Download Advanced Mathematical Concepts Extra Examples

Advanced Mathematical Concepts
Chapter 2
Lesson 2-2
Example 1
Solve the system of equations by elimination.
x + 4y – z = 20
3x + 2y + z = 8
2x – 3y + 2z = -16
One way to solve a system of three equations is to choose pairs of equations and then eliminate one of the
variables. Because the coefficient of x is 1 in the first equation, it is a good choice for eliminating x from
the second and third equations.
To eliminate x using the first and second
To eliminate x using the first and third
equations, multiply each side of the first
equations, multiply each side of the first
equation by –3.
equation by –2.
-3(x + 4y – z) = -3(20)
-3x – 12y + 3z = -60
-2(x + 4y – z) = -2(20)
-2x – 8y + 2z = -40
Then add that result to the second equation.
-3x – 12y + 3z = -60
3x + 2y + z = 8
-10y + 4z = -52
Then add that result to the third equation.
-2x – 8y + 2z = -40
2x – 3y + 2z = -16
-11y + 4z = -56
Now you have two linear equations in two variables. Solve this system. Eliminate z by multiplying each
side of the second equation by –1 and adding the two equations.
-10y + 4z = -52
-1(-11y + 4z) = -1(-56)
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-10y + 4z = -52
11y – 4z = 56
y
= 4
The value of y is 4.
By substituting the value of y into one of the equations in two variables, we can solve for the value of z.
-10y + 4z = -52
-10(4) + 4z = -52
y=4
4z = -12
z = -3
The value of z is –3.
Finally, use one of the original equations to find the value of x.
x + 4y – z = 20
x + 4(4) – (-3) = 20
y = 4, z = -3
x =1
The solution is x = 1, y = 4, and z = -3. This can be written as the ordered triple (1, 4, -3). Check by
substituting the values into each of the original equations.
Example 2
Solve the system of equations by substitution.
2x = -6y
x + y + z = 10
-4x – 4y – z = -4
You can easily solve the first equation for x.
2x = -6y
x = -3y
Divide each side by 2.
Then substitute –3y for x in each of the other two equations. Simplify each equation.
x + y + z = 10
-4x – 4y – z = -4
-3y + y + z = 10
x = -3y
-4(-3y) – 4y – z = -4
x = -3y
-2y + z = 10
8y – z = -4
Solve –2y + z = 10 for z.
-2y + z = 10
z = 10 + 2y
Add 2y to each side.
Substitute 10 + 2y for z in 8y – z = -4. Simplify.
8y – z = -4
8y – (10 + 2y) = -4
z = 10 + 2y
Advanced Mathematical Concepts
Chapter 2
6y – 10 = -4
y =1
Now, find the values of z and x. Use z = 10 + 2y and x = -3y. Replace y with 1.
z = 10 + 2y
z = 10 + 2(1)
z = 12
x = -3y
x = -3(1)
x = -3
y=1
y=1
The solution is x = -3, y = 1, and z = 12. Check each value in the original system.
Example 3
MANUFACTURING A manufacturer of golf balls supplies three driving ranges with balls. The
output from a week’s production of balls is 320 cases. The manufacturer must send driving range A
three times as many cases as are sent to driving range B, and must send driving range C 160 cases
less than ranges A and B together. How many cases should be sent to each driving range to
distribute the entire week’s production to them?
Write a system of equations. Define the variables as follows.
x = the number of cases sent to driving range A
y = the number of cases sent to driving range B
z = the number of cases sent to driving range C
The system is:
x + y + z = 320
x = 3y
z = x + y – 160
total number of cases produced
cases to A equal three times cases to B
cases to C equal 160 less than A and B together
The second equation tells us that 3y can be substituted for x in the other two equations. Simplify each
equation.
x + y + z = 320
3y + y + z = 320
4y + z = 320
z = x + y – 160
z = 3y + y – 160
z = 4y - 160
x = 3y
x = 3y
Substitute 4y – 160 for z in 4y + z = 320.
4y + z
4y + 4y – 160
8y – 160
y
= 320
= 320
= 320
= 60
z = 4y - 160
Now, find the values of x and z. Use x = 3y and z = 4y – 160. Replace y with 60.
x = 3y
x = 3(60)
x = 180
y = 60
z = 4y – 160
z = 4(60) – 160
z = 80
y = 60
The manufacturer should send 180 cases of golf balls to driving range A, 60 cases to driving range B, and
80 cases to driving range C.
Assignment: Pages 76-77
8, 11, 14, 17, 24, 25, 28