Example 1 Solve each of the following
equations.
Solution
In the following problems we will describe in detail the first problem and the leave most of the explanation out of the following problems.
a. 3 (x + 5) = 2 (– 6 – x) – 2x
For
this problem there are no fractions so we don’t need to worry about the first
step in the process. The next step tells to simplify both sides. So, we will
clear out any parenthesis by multiplying the numbers through and then combine
like terms.
3(x + 5) = 2(– 6
– x) – 2x
3x + 15 = – 12 –
2x – 2x
3x + 15 = –
12 – 4x
The
next step is to get all the x’s on one side and all the numbers on the
other side. Which side the x’s go on is up to you and will probably vary
with the problem. As a rule of thumb we will usually put the variables on the
side that will give a positive coefficient. This is done simply because it is
often easy to lose track of the minus sign on the coefficient and so if we make
sure it is positive we won’t need to worry about it.
So, for
our case this will mean adding 4x to both sides and subtracting 15 from
both sides. Note as well that while we will actually put those operations in this time we
normally do these operations in our head.
3x + 15 = – 12 – 4x
3x + 15 – 15 + 4x
= –
12 – 4x –
4x – 15
7x = – 27
The next step says to get a coefficient of 1 in front of the x. In this case we can do this by dividing both sides by a 7.
Now,
if we’ve done all of our work correct
is the solution to the equation.
is the solution to the equation.
The
last and final step is to then check the solution. As pointed out in the
process outline we need to check the solution in the original equation.
This is important, because we may have made a mistake in the very first step
and if we did and then checked the answer in the results from that step it may
seem to indicate that the solution is correct when the reality will be that we
don’t have the correct answer because of the mistake that we originally made.
The problem of course is that, with this solution, that checking might be a little messy. Let’s do it anyway.
So,
we did our work correctly and the solution to the equation is,
Note
that we didn’t use the solution set notation here. For single solutions we will
rarely do that in this class. However, if we had wanted to the solution set
notation for this problem would be,
Before
proceeding to the next problem let’s first make a quick comment about the
“messiness’ of this answer. Do NOT expect all answers to be nice simple
integers. While we do try to keep most answer simple often they won’t be so do
NOT get so locked into the idea that an answer must be a simple integer that you immediately assume that you’ve made a mistake
because of the “messiness” of the answer.
Okay, with this one
we won’t be putting quite as much explanation into the problem.
In this case we have
fractions so to make our life easier we will multiply both sides by the LCD,
which is 21 in this case. After doing that the problem will be very similar to
the previous problem. Note as well that the denominators are only numbers and
so we won’t need to worry about division by zero issues.
Let’s first multiply
both sides by the LCD.
Be careful to
correctly distribute the 21 through the parenthesis on the left side. Everything inside the parenthesis needs to be multiplied by the 21 before we simplify. At
this point we’ve got a problem that is similar the previous problem and we won’t bother with all
the explanation this time.
7(m – 2) + 21 = (2m)
(3)
7m – 12 + 21 = 6m
7m + 7 = 6m
m =
– 7
So,
it looks like m = -7 is the solution. Let’s verify it to make sure.
So,
it is the solution
This
one is similar to the previous one except now we’ve got variables in the
denominator. So, to get the LCD we’ll first need to completely factor the
denominators of each rational expression.
So,
it looks like the LCD is 2(y – 2). Also note that we will need to avoid y
= 3 since if we plugged that into the equation we would get division by
zero.
Now,
outside of the y’s in the denominator this problem works identical to
the previous one so let’s do the work.
5(y – 3) = 2(10 – y)
5y – 15 = 20 – 2y
7y = 35
y = 5
Now
the solution is not y = 3 so we won’t get division by zero with the
solution which is a good thing. Finally, let’s do a quick verification.
So
we did the work correctly.
In
this case it looks like the LCD is (z + 3) (z – 10) and it also
looks like we will need to avoid z = – 3 and z =10 to make sure
that we don’t get division by zero.
Let’s
get started on the work for this problem.
2z (z – 10) = 3 (z
+ 3) + 2 (z – 10)
2z2 – 20z
= 3z + 9 + 2 (z2 – 7z – 30)
At
this point let’s pause and acknowledge that we’ve got a z2 in the work
here. Do not get excited about that. Sometimes these will show up temporarily
in these problems. You should only worry about it if it is still there after we
finish the simplification work.
So,
let’s finish the problem.
2z2
– 20z = 3z + 9 + 2z2 – 14z – 60
– 20z = – 11z – 51
51 = 9z
Notice
that the z2 did in fact cancel out. Now, if we did our work correctly should
be the solution since it is not either of the two values that will give
division by zero. Let’s verify this.
The
checking can be a little messy at times, but it does mean that we KNOW the
solution is correct.
Example 2 Solve
each of the following equations.
Solution
The
first step is to factor the denominators to get the LCD.
So,
the LCD is (x + 2) (x + 3) and we will need to avoid x = –
2 and x = – 3 so we don’t get division. by zero.
Here
is the work for this problem.
2(x + 3) = – x
2x = – x
x = –
2
So,
we get a “solution” that is in the list of numbers that we need to avoid so we
don’t get division by zero and so we can’t use it as a solution. However, this
is also the only possible solution. That is okay. This just means that this
equation has no solution.
The
LCD for this equation is x + 1 and we will need to avoid x = – 1 so
we don’t get division by zero. Here is the work for this equation.
2 = 4(x + 1) – 2x
2 = 4x + 4 – 2x
2 = 2x + 4
– 2 = 2x
– 1 = x
So,
we once again arrive at the single value of x that we needed to avoid so
we didn’t get division by zero. Therefore, this equation has no solution.
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