College Algebra Tutorial 53


College Algebra
Answer/Discussion to Practice Problems  
Tutorial 53: Practice Test on Tutorials 49 - 52


WTAMU > Virtual Math Lab > College Algebra > Tutorial 53: Practice Test on Tutorials 49 - 52

 


Problems 1a - 1b:  Solve each system by either the substitution or elimination by addition method.

 
1a. 
problem 1a
 
checkAnswer:

I'm going to chose to use the elimination by addition method, however it would be perfectly fine for you to use the substitution method.  Either way the answer will be the same.

Multiplying each equation by it's respective LCD we get:

ad1a1
 

Multiplying the first equation by -2 and then adding the resulting equations together we get:

ad1a2
 

Solving for x we get:

ad1a3
 

Plugging in -6 for x into the first simplified equation to find y's value, we get:

ad1a4
 

(-6, 3) is the solution to this system.
 
 

1b. 
problem 1b
 
checkAnswer:

I'm going to chose to use the substitution method, however it would be perfectly fine for you to use the elimination by addition method.  Either way the answer will be the same.

Second equation solved for x:

ad1b1
 

Substituting the expression 3y + 5 for x into the first equation and solving for y we get:

ad1b2
 

Since the variable dropped out AND we have a TRUE statement, then when have an infinite number of solutions.  They end up being the same line.

Two ways to write the answer are {(x, y)| 2x - 6y = 10} OR {(x, y) | x = 3y + 5}.
 
 

Problem 2a:  Solve the system by either the substitution or elimination by addition method.

 
2a. 
problem 2a
 
checkAnswer:

I'm going to chose to use the elimination by addition method, however it would be perfectly fine for you to use the substitution method.  Either way the answer will be the same.

Eliminating z from equations (1) and (2) we get:

ad2a1
 

Eliminating z from equations (1) and (3) we get:

ad2a2
 

Putting equations (4) and (5) together into a system of two equations and two unknowns we get:

ad2a3
 

Eliminating y from equations (4) and (5) we get:

ad2a4

Solving for x we get:

ad2a5
 

Plugging 2 in for x in equation (4) and solving for y we get:

ad2a6
 

Plugging in 2 for x and -1 for y in equation (1) and solving for z we get:

ad2a7
 

(2, -1, 3) is the solution to our system.
 
 

Problems 3a - 3b:  Solve the given word problems using systems of equations.

 
3a. It took a boat 4 hours to travel downstream 120 miles.  Upstream, the same trip took 6 hours.  Find the rate of the boat in still water and the rate of the current.
 
 
 
checkAnswer:

Let x = the rate of the boat in still water

Let y = the rate of the current.
 
    (Rate) (Time) =  Distance With current x + y 4 120 Against current x - y 6 120

Using the fact that (Rate)(Time) = Distance, we get the system:

ad3a1
 

Simplifying this system we get:

ad3a2
 

Eliminating the x's we get:

ad3a3
 

Solving for y we get:

ad3a4
 

Plugging 5 in for y in the first simplified equation and solving for x we get:

ad3a5
 

The speed of the boat in still water is 25 mph and the rate of the current is 5 mph.
 
 

3b. A piggy bank contains only nickels, dimes and quarters.  The value of the coins is $5.50.  The number of nickels is three times that of the dimes. The number of nickels is six more than twice the number of quarters.  Find the number of nickels, dimes, and quarters in the piggy bank.
 
 
 
checkAnswer:

Let n = the number of nickels
Let d = the number of dimes
Let q = the number of quarters

The value of the coins is $5.50:

ad3b1
 

The number of nickels is three times that of the dimes:

ad3b2
 

The number of nickels is six more than twice the number of quarters:

ad3b3
 

Putting this together in a system we get:

ad3b4
 

Simplifying the system we get:

ad3b5
 

Eliminating n from equations (1b) and (2b) we get:

ad3b6
 

Eliminating n from equations (2b) and (3b) we get:

ad3b7
 

Putting equations (4) and (5) together into a system of two equations and two unknowns we get:

ad3b8
 

Eliminating d from equations (4) and (5) we get:

ad3b9
 

Solving for q we get:

ad3b10
 

Plugging in 12 for q in equation (3) and solving for n we get:

ad3b11
 

Plugging in 30 for n in equation (2) we get:

ad3b12
 

There are 30 nickels, 10 dimes and 12 quarters.
 

Problems 4a - 4b:  Solve each system by either the substitution or elimination by addition method.

 
4a. 
problem 4a
 
 
checkAnswer:

I'm going to chose to use the elimination by addition method.  Note that this one would be very difficult to try to solve it by the substitution method.

Eliminating the x squared terms we get:

ad4a1
 

Solving for y we get:

ad4a2
 

Plugging 2 in for y in the first equation and solving for x we get:

ad4a3

Two solutions are (3, 2) and (-3, 2).
 

Plugging -2 in for y in the first equation and solving for x we get:

ad4a4

Two more solutions are (3, -2) and (-3, -2).
 

The four solutions are (3, 2), (-3, 2), (3, -2), and (-3, -2).
 
 

4b. 
problem 4b
 
 
checkAnswer:

I'm going to chose to use the substitution method, however it would be perfectly find for you to use the elimination by addition method.  Either way the answer will be the same.

Second equation solved for y:

ad4ab1
 

Substituting the expression x for y into the first equation and solving for x

ad4b2
 

This will not factor so we will need to use the quadratic formula to solve:

ad4b3
 

Since, we have a negative in the square root, this means we did not get real number answer for x.

The answer is no solution.


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WTAMU > Virtual Math Lab > College Algebra >Tutorial 53: Practice Test on Tutorials 49 - 52


Last revised on March 25, 2011 by Kim Seward.
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