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