Let's take a look at linear equations.
A linear equation in one variable is an equation that can be written in the form
a x + b equals 0,
where a and b are real numbers and a cannot equal zero.
What we want to notice about this equation is the variable x is raised the first power
and the other numbers a and b are real numbers. Remember a can't be zero.
Let's look at an example and determine whether each equation is linear or non-linear.
Let's look at this first example. We have 3 x minus 1/14 equals 1.
Well this is a linear equation. Notice the variable is x and it's raised to the first power.
The other numbers are real numbers, so this is a linear equation.
Let's look at another one. Let’s look at this example.
We see 3 over x plus 7 equals 8. This is not a linear equation
because the variable x is in the denominator.
Recall that you could write this as 3x to the negative 1 plus 7 equals 8.
We can see that this exponent is not a one. It's not a first-degree or linear equation.
So this one is not linear.
Let's go back over here. This one was linear.
Okay, let's look at a couple other examples. Let's look at this one.
While we see the variable in two terms, both of those variables x
are raised to the first power, so this is a linear equation.
The other numbers are real numbers, so this is a linear equation.
Let's look at the next one. We have 5 x squared plus 13 x equals 7. Well this one is not linear.
Notice that the variable here x is raised to the second power.
It is not a first degree equation, so this one is not linear.
Alright, let's look at this next one. We have 7 minus 0.5x equals 0.5 minus 7x.
Again we have variables in two terms, but both of those variables are raised
to the first power. It is a first degree. The other numbers are real numbers,
so this one is linear.
Now, let's look at this last one. We have 7 x minus 1 over x plus 15 equals 2.
Notice we have variables in the denominator, so this is not a linear equation.
Let's take a look at solving a linear equation with integer coefficients.
Let's look at this example.
We want to solve this equation. Negative 2, parenthesis 6 minus x, plus 6 equals 6 minus 4, times x plus 2.
The first thing we need to do is distribute, so we are going to distribute on the left.
Negative 2 times 6, negative 12. Negative 2 times negative x, plus 2x. Plus 6 again.
On the right we are going to distribute the negative 4, so negative 4x minus 8.
Now on each side we are going to add like terms. So on the left
we will add the negative 12 and the 6 to get negative 6.
On the right we're going to add the 6 and minus 8 to get negative 2.
Okay, so now I want to bring the x terms to whichever side. I can bring them to either side
but my preference is to bring the x terms to whichever side the coefficient is positive.
So I'm going to add 4x to both sides so that's going to give me 6x.
At the same time I'm going to add 6 to both sides, so that is going to be 6 minus 2 equals 4.
Then the last step we're going to divide both sides by 6.
Then of course we always reduce fractions. 4/6 is the same thing as 2/3.
So our solution to this linear equation is 2/3.
Let's take a look at solving linear equations with fractions.
These can be difficult for students, so let's go through a process we can use to solve them.
Instead of adding and subtracting fractions, what we can do
is we can multiply both sides of the equation by the same nonzero number
and that's going to get rid of fractions.
The number that we want to multiply both sides of the equation by
is the least common denominator. That is the smallest number
that each of the denominators go into.
So let's look at this example, we have 1/3 y minus 2 equals 3 minus 7/2 y.
Well we've got a denominator of 3 and a denominator of 2.
So, the least common denominator here is going to equal 6.
We want to multiply both sides of the equation by 6
and that's going to get rid of fractions. So let's see
over here on the left 6 times 1/3, 3 into 6 is 2, so we have 2 y.
Don't forget to multiply your constant term 6 times 2, is 12.
On the right hand side, 3 times 6 gives me 18
and then here when I multiply 7/2 y times 6, 2 goes into 6, 3, 3 times 7 is 21, so I have 21 y.
Now it's just like an equation with integer coefficients.
Again we want to get the y terms on one side. Doesn't matter which side.
I'm going to add 21y to both sides. That'll give me 23 y on the left
and then I'm going to add 12 to both sides. That’s going to give me 30.
Then the last step, divide both sides by 23.
So, I end up with y equal 30/23 is my final answer.
Let's take a look at this example. We want to solve the linear equation.
