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Section 4.5, Part A
Solving Problems with
Systems of Linear Equations 1
2
Steps in Solving Problems Involving Systems
of Two Linear Equations in Two Variables:
1) Understand the problem.
• Read and reread the problem.
• Choose two variables to represent the two
unknowns.
2) Translate the problem into two equations.
3) Solve the system of equations.
4) Interpret the results.
• Check proposed solution in the problem.
problem !!!!!
• State your conclusion.
Example
One number is 4 more than twice the second number.
Their total is 25. Find the numbers.
1. UNDERSTAND
Read and reread the problem.
Since we are looking for two numbers, we let
x = first number
y = second number
continued
continued
2. TRANSLATE
One number is 4 more than twice the second number.
x = 2y + 4
Their total is 25.
x + y = 25
continued
continued
3. SOLVE
We are solving the system
x = 2y +4
x + y = 25
Using the substitution method, we substitute the
solution for x from the first equation into the second
equation.
x + y = 25
(2y +4) + y = 25
Replace x with 4 + 2y.
3y + 4 = 25
Simplify.
3y = 21
Subtract 4 from both sides.
y=7
Divide both sides by 3.
continued
continued
Now we substitute 7 for y into the first equation.
x = 4 + 2y = 4 + 2(7) = 4 + 14 = 18
4. INTERPRET
Check: Substitute x = 18 and y = 7 into both of
the equations.
First equation:
x = 4 + 2y
18 = 4 + 2(7)
True
Second equation:
x + y = 25
18 + 7 = 25
True
State: The two numbers are 18 and 7.
Example
Hilton University Drama club sold 311 tickets for
a play. Student tickets cost 50 cents each; nonstudent tickets cost $1.50. If the total receipts
were $385.50, find how many tickets of each
type were sold.
1.
UNDERSTAND
Read and reread the problem.
Since we are looking for two numbers, we let
s = the number of student tickets
n = the number of non-student tickets
continued
continued
2. TRANSLATE
Hilton University Drama club sold 311 tickets for a play.
s + n = 311
total receipts were $385.50
Admission for
students
0.50s
Admission for
non-students
+
1.50n
Total
receipts
=
385.50
continued
continued
3. SOLVE
We are solving the system
s + n = 311
0.50s + 1.50n = 385.50
Since the equations are written in standard form (and we
might like to get rid of the decimals anyway), we’ll solve by
the addition/elimination method. (Substitution could be used
instead, if you prefer to do it that way.)
Question: If we wanted to eliminate s, what would we
multiply the 0.50s in the second equation by to make it
become -1s?
Answer: Multiply the second equation by –2.
s + n = 311
–2(0.50s + 1.50n) = –2(385.50)
s + n = 311
–s – 3n = –771
–2n = –460
n = 230
continued
continued
Now we substitute 230 for n into the first equation to solve
for s.
s + n = 311
s + 230 = 311
s = 81
4. INTERPRET
Check: Substitute s = 81 and n = 230 into both of the
s + n = 311
First Equation
equations.
81 + 230 = 311
0.50s + 1.50n = 385.50
0.50(81) + 1.50(230) = 385.50
40.50 + 345 = 385.50
True
Second Equation
True
State: There were 81 student tickets and 230 non student
tickets sold.
How could we set this problem up using two variables?
How would you
x=?
Ounces of 16% solution
y=?
Ounces of 8 % solution
check
these answers?
Equation 1?
x + y = 32
Equation 2?
0.16 ● x + 0.08 ● y = 0.11●32
The assignment on this material (HW 4.5A)
Is due at the start of the next class session.
You may now OPEN
your LAPTOPS
and begin working on the
homework assignment.