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Excluded Values For Equations

We've used the LCD to add and subtract fractions and even simplify complex fractions. Let's use this LCD one last time to solve equations like this one. This is an example of an equation with a rational expressions. These two fractions in particular are the rational expressions. We have a polynomial denominator here x. And a polynomial in the denominator here, 2 x. Before we get started solving this equation, let's think about what values of x would not make this equation true. If x equaled one of these numbers, it wouldn't make the equation true. Which one is it?

Excluded Values for Equations

Well, we know that x cannot equal 0. If x equaled 0, then this first fraction would have a denominator of 0, and the second fraction would also have a denominator of 0. We know denominators can never equal 0 for fractions, they would be undefined. So, even when we attempt to solve this equation, if we were to get an answer of x equals 0, we would have to exclude it from our solution set. We know x can never be equal to 0.

Finding the LCD to Solve Equations

We know the lowest common denominator is helpful whenever we want to solve this type of equation. So in order to solve this equation we're actually going to use the LCD. We're going to multiply every term in the equation by the lowest common denominator to simplify the equation. So let's start by figuring out the LCD. What is it for this equation?

Finding the LCD to Solve Equations

When finding the LCD we want to look at the denominators of all of our fractions. So that would be x, 2 and 2x. Since x is a factor of 2x and 2 is a factor of 2x we really just want this as our LCD. We want to take the product of all of the different factors And their highest powers of any denominator. Great thinking if you got 2x.

Solving Equations Using the LCD

So if we did multiply each of these terms by 2 x what new equation would we get here? Keep in mind that when we multiply through by the LCD we're really going to clear each of these fractions. That's the goal. We want to get something that's more simple to work with.

Solving Equations Using the LCD

If we multiply each term by 2x, we'll get 4 minus 3x equals 7. Nice algebraic thinking if you got that one correct. To get the answer, we simply simplify the factors in the numerator and denominator. For this first term, the x's would cancel to equal 1. This would leave us with 2 times 2, which equals 4. For the second term, the factors of 2 would cancel it to leave us with negative 3x, here. And for our last fraction or term, we would simply have 7.

Check Against Excluded Values

Since this equation is linear, we can use our skills from before. We solve the equation by grouping the variable terms on one side, and the constant terms on the other. We subtract 4 from both sides first to get negative 3x equals positive 3. Then we divide both sides by negative 3 to get x is equal to negative 1. This is our answer, and for these types of problems we're going to write the answer as a solution set. So this would be the solution set to our original equation. Once we find our solution, we want to make sure we always check it against the excluded values. For this equation, we knew x cannot equal our other equations, we can check to make sure that we're right. We plug in a value of negative 1 for x here and here. When I simplify the equation, I'll have negative 3 and a half on the left and negative 3 and a half on the right. So yes, this is correct.

Multiplying Equations by the LCD

Let's try another equation containing rational expressions. What equation would we get if we multiplied this through by the LCD? Write your answer in this box.

Multiplying Equations by the LCD

We would get 5x plus 6 equals 6x squared. Great algebra work for getting that one correct. The LCD, or the lowest common denominator, is x squared here. We only need to look at these two denominators, and since x is already a factor of x squared, we use this as the LCD. Multiplying each term by the LCD, we would get 5x for the first term, 6 for the second term and 6x squared for this term. We get 5x here since we will have one more x in the numerator than we would in the denominator. This one power of x cancels leaving us with 5 times x, or 5x. Here, the x squared's reduced to 1, so we're left with positive 6. And here, we just multiply these two expressions together to get 6x squared.

Factoring Quadratics for the Solution Set

So we've changed this equation that had rational expressions to an equation that contains a quadratic, x squared. Since this equation is quadratic, we want to bring all the terms to one side of the equation, factor the expression, and then set each factor equal to zero. So I want you to try and rearrange this equation, so that way you can factor it. Then solve it just like you've done before. When you think you've the values of x, enter them here as a set, and be sure to separate the values by a comma.

Factoring Quadratics for the Solution Set

The correct answers were negative 2 3rds and positive 3 halves. Fantastic work if you found these two answers. Factoring is one of those important skills that will come up again and again, so let's make sure that we know how to do this. I'm going to subtract 5x from both sides and 6 from both sides. I perform this same operation on both sides and I'm just showing these two steps I want. This leaves me with 0 on the left and 6x squared minus 5x minus 6 on the right. I also know I want to subtract on both sides here since I want the x squared term to have a positive coefficient. It's generally easiest to factor when this term is positive. So, I've taken my equation and I've rewritten it up here to factor this part of the equation we need to find the factors of negative 36 that sum to negative 5. These numbers are negative 9 and positive 4, so we can rewrite our middle term and factor it by grouping. This gives us factors of 3x plus 2 and 2x minus 3. Now that the right-hand side of our equation is factored, we can use the zero product property and set each factor equal to 0. Solving for x in this equation will get x is equal to negative 2 3rds. And summing for x in this equation will get x is equal to 3 halves. The last thing we want to do for our solution set is to make sure we check the excluded values. We know for these denominators, x cannot equal 0 or else we would have fractions that would be undefined. Since both of these values are not 0, we can rest assured that this is our correct answer.