We see we've got fractions in it.
We find the least common denominator. We've got denominators of 6 and 12.
So, the least common denominator in this example is going to be 12.
Now we're going to multiply both sides of the equation by 12.
Okay, let's start with the left hand side. When we multiply the left hand side by 12,
6 goes into 12, 2. So I have 2 times 6 x minus 5 on the left.
Now on the right I am going to go ahead and distribute that 12.
So, we have 12 times 7/6 x minus 12 times x plus 5 over 12.
So, let's just go and distribute on the left. We have 12 x minus 10.
When we’re multiplying 12 times 7/6 x,
6 goes into 12, 2, 2 times 7 gives me 14 x.
Now when we multiply these two 12s are going to divide out,
but that negative gets distributed through the entire numerators so be careful there.
It's not this. It’s common a mistake you need that parentheses.
You need to go ahead and distribute the negative through the entire numerator.
Now let’s go ahead and distribute on the right. We have 14 x minus x minus 5.
Now we’ll add the x terms on the right, so we have 13 x minus 5.
Now we're going to subtract 12 x from both sides, so that’s going to give me x on the right.
Now we're going to add 5 to both sides, so that's going to give me negative 5.
My solution to this example is x equals negative 5.
Be careful, I want to warn you. Every time you have a negative in front of an expression,
you have to distribute that negative through the entire numerator.
Be careful in this type of example.
Before we look at rational equations, we need to look at two definitions.
The first is a polynomial expression. It's an algebraic expression
in which the exponents of all variable factors are non-negative integers.
That's another way of saying whole numbers, like 0, 1, 2, 3, 4.
Let's look at some examples of polynomial expressions.
This is a polynomial expression. Notice the variables are all raised to non-negative integers.
We have 3, 2, 1 and of course we have a constant term.
Here's another example of a polynomial expression.
3 x plus 4. The variable x is raised to the first power.
Here's another polynomial expression. 4 x squared minus 6 x plus 7.
Again, the variables are all raised to non-negative integers and of course we have a constant.
The expression 2/5 x is also a polynomial expression.
The variable x is also raised to the first power.
It is ok if you have fractions. Coefficients of variables can be any real number.
The last example 3
this constant is also a polynomial. You can actually think of 3
as 3 x to the 0, x to the 0 is 1, so this constant is also a polynomial expression.
Now let's look at what a rational expression is.
A rational expression is an algebraic expression that can be written in the form
P over Q, where P and Q are polynomials
such that the degree of Q is greater than or equal to one.
That's another way of saying Q can't be a constant.
Let's look at some examples of rational expressions.
Here is a rational expression. Notice the numerator and denominator are both polynomials
and this degree is of course one so that of course makes it degree one.
So it’s a rational expression.
The next example here is a rational expression. We’ve got a polynomial over a polynomial,
and the degree here is at least one. It’s actually 2.
This is another polynomial expression.
We've got a polynomial in the numerator and we've got a polynomial the denominator x,
which is degree one, so it is a rational expression.
Let me show you one that would not be rational.
If it had something like this
x cubed plus 5 x over 3.
So again notice that the denominator is a constant.
That's not degree one, so this would not be a rational expression.
Let's look at the definition of a rational equation.
A rational equation is an equation consisting of one or more rational expressions
with any other expressions of the equation being polynomials.
So, let's look at this example and determine which are rational equations.
Let’s look at the first one.
This one is a rational equation. This is a rational expression.
This is a rational expression, and here's my constant polynomial.
So, this one is a rational equation.
Alright, let's look at the next example.
So, in the next example I see some fractions,
but this one is actually not rational.
The variables here x are both raised to the first power.
I've got these denominators of 2 and 3,
but those are not polynomials of degree 1. They’re constants so this is not a rational equation,
but I want you to notice that it's actually a linear equation.
So notice that the variables x are raised to the first power, so this is not rational
because these denominators are not a degree one polynomial.
But remember it is actually linear. We saw those earlier.
Alright let’s look at some more examples. Let’s look at this example.
This one is a rational expression.
This is a rational expression.
In fact all three of these are rational expressions.
So, this one is a rational equation.
