Download 3-2 Solving Systems of Equations Algebraically

3-2 Solving Systems of Equations Algebraically
Substitution: one Equation is solved for one variable in terms of the other. This expression is substituted
for the variable in the other equation.
(ex) 3x + y = 10
(1)
2x + y = 8
Steps to follow:
1. get x or y by itself
2. put into other equation
3. solve for the variable
4. put answer into an original equation
y = -3x = 10
3(2) + y = 10
6 + y = 10
y=4
2x + (-3x + 10) = 8
-x + 10 = 8
-x = -2
x=2
Answer: ( 2, 4)
(ex) x = 2y + 5
x = 2(2) + 5
x=9
x – 3y = 3
(2y + 5) -3y = 3
-y + 5 = 3
-y = -2
y=2
Answer: ( 9, 2)
Elimination: you eliminate one of the variable by + or – to the other equation.
(ex)
5a + 2b = 16
4a – 2b = 2
9a
= 18
a
=2
5(2) + 2b = 16
10 + 2b = 16
2b = 6
b=3
Answer: ( 2, 3)
Sometimes you will have to multiply first before you eliminate.
( a – b = 2) (x2) 2a – 2b = 4
-2a + 3b = 3
-2a + 3b = 3
b=7
Answer: ( 9, 7)

a–7=2
a=9
(24)
3a+ 4b = 9
-3a-2b = -3
2b = 6
2
2
b=3

3a + 4(3)= 9
3a +12 = 9
-12 -12
3a = -3
3
3
a = -1
Answer: (-1, 3)
(34)
(2x -3y = 6) (x)(-3) 
6x-9y = 9
6x+9y = -18
6x-9y = 9
0 = -9 no solutions