1. Warm Up #4 1. Evaluate –3x – 5y for x = –3 and y = 4. –11 ANSWER
2. Solve the system by graphing.
x + y = 2
2x + y = 3
ANSWER
(1, 1)

2. Solving a system of equations using Substitution
Step 1: Get one of the 4 variables by itself on one side of the equals sign
Step 2: Substitute its value into the other equation and solve
Step 3: Substitute the answer back into the equation with the other variable by itself and solve
Step 1: Already done
Solve:
3y - 2x = 11 y = 9 - 2x
Step 2: 3(9 – 2x) – 2x = 11
27 – 6x – 2x = 11
Step 3:
y = 9 – 2(2)
27 – 8x = 11
- 27
y = 9 – 4
-8x = -16
y = 5 -8
These 2 lines intersect at the point ( 2 , 5 )
x = 2

3. Step 1: Rewrite the first equation as y = 3x + 5 Solve:
Step 1: Get one of the 4 variables by itself on one side of the equals sign
Step 2: Substitute its value into the other equation and solve
Step 3: Substitute the answer back into the equation with the other variable by itself and solve
Step 1: Rewrite the first equation as y = 3x + 5
Solve:
y – 3x = 5 y + x = 3
Step 2: (3x + 5) + x = 3
4x + 5 = 3
- 5
Step 3:
y = 3(-½) + 5
4x = -2 4
y = -3/2 + 5
x = -½
y = 3½
These 2 lines intersect at the point ( -½ , 3½ )

4. Use the substitution method
EXAMPLE 3
Use the substitution method
Solve the system using the substitution method.
2x + 5y = –5
STEP 1
Solve Equation 2 for x.
x + 3y = 3
x = –3y + 3
2x +5y = –5
Write Equation 1.
2(–3y + 3) + 5y = –5
Substitute –3y + 3 for x.
y = 11
Solve for y.
Solution (-30 , 11)
x = –3y + 3
Write revised Equation 2.
x = –3(11) + 3
Substitute 11 for y.
x = –30
Simplify.

5. Elimination (Linear Combination)
The goal when using Elimination is to add the two equations together to Eliminate one of the variables
Ex x – y = 8
-5x + y = 4
You need to have the coefficients of either the x or the y add up to zero (opposites)
-3x = 12
Notice the coefficients of the y terms -3 -3
They add up to zero, so we can add the two equations together
x = -4
-5(-4) + y = 4
Then solve for the other variable
20 + y = 4
- 20
y = -16
Then substitute it in either equation to find the other variable
Solution (-4 , -16)

6. What if they don’t add up to zero to start with?
You can multiply either or both of the equations by something to get them to be opposites
3x + 6y = -6
5x - 2y = 14
3( )
Look at the y coefficients in this problem
3x + 6y = -6
15x - 6y = 42
We can multiply the bottom equation by 3 to make them opposites
18x = 36 18 18
Now add the equations together and solve
x = 2
Substitute in one of the equations
3(2) + 6y = -6
6 + 6y = -6
- 6
Solution (2 , -2)
6y = -12
y = -2

7. 2( ) 3x + 4y = -25 2x - 3y = 6 -3( ) 6x + 8y = -50 -6x + 9y = -18
Needing to multiply both equations
2( )
3x + 4y = -25
2x - 3y = 6
We can eliminate either the x or the y terms. Let’s eliminate the x this time. If they were 6x and -6x they would add to zero.
-3( )
6x + 8y = -50
-6x + 9y = -18
Now solve as before
17y = -68 17 17
y = -4
2x – 3(-4) = 6
Solution (-3 , -4)
2x + 12 = 6
- 12
2x = -6
x = -3

8. Special cases -3( ) x + 4y = 1 3x + 12y = 3 -3x - 12y = -3
Multiply top equation by -3
-3( )
x + 4y = 1
3x + 12y = 3
-3x - 12y = -3
3x + 12y = 3
0 = 0
Since 0 always equal 0 then this system has infinite solutions.

9. -6x - 12y = -3 3x + 6y = 4 2( ) -6x - 12y = -3 6x + 12y = 8 0 = 5
Multiply the bottom equation by 2
2( )
-6x - 12y = -3
6x + 12y = 8
0 = 5
Since 0 can never equal 5, there are no solutions.

10. Classwork Assignment
WS 3.2 (1-23 odd)