Solving Equations with Rational Expressions Practice 1

Try solving this equation. Multiply through by the LCD, and then see if you need to solve a linear equation or a quadratic equation by factoring. Write your answer here as a set. Keep in mind that not every equation will have solutions. If ever we solve our equation and we get an exclued value, then we need to remove it from our solution set. So if there isn't a solution I want you to type in ns for no solution. Good luck on this one.

Solving Equations with Rational Expressions Practice 1

Negative 3 is correct, nice equation solving if you got that one correct. The lowest common denominator for our fractions is 9x. We know x is already a factor of 3x, so really we just need to find the LCD for 9 and 3x. When we use our factor tree method we can see that we get a final product of 9x. And our first fraction, the x terms cancel to 1, leaving us with 9 times 1 or 9. For our 2nd fraction the factors of 9 canceled the 1 leaving us with -1 times x, or -x. And finally for our last fraction 9 x divided by 3 x equals 3. So we're left with 4 times 3 which equals 12. Solving this linear equation will get a access equal to -3. Remember, we just reverse the signs here, since negative x equals 3, so positive x would equal negative 3. For this original equation, x cannot equal 0. It's an excluded value. And since our answer was x equals negative 3, we're okay. We can keep this as our solution set.

Solving Equations with Rational Expressions Practice 2

Try solving this equation. Write your answers as set here and remember to type ns if there's no solution. Also, keep in mind that the excluded value here is not x equals 0. There are other excluded values for these fractions. What do you think they are?

Solving Equations with Rational Expressions Practice 2

Well, it turns out there's no solution for this equation, let's see why. We know each of these denominators cannot equal zero, so x cannot equal positive 3 and x cannot equal negative 3. Remember, we don't want these fractions to be undefined. This number would make this denominator equal to zero. And this number would also make this denominator equal zero as well. This last denominator as factors of x plus 3 and x minus 3. So really we've already have take in account of numbers that x can not be for it since we solved these two equations already. If we factor this last denominator we can see that this LCD would be x minus 3 times x plus 3. We know this since this is a factor of this denominator. And this is also a factor of this denominator. Next, we multiply each term by the LCD. We cancel the factors that appear in the numerators and in the denominators. We're left with 2 times x plus 3 for the first term. And negative 3 times x minus 3 for the second term. Our last term is just negative familiar with. We simply want to distribute this 2 and this negative 3 first and then combine our like terms and solve for x. Positive 2 x and negative 3 x equals negative x. And positive 6 and positive 9 equals positive 15. Make sure you're careful on this distribution. We have negative 3 times negative 3, which equals positive 9. It's really easy to make that sort of mistake. I'm going to rewrite this equation up here. Then we subtract 15 from both sides to get negative x is equal to negative 3. If we multiply through by a negative 1, positive x would equal positive 3. But we want to be careful here. We know x equals 3 is an excluded value. It makes this denominator and this denominator equal to 0. So these fractions would be undefined. Since x equals 3 as an excluded value up here, we know that there's no solution. There's no real number that exists that can satisfy this equation.

Solving Equations with Rational Expressions Practice 3

Try finding the solution set for this equation. And remember, if there is no solution type ns here.

Solving Equations with Rational Expressions Practice 3

want to start out by finding the LCD for these three fractions. Since this is a quadratic trinomial, we should factor it first to determine the LCD. This denominator factors to x minus 3 times x minus 4. Since we have an equation, we can multiply each term by the same factor. This keeps our equation balanced. When we multiply through by the LCD, we'll get 1 times x minus 4, minus 4 times x minus 3, which equals x. We distribute the negative 4 to get negative 4 x and positive 12. Next we combine all like terms to get negative 3 x plus 8 equals x. To solve this part we want to get our variables on one side of the equation and our constants on the other. This is why we add positive 3 x to each side. And finally dividing both sides by 4 we'll get x is equal to 2. Our solution set. We know x cannot equal positive 3 and x cannot equal positive 4. These are excluded values. We check our solution set against these to be sure this is correct. Otherwise it, we would have no solution.

