Intermediate Algebra
Answer/Discussion to Practice Problems
Tutorial 21: 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:

No simplification is needed for this problem.
I'm going to go ahead and have us use the substitution
method as shown in Tutorial 19: Solving Systems of Linear Equations in
Two Variables to solve this.
Solve one equation for either variable
Since equation (1) is nicely solved for y, we will use that to substitute
into the other equation.
y = 5 + 2x (1)
Substitute what you get above into the other
equation AND solve for the remaining variable.
Substituting in 5 + 2x for y into equation (2) and solving for x 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.
(return to
problem 1a) 
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:

Simplify if needed.
We can simplify this by dividing both sides of equation (1) by 3
and equation (2) by 2: 

*Div. both side of eq. (1) by 3
*Div. both sides of eq. (2) by 2

Multiply one or both equations by a number
that will create opposite coefficients for either x or y if needed AND
add the equations.
Since we already have opposite coefficients
on y, we can go right into adding equations
(3) and (4) together: 

*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.
(return to
problem 1b) 
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:

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 
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%.
(return to
problem 1c) 
Last revised on July 10, 2011 by Kim Seward.
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