Let's look at the last one, we have x squared minus 2 x equals ½.
This is not rational. None of these are rational expressions.
This at the constant, and of course these are polynomials
So this is not a rational equation.
Alright, so it is important that you can determine which equations are rational
and which are not. Next let’s look at solving rational equations.
Let's look at solving this rational equation.
The process of solving the rational equation
is very similar to the process of solving a linear equation with fractions.
That is we get the least common denominator,
and we multiply both sides of the equation by the least common denominator.
However for rational equations you have to be aware of any restricted values.
Those are values that make the denominator zero. Remember division by zero is undefined.
So in this example, we have a denominator of x minus 7 twice.
So the restricted value here would be 7.
So x cannot be 7, that's my restricted value.
Now we're going to go ahead and find the least common denominator,
which of course is x minus 7. It's the only denominator there.
We're going to multiply both sides of the equation by x minus 7.
Let's see what we get.
Alright, so on the left hand side when we multiply the first term by x minus 7,
we're going to get an x plus 8.
Then we've got to multiple the x minus 7 times 1. We get x minus 7.
On the right hand side the x minus 7 divided out and we are left with 15.
So, again multiplying by the least common denominator gets rid of the fraction.
So, we're going to go ahead and solve this equation. Combine like terms.
2 x plus 1 on the left equals 15. Subtract the 1 from both sides.
I noticed that we get x equals 7, but remember we have to check
because this is actually a restricted value.
Whenever we get a solution to a rational equation and it's a restricted value,
we call this an extraneous solution.
So, this solution solves linear equation but it does not solve the original equation.
So in this case, x cannot equal 7, so I actually have no solution here.
So what's important whenever you're solving a rational equation,
remember to check. Always check your solutions to rational equations
to be sure they are not a restricted value or extraneous solution.
It is important that you remember that you can identify if it’s a rational equation
and then you remember to check it.
Let’s look at another example.
Alright, let’s take a look at solving this rational equation.
We've got three rational expressions. Note that this denominator is not factored.
Our first step is we need to factor this denominator. Let's do that.
Alright, let's see we need to get 3 w squared, so I need a 3 w and a w. You need to get a plus 20.
I'm going to go ahead and try the factors I see here to see if that works
because I know that's going to give me positive 20.
Notice the outer is 15 and negative 4 gives me the negative 19.
This is how this particular denominator factors.
Now we can find the restricted value.
Remember those are the values that make the denominator zero.
So w minus 5 can't equal zero, so that’s w can’t be 5.
3 w minus 4 can't be zero, so w can't be 4/3. These are the restricted values.
We have to remember to check, and make sure they're not either one of those at the end.
Now that we've got this denominator factored,
we need to go ahead and determine what the least common denominator is.
It's going to be the product of both. It's going to be w minus 5 and 3 w minus 4.
So, we're going to multiply both sides of the equation by the least common denominator.
We're going to have w minus 5, times 3 w minus 4, times w over w minus 5.
Then we’ve got the right hand side. We’ve got to multiply by w minus 5 and 3 w minus 4.
Alright, let's go look on the left hand side. When we multiply these two together,
the w minus 5s are going to divide out. You're going to have w times 3 w minus 4.
When you multiply this term times the second term, on the left hand side,
the 3 w minus 4s are going to divide out. You’re going to have 3 w times w minus 5.
On the right hand side, both of the factors are going to divide out.
That just leaves you with the numerator, w minus 5.
Remember when you multiply by the least common denominator, no more fractions.
Now we’ll go ahead and distribute, so we get 3 w squared minus 4 w minus 3 w squared plus 15 w.
Notice the 3 w squareds subtract so then you end up with negative 4 w plus 15 w, 11 w.
Now we'll go ahead and subtract the w from both sides.
Finally divide both sides by 10. Remember you always reduce your fractions.
Remember it’s a rational equation, so I have to check my solution.
We did not get one of those restricted values.
This value does not make the denominator zero,
so it is a valid solution. It's not an extraneous solution.
You need to practice working all the examples here
because throughout this course you’re going to solve these types of equations
over and over. It's very important that you can solve them on your own.