College Algebra
Answer/Discussion to Practice Problems
Tutorial 49: Solving Systems of
Linear Equations in Two Variables
Answer/Discussion
to 1a

I’m going to chose to use the elimination by addition
method, however
it would be perfectly find for you to use the substitution
method.
Either way the answer will be the same. 
This equation is full of those nasty fractions.
We can simplify
both equations by multiplying each separate one by it’s LCD, just like
you can do when you are working with one equation. As long as you
do the same thing to both sides of an equation, you keep the two sides
equal to each other.
Multiplying each equation by it's respective LCD we
get: 

*Mult. by LCD of 6
*Mult. by LCD of 15

Again, you want to make this as simple as
possible.
Note how the coefficient on y in the first
equation is 2 and in the second equation it is 3. We need to have
opposites, so if one of them is 6 and the other is 6, they would
cancel
each other out when we go to add them. If we added them together
the way they are now, we would end up with one equation and two
variables,
nothing would drop out. And we would not be able to solve it.
So I proposed that we multiply the first equation by
3 and the second
equation by 2, this would create a 10 and a 10 in front of the y’s
and we will have our opposites.
Multiplying the first equation by 3 and the second
equation by 2
we get: 

*Mult. both sides of 1st eq.
by 3
*Mult. both sides of 2nd eq. by 2
*y's
have opposite
coefficients


*Note that y's
dropped out

You can choose any equation used in this problem to
plug in the found x value.
I choose to plug in 2 for x into the
first simplified equation (found in step 1) to find y’s
value. 

*Plug in 2 for x
*Inverse of add 6 is sub. 6
*Inverse of mult. by 2 is div.
by 2

You will find that if you plug the ordered pair (2, 1)
into BOTH equations
of the original system, that this is a solution to BOTH of them.
(2, 1) is a solution to our system. 
Answer/Discussion
to 1b

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. 
Both of these equations are already simplified.
No work needs
to be done here. 
Note how the first equation is already solved for y.
We can use that one for this step.
It does not matter which equation or which variable you
choose to solve
for. But it is to your advantage to keep it as simple as
possible.
First equation solved for y: 
Substitute the expression 2x 
5 for y into the second equation and solve for x:
(when you plug in an expression like this, it is just like you plug
in a number for your variable) 

*Sub. 2x  5 for y
*Variable dropped out AND true 
Wait a minute, where did our
variable go????
As mentioned above, if the variable drops out AND we
have a TRUE statement,
then when have an infinite number of solutions. They end up being
the same line. 
Since we did not get a value for x,
there
is nothing to plug in here. 
There is no value to plug in here.
When they end up being the same equation, you have an
infinite number
of solutions. You can write up your answer by writing out either
equation to indicate that they are the same equation.
Two ways to write the answer are {(x, y) y = 2x  5} OR {(x, y)
 4x  2y =
10}. 
Answer/Discussion
to 1c

I’m going to chose to use the elimination by addition
method, however
it would be perfectly find for you to use the substitution
method.
Either way the answer will be the same. 
Both of these equations are already simplified and in
the right form.
No work needs to be done here. 
Again, you want to make this as simple as
possible.
Note how the coefficient on x in the first
equation is 2 and in the second equation it is 4. We need to
have
opposites, so if one of them is 4 and the other is 4, they would
cancel
each other out when we go to add them. If we added them together
the way they are now, we would end up with one equation and two
variables,
nothing would drop out. And we would not be able to solve it.
So I proposed that we multiply the first equation by
2, this
would create a 4 and a 4 in front of the x’s
and we will have our opposites.
Multiplying the first equation by 2 we get: 

*Mult. both sides of 1st eq.
by 2
*x's
and y's
have opposite coefficients


*Note that x's
and y's both dropped out

Wait a minute, where did our
variable go????
As mentioned above if your variable drops out and you
have a FALSE statement,
then there is no solution. If we were to graph these two,
they would be parallel to each other. 
There is no value to plug in here. 
There is no value to plug in here.
The answer is no solution. 
Answer/Discussion
to 2a


*Plug in 0 for y for xint
*Inverse of mult. by 5 is div.
by 5
*xintercept 
The xintercept is (2, 0).
yintercept 

*Plug in 0 for x for yint
*Inverse of mult. by 2 is div.
by 2
*yintercept 
The yintercept is (0, 5).
Find another
solution by letting x = 1. 

*Plug in 1 for x
*Inverse of add 5 is sub. 5
*Inverse of mult. by 2 is div.
by 2

Another solution is (1, 5/2).
Solutions:
x

y

(x, y)

2

0

(2, 0)

0

5

(0, 5)

1

5/2

(1, 5/2)

Plotting the ordered pair solutions and drawing the
line:


*Plug in 0 for y for xint
*xintercept

The xintercept is (1, 0).
yintercept 

*Plug in 0 for x for yint
*Inverse of mult. by 1 is div.
by 1
*yintercept 
The yintercept is (0, 1).
Find another
solution by letting x = 1. 

*Plug in 1 for x
*Inverse of add 1 is sub. 1
*Inverse of mult. by 1 is div
by 1

Another solution is (1, 2).
Solutions:
x

y

(x, y)

1

0

(1, 0)

0

1

(0, 1)

1

2

(1, 2)

Plotting the ordered pair solutions and drawing the
line:

We need to ask ourselves, is there any place that the
two lines intersect,
and if so, where?
The answer is yes, they intersect at (4, 5). 
You will find that if you plug the ordered pair (4, 5)
into BOTH equations
of the original system, that this is a solution to BOTH of them.
The solution to this system is (4, 5).
(return
to problem 2a) 
Last revised on March 26, 2011 by Kim Seward.
All contents copyright (C) 2002  2011, WTAMU and Kim Seward. All rights reserved.

