Download 10 pts possible 2 pts 1. Evaluate –3x – 5y for x = –3 and y = 4. 2

Bell Quiz 3-2
2 pts
1. Evaluate –3x – 5y for x = –3 and y = 4.
3 pts
2. Solve the system by graphing.
x+y=2
2x + y = 3
5 pts
3. Twice a number x plus a number y is 3. The number y
subtracted from three times the number x is 7. Find x
and y by graphing.
10 pts
possible
Section 3-2A
1
3-2
Chapter 3: Linear Systems
Section 3-2A
2
3-2A
Chapter 3: Linear Systems
Section 3-2A
3
Key Concept
The Substitution Method
Step 1: Solve one of the equations for one of its variables
Step 2: Substitute the expression from Step 1 into the
other equation and solve for the other variable
Step 3: Substitute the value from Step 2 into the revised
equation from Step 1 and solve.
Step 4: Write solution as an Ordered Pair
Section 3-2A
4
EXAMPLE 1
Use the substitution method
Solve the system using the substitution method.
2x + 5y = –5
x + 3y = 3
(– 30, 11)
Section 3-2A
5
EXAMPLE 1
Use the substitution method
Solve the system using the substitution method.
2x + 5y = –5
x + 3y = 3
Equation 1
Equation 2
SOLUTION
STEP 1 Solve Equation 2 for x.
x = –3y + 3
Revised Equation 2
EXAMPLE 1
Use the substitution method
STEP 2
Substitute the expression for x into Equation 1 and solve
for y.
2x +5y = –5 Write Equation 1.
2(–3y + 3) + 5y = –5
y = 11
Substitute –3y + 3 for x.
Solve for y.
STEP 3
Substitute the value of y into revised Equation 2 and
solve for x.
x = –3y + 3
Write revised Equation 2.
x = –3(11) + 3 Substitute 11 for y.
x = –30
Simplify.
EXAMPLE 1
Use the substitution method
ANSWER
The solution is (– 30, 11).
CHECK Check the solution by substituting into the
original equations.
2(–30) + 5(11) =? –5
–5 = –5
Substitute for x and y. –30 + 3(11) =? 3
Solution checks.
3=3
GUIDED PRACTICE
for Example 1
Solve the system using the substitution method.
1. 4x + 3y = –2
x + 5y = –9
ANSWER
The solution is (1,–2).
Section 3-2A
9
EXAMPLE 4
Solve linear systems with many or no solutions
Solve the linear system.
a. x – 2y = 4
3x – 6y = 8
a. no solution
b. 6x – 3y = 15
– 2x + y = – 5
b. infinitely many solutions
Section 3-2A
10
EXAMPLE 4
Solve linear systems with many or no solutions
Solve the linear system.
a. x – 2y = 4
3x – 6y = 8
SOLUTION
a. Because the coefficient of x in the first equation is 1,
use the substitution method.
Solve the first equation for x.
x – 2y = 4
x = 2y + 4
Write first equation.
Solve for x.
EXAMPLE 4
Solve linear systems with many or no solutions
Substitute the expression for x into the second equation.
3x – 6y = 8
3(2y + 4) – 6y = 8
12 = 8
Write second equation.
Substitute 2y + 4 for x.
Simplify.
ANSWER
Because the statement 12 = 8 is never true, there is
no solution.
GUIDED PRACTICE
for Example 4
Solve the linear system using the substitution method.
2. 12x – 3y = – 9
3. 6x – 2y = 5
–4x + y = 3
– 3x + y = 7
2. infinitely many solutions
Section 3-2A
3. no solution
13
HOMEWORK
Sec 32A (pg 164)
314
Section 3-2A
14