = 3
Solving for y
Substitution will show that (2, 3) checks. The solution of the system is (2, 2).

20. TRY THIS… Use linear combinations to solve these systems:
5x + 3y = x + 2y = -16
-5x + 2y = x – 5y = 31
(1, 4)
(-3, 1)

21. EXAMPLE 4 Use linear combinations to solve this system: 5x + 4y = 11
We can multiply both sides of the first equation by -3 and both sides of the second equation by 5, in order to make the x-terms additive inverses of each other. Then solve for y.
-3(5x + 4y) = -3(11)
-15x – 12y = -33
5(3x – 5y) = 5(23)
15x – 25y = -115
-37y = -138
Adding
y = 4
Solving for y
When we substitute 4 for y in either of the original equations, we find x = -1. The ordered pair (-1, 4) checks and is the solution of the system.

22. EXAMPLE 5 Use linear combinations to solve this system:
0.3x + 0.5y = -0.1
0.01x + 0.4y = -0.38
We multiply the first equation by 10 and the second by 100 to clear the decimals.
-3x + 5y = -1
10(-0.3x + 0.5y) = 10(-0.1)
100(0.01x + 0.4y) = 100(-0.38)
x – 40y = -38
-3x + 5y = -1
-3x + 5y = -1
3(x – 40 y) = 3(-38)
3x – 120y = -114
-115y = -115
Substituting 1 for y and solving for x, we obtain (2, 1) as a solution.
y = 1

23. EXAMPLE 6 Use linear combinations to solve this system: 3x + 4y = 6
To clear fractions, we multiply both sides of the first equation by 6 and the second by 12.
3x + 4y = 6
9x – 4y = 24
12x = 30
Solution is

24. TRY THIS… Use linear combinations to solve these systems:
3x + 5y = x + 0.3y = 0.1
5x + 3y = x – 0.01y = 0.07

25. TRY THIS…
Use linear combinations to solve these systems:

26. STEPS FOR USING LINEAR COMBINATIONS
1. Write both equations in the form Ax + By = C
2. Clear any decimals or fractions.
3. Choose a variable to eliminate.
4. Make the chosen variable’s terms additive inverses by multiplying one or both equations by a number.
5. Eliminate the variable by adding the equation.
6. Substitute to solve for the remaining variable.

27. CRAMER’S RULE FOR TWO EQUATIONS
Gabriel Cramer ( ) developed an algorithm to solve large systems of equations. The general system of two equations,
ax + by = c
dx + ey = f can be solved
Since the denominators are the same, ac – bd, it is helpful to find this value first and store the result.

28. EXAMPLE Let’s look at an example of Cramer’s Rule: Solve: 5x – 2y = 10
ax + by = c
dx + ey = f
The denominator is ae – bd.
5 x – /- x /- = STO
Solving for x,
10 x – /- x = / RCL = -6

29. EXAMPLE CONT… Now let’s solve for y: Solve: 5x – 2y = 10
ax + by = c
dx + ey = f
The denominator is ae – bd.
5 x – /- x /- = STO
Solving for y,
5 x – /- x = / RCL = -20
The solution is (-6, -20)

30. TRY THIS… Solve these systems using the algorithm shown:
0.2x + 1.2y = 3.1 b x + 10y = -66.2
2x + y = x – 36.5 = 5.7 y
x y = 3.6
y x = 4.484