Download Solving Systems of Equations By Substitution

Solving Systems of Equations
using Substitution
HCPS
Henrico County Public Schools
Designed by Vicki Hiner- Godwin High School
Lesson Objective:
 Solve systems of equations by substitution
method.
Assignment:
 pp. 450-451 #5-30 Skip 9
 Fractions on some of these, so don’t freak
out.
Solving Systems of Equations
using Substitution
Steps:
1. Solve One equation for One variable( y= ; x= ; a=)
2. Substitute equation from step one into other equation
(get an equation with only one variable)
3. Solve for the first variable.
4. Go back and use the found variable in step 3 to find
second variable.
5. Check the solution in both equations of the system.
WHAT DO WE DO???? CLICK TO SEE AN EXAMPLE
GIVEN EXAMPLE:
y= 4x
3x+y=-21
STEP1:
y=4x (Already solved for y)
STEP 2:
Substitute into second equation:
3x + y = -21 becomes:
GIVEN EXAMPLE:
y= 4x
3x+y=-21
STEP1:
y=4x (Already solved for y)
STEP 2:
Substitute into second equation:
3x + y = -21 becomes:
3x +4x =-21
GIVEN EXAMPLE:
y= 4x
3x+y=-21
STEP 3: Solve for the variable
3x + 4x=-21
7x=-21
x=-3
GIVEN EXAMPLE:
y= 4x
3x+y=-21
STEP 4: Solve for the other
variable use x=-3 and y=4x
y=4x and x = -3 therefore:
y=4(-3) or y = -12
Solution to the system is (-3,-12)
GIVEN EXAMPLE:
y= 4x
3x+y=-21
Check solution ( -3,-12)
y=4x
3x+y=-21
-12=4(-3)
3(-3)+(-12)=-21
-12=-12
-9+(-12)=-21
-21=-21
Solving Systems of Equations
using Substitution
Review Steps --Questions?
Steps:
1. Solve One equation for One variable( y= ; x= ; a=)
2. Substitute equation from step one into other equation
(get an equation with only one variable)
3. Solve for the first variable.
4. Go back and use the found variable in step 3 to find
second variable.
5. Check the solution in both equations of the system.
GIVEN EXAMPLE:
x + y=10
5x - y=2
STEP1: Solve for y
x + y = 10
STEP 2:
y = -x +10
Substitute into second equation: 5x - y = 2
becomes:
GIVEN EXAMPLE:
x + y=10
5x - y=2
STEP1: Solve for Y
x + y = 10
STEP 2:
y = -x +10
Substitute into second equation: 5x - y = 2
becomes:
5x - (-x+10) = 2
GIVEN EXAMPLE:
x + y=10
5x - y=2
STEP 3: Solve for the variable
5x-(-x+10)=2
5x+x-10=2
6x-10=2
6x=12
x=2
GIVEN EXAMPLE:
x + y=10
5x - y=2
STEP 4: Solve for the other
variable use x=2 and x+y=10
x=2 and x+y = 10 therefore:
2+y=10 and y = 8
Solution to the system is (2,8)
GIVEN EXAMPLE:
x + y=10
5x - y=2
Check solution (2,8)
x + y=10
2+8=10
10=10
5x-y=2
5(2)-(8)=2
10-8=2
2=2
THE END OF NOTES!
You TRY:
1.
y  2x  2
2x  3y  10
2.
2a  3b  7
2a  b  5