Welcome to Georgia Highland College Math 97 and Math instructional videos.
In this video segment, we’ll be answering the question, how do you solve a
linear equation that involves distribution.
Well if you watched our previous video on how to solve linear equations will
notice that these are the exact same steps that we go through because
distribution, when it shows up in the original problem, is part of the
simplifying process.
So just to remind you of the steps to solving a linear equation, you simplify
both sides whether that's distribution, combining like terms, anything like
that. Collect the variable terms on one side and the constant terms on the other
side. Isolate your variable to solve, and then finally check by substituting
into the original equation.
Let’s take a look at an example that involves distribution.
All right, so you see we have here a linear equation involving distribution.
On the left-hand side, that needs to distribute to the X and the negative 3 and
on the right-hand side the negative 3 needs to distribute to the X and the 2.
So we’ll begin the simplification process by doing just that. I'm going to
distribute my 2 to the X and to the negative 3 and I’m also going to distribute
the negative 3 to the X and the 2 here.
So let’s go ahead and do that process.
2 times X is 2X, 2 times negative 3 is negative 6, and then I have a negative 17
hanging on to the end of that left hand side.
The right-hand side, I have a 13 out front and when I distribute my negative 3,
negative 3 times X is negative 3X, and negative 3 times positive 2 is negative
6.
Now I have more simplification to go through because after I distributed, I now
have like terms on both sides of the equation that need to be combined and they
happen to all be constants.
So I'll go ahead and do that.
So my 2X hangs out by itself and when I combine negative 6 and negative 17 I get
negative 23. On the right hand side, I can combine my 13 and my negative 6 to
give 7, and then I have a negative 3X hanging on there.
Now I’m ready to move to the next step of the process which is to collect all of
the variable terms on one side of the equation and all of the constants on the
other.
So like I said in the last video, I like to bring my variable to the left-hand
side. For some reason that just feels right to me. So I'll go ahead and add 3X
to both sides of this equation.
Remember, whatever you do to one side you have to do to the other when you're
talking about an equation.
And that gives us 5X minus 23 is 7.
By the way, the property that tells us to do the same thing to both sides of the
equation is the equality property. So if you're not familiar with that, you can
go back in your textbook and read that property that gives us the ability to add
3X to both sides there.
Now I'm ready to collect my constants on the other side.
So I will add 23 to both sides to make 0 there and I'm left with 5X equals 30,
when I combined 7 and 23. Now I need to isolate the variable by dividing 5 but
whatever I do to one side I’ve got to do to the other.
5 divided by 5 is just 1, so on that left hand side I actually have1 being
multiplied with X which just leaves us with X, and 3 divided by 5 is 6.
So my potential solution right now is 6 but I can't say that for certain until I
actually check it. So you always want to use the original equation to check
2 times X minus 3 minus 17 equals 13 minus 3 times X plus 2 and wherever I have
an X I’m going to test my potential solution by substituting it in 4X.
So, I have 2 times, well 6 minus 3 is 3, and then I’m going to subtract 17 from
that value. On the right hand side, I have 13 minus 3 times 6 plus 2 which is 8.
And let me just remind you when you’re checking these equations, you never want
to go from side to side, you just want to simplify down both sides and then see
if they match up at the bottom to make that true statement. So that’s why I’m
not taking things from side to side here. I’m just simplifying down both sides
of the equation.
So I have 6 minus 17 equals 13 minus 24 and 6 minus 17 is negative 11 and 13
minus 24 is also negative 11. So there’s the true statement we’re looking for
which allows us to say that 6 is the solution to this equation.
In other words, when I plug 6 in it makes the statement true.
So I hope that this example has been helpful for you in understanding that
distributing in an equation like this is simply part of this first step of
simplifying both sides of the equation before you begin moving terms from side
to side, collecting variables on one side and constants on the other.
If you have any other questions about this type of equation, please make sure to
contact your Highlands instructor.
Thank you.