Solving Equations with Rational Expressions Practice 4

Try solving this fourth equation. Write your answer as a set here, and remember to type NS if you think there's no solution. Good luck.

Solving Equations with Rational Expressions Practice 4

Now the lowest common denominator might of been tricky here. You want to make sure that it's n times 6 minus 3n. This is one factor in these denominators and n is a completely different factor. So we want to multiply those two together to get our LCD. For the first term we'll have 5 times 6 minus 3 n. And for the second term we'll have plus 4 times n, or positive 4 n. The right side of our equation is simply 2 n squared. Since these factors reduce to one. If we distribute positive 5, and combine our like terms, we'll get 30 minus 11 n equals 2 n squared. Now that we're at this part we have a quadratic equation, since we have a 2 N squared. We want to set this equation equal to zero by moving or rearranging these two terms. If we move these terms to the right side of the equation we'll get zero is equal to 2 N squared, plus 11 minus 30. This is something that we can factor and solve. So here is our equation and, to factor it, we want to find the factors of negative 60 that's sum to positive 11. When we use factoring by grouping, we'll get 2n plus 15 as 1 factor. And n minus remaining equations. N could equal negative 15/2, or n could equal positive 2. And here's where we want to be careful. We need to check for the excluded values for the original equation. We know n cannot equal zero, since this denominator would equal zero. And the fraction would be undefined. We also know that n cannot equal 2. If n was equal to 2 both of these denominators would be zero and these two fractions would be undefined. This means that this solution is not valid. We can only take this as our actual solution. So here's our solution set, negative 15 halves. Fantastic algebraic thinking for getting that one correct.

Solving Equations with Rational Expressions Practice 5

Here's another rational expression. This time the right hand side is set equal to 0. What do you think would be the solution set for x? Write your answer here as a set, and type ns if you think there's no solution.

Solving Equations with Rational Expressions Practice 5

Here are the correct answers negative 1. Great work if found that one. We start solving this equation by multiplying both sides by the LCD this leaves us with 2 times x minus 8 minus 6 times 2x minus 1 on the left hand side of the equation. On the right hand side we'll still have 0 since 0 times any factor would still be 0. Next we distribute and combine our like terms to get negative 10x. Minus both sides, and then we divide both sides by negative 10. This gives us our final value of negative 1. For the excluded values, we want to look at these 2 denominators to figure out what x cannot equal. We know x cannot equivalent half or this denominator here equals 0. Likewise we know x cannot equal positive 8 or this denominator would be 0 as well. Our final answer is not one of these excluded values. So here is our only solution, negative 1. Notice that there is another way to solve this equation. If we move this second fraction to the right side of the equation we would have one fraction equal to another. We would just add this to both sides of our equation. Next we would multiply by the LCD or simply cross-multiply to get this next equation. After distributing and solving for x, we would find x would be equal to negative 1. If you use this method, that's great.

Solving Equations with Rational Expressions Practice 6

Here's the last equation we'll try together. What do you think the solution set is for it? Type NS if you think there's no solution, and good luck.

Solving Equations with Rational Expressions Practice 6

Negative 7 5ths and 2 are the correct answers. Amazing work. You got this one correct. When we start solving this problem, we want them multiplied by the LCD which is x times x plus 1. It's the product of these two denominators. We cancel the factors that reduce to 1 and then we distribute the remaining factors here on x terms. So we'll have 5x times x plus 1 Plus 6 times x, which equals 14 times x plus 1. Next, we distribute the 5x and the 14 to get our new equation. I have a quadratic term here, so I know I need to factor. I'm going to move these terms to the left side, so that way I can set this equation, or this left hand side of my equation, equal to 0. We'll combine like terms first by adding 5x and equation. The 14 doesn't have a like term on this side of the equation, so I put it last. This gives us five x squared minus three x minus 14 which equals zero. Next we want to factor this quadratic by find the factors of negative 70 That sum to negative 3. These factors are negative 10 and positive 7. We use these 2 factors to rewrite our middle term, and factor by grouping. So this quadratic can be factored as 5x plus 7 times x minus 2. We want to make sure that we set it equal to zero. So we can set each factor equal to zero. And solve for x. We set this first factor equal to 0. And the second factor equal to 0. Doing so we can solve and get x equal to negative 7 5ths, and x is equal to positive 2. But we're not done yet. We want to make sure we check the excluded values to see if either of these are not part of the solution. Well, the only excluded values are x equals 0 and x equals negative 1. If x were 0 this fraction would be undefined. And if x is equal to negative one, this fraction would be undefined. So our two values of negative 7 fifths and positive 2 are correct.