I want to work a variety of problems here that involve similar expressions,
but each problem has a different instruction. Let's take a look at our first one.
Our first problem here we want to solve the equation 8x - 5 = 0.
Okay, no problem. I'll add 5 to each side using the Addition Property of Equality,
8x = 5,
then I'll divide both sides by 8, or multiply both sides by 1/8,
either way I look at, I end up with x =
5/8.
So, there's a linear equation with one variable, and I solved that equation
to get my solution.
Next problem: I want to add (8x - 5) and (2x - 3). So this isn't an equation,
this is just two expressions, and I want to find their sum.
Alright, so
8x + 2x = 10x,
(-5) + (-3)
= -8.
So there is the sum of these two expressions right here.
Next problem,
solve 8x - 5 = 2x - 3.
Alright, so I have a linear equation and one variable, I'll collect all the
variable terms on one side, all the constant terms on the other.
I use my Addition Property of Equality and my Multiplication Property
of Equality.
Adding -2x to both sides, I get 6x
on the left side,
and I'll add positive 5 to both sides, and get 2 on the right side.
Multiply both sides by 1/6,
and end up with x = 2/6
which is
1/3.
So there's my solution
to that equation.
Next, let's multiply (8x - 5) into (2x - 3). Okay, so here I have
the product now of these two expressions.
So, I'll multiply (8x)(2x) = 16x^2,
I'll multiply the last two terms, (-5)(-3)
= 15, that's the easy part.
I multiply the two inside terms (-5)(2x) = -10x,
multiply the two outside terms, (8x)(-3) = -24x.
-10x + -24x = -34x.
So there's the product of
these two binomial is right here.
So, I'm paying attention to the instructions with these problems. I don't know
what to do with these expressions,
unless I'm told
by these instructions right here.
So, last problem: solve 16x^2 - 34x + 15 = 0.
Okay, it's quadratic with x^2,
so I'm going to try to solve it by factoring. Does it factor? Well,
sure it does, because look at these two binomial I multiplied right here. They multiplied
to give me 16x^2 - 34x +15, that means
that this expression factors into this product.
So I'll use my results from part d. here to just write this as:
(8x
(8x - 5
(8x - 5)(2x
(8x - 5)(2x - 3)
(8x - 5)(2x - 3) = 0.
Now I'll use my Multiplication Property of Equality, oh I take that back,
I'll use my Zero Factor Property to set the each of these factors equal to 0,
because the only way I can multiply two expressions and get 0 is if one or
the other, or both, are 0.
In any of those cases, I get a solution to my equation.
So, 8x
8x - 5
8x - 5 = 0, or
2x - 3
2x - 3 = 0
8x - 5 = 0, well I solved that right here, that's 5/8
So let's put that over here:
x = 5/8, or
2x - 3 = 0, I'll add 3 to both sides, divide by 2 or
multiply by 1/2, either way,
and I get x =
x = 3/2. There's my other solution.
So, two solutions to this quadratic equation right here.
So, the point of these problems is this:
you want to get used to paying attention to the instructions that go with the problem.
I see a lot of people in my classes that
end up being able to solve equations like this like this very easily on their homework,
because on their homework, all the problems are like this, they're all "solve."
They can do problems like this because we've been doing them on their homework
they're all "add this and this."
But when they get on a test, and the problems are all mixed up and have
different instructions, sometimes they get confused.
So, problems like this are really good practice at paying attention to the
instructions,
and looking at the kind of problem that goes with those instructions,
so when you're on the test or a quiz your confusion doesn't happen there.