College Algebra
Answer/Discussion to Practice
Problems
on Systems of Linear Equations
and Problem Solving
Answer/Discussion
to 1a
The larger of two numbers is 5 more than twice the smaller. If
the smaller is subtracted from the larger, the result is 12. Find
the numbers. |
| Make sure that you read the question carefully several times.
Since we are looking for two numbers, we will let
| x = the smaller number
y = the larger number |
|
| Since we have two unknowns, we need to build
a system with two equations.
Equation (1):
Equation (2):
|
| Putting the two equations together in a system we get:

|
 |
*Substitute 5 + 2x in for y
*Inverse of add 5 is sub. 5
|
| Solve for second variable.
Using equation (1) to plug in 7 for x
and
solving for y we get: |
 |
*Equation (1)
*Plug in 7 for x
|
| You will find that if you plug the ordered pair (7, 19) into BOTH equations
of the original system, this is a solution to BOTH of them.
Final Answer:
The numbers are 7 and 19. |
Answer/Discussion
to 1b
It takes a boat 2 hours to travel 24 miles downstream and 3 hours to
travel 18 miles upstream. What is the speed of the boat in still
water and of the current of the river? |
| Make sure that you read the question carefully several times.
Since we are looking for two different rates, we will let
| x = rate of the boat
y = the rate of the current |
Since this is a rate/distance problem, it might be good to organize
our information using the distance formula.
Keep in mind that the rate of the current is affecting the overall speed.
When the boat is going upstream, it will be going slower. That
rate will be x - y.
When the boat is going downstream, it will be going faster.
That rate will be x + y.
|
|
(Rate)
|
(Time)
|
= Distance
|
|
Upstream
|
x - y
|
3
|
18
|
|
Downstream
|
x + y
|
2
|
24
|
|
| Since we have two unknowns, we need to build
a system with two equations.
Equation (1):
Equation (2):
|
| Putting the two equations together in a system we get:

|
 |
*Div. both side of eq. (1) by 3
*Div. both sides of eq. (2) by 2
|
 |
*y's have opposite
coefficients
*y's dropped out |
| Solve for remaining variable.
Solving for x we get: |
 |
*Inverse of div. by 2 is mult. by 2
|
| Solve for second variable.
Using equation (4) to plug in 9 for x and
solving for y we get: |
 |
*Equation (4)
*Plug in 9 for x
*Inverse of add 9 is sub. 9
*Inverse of mult. by -1 is div. by -1
|
| You will find that if you plug the ordered pair (9, 3) into BOTH equations
of the original system, this is a solution to BOTH of them.
Final Answer:
The boat speed is 9 mph and the rate of the current is 3 mph. |
Answer/Discussion
to 1c
A student has money in three accounts that pay 5%, 7%, and 8%, in annual
interest. She has three times as much invested at 8% as she does
at 5%. If the total amount she has invested is $1600 and her interest
for the year comes to $115, how much money does she have in each account? |
| Make sure that you read the question carefully several times.
Since we are looking for three different amounts, we will let
| x = the amount invested at 5%
y = the amount invested at 7%
z = the amount invested at 8% |
|
| Since we have three unknowns, we need to build
a system with three equations.
Equation (1):
Equation (2):
Equation (3):
|
| Putting the three equations together in a system we get:

|
This is a system of linear equations with three variables, which can
be found in Tutorial 50:
Solving Systems of Linear Equations in Three Variables.
Simplify and put all three equations in the
form Ax + By + Cz = D if needed.
Equation (1) needs to be put in the correct form and equation (3)
needs to be multiplied by 100 to get rid of the decimals: |
 |
*Inverse of add 3x
is sub. 3x
*Mult. both sides of eq. (3)
by 100
|
| Choose to eliminate any one of the variables
from any pair of equations.
Since y is already eliminated in equation (4), it would be quickest
and easiest to eliminate y.
We can use equation (4) as one equation with y eliminated |
 |
*y is already
eliminated |
Eliminate the SAME variable chosen in step
2 from any other pair of equations, creating a system of two equations
and two unknowns.
Multiplying equation (2) by -7 and adding it to equation (5) we get: |
 |
*Mult. both sides of eq. (2) by -7
*y's have opposite
coefficients
*y's dropped out |
| Solve the remaining system found above.
Putting those two equations together we get: |
| I'm going to choose to eliminate z.
Multiplying equation (6) by -1 and then adding that to equation (4)
we get: |
 |
*Mult. both sides of eq. (6) by -1
*z's have opposite
coefficients
*z's dropped out |
 |
*Inverse of mult. by -1 is div. by -1 |
| Using equation (4) to plug in 300 for x
and solving for z we get: |
 |
*Equation (4)
*Plug in 300 for x
*Inverse of sub. 900 is add 900
|
| Solve for the third variable.
Using equation (2) to plug in 300 for x
and 900 for z we get: |
 |
*Equation (2)
*Plug in 300 for
x
and 900 for z
*Inverse of add 1200 is sub. 1200
|
You will find that if you plug the ordered triple (300, 400, 900) into
ALL THREE equations of the original system, this is a solution to ALL THREE
of them.
Final Answer:
$300 is invested at 5%, $400 is invested at 7% and $900 is invested
at 8%. |
All contents copyright (C) 2002 - 2008, WTAMU and Kim Seward. All rights reserved. Last revised on November 13, 2002 by Kim Seward. |