In Exercises 1–34, solve each rational equation. If an equation h...

Question

In Exercises 1–34, solve each rational equation. If an equation has no solution, so state.3y/(y²+5y+6) + 2/(y²+y−2) = 5y/(y²+2y−3)

Accepted Answer

Hello, today, we're going to be finding the solution to the following rational equation. So what we are given is six M over the quantity M squared plus nine M plus 20 plus three over M squared plus M minus 12, equal to four M over M squared plus two M minus 15. Now, there's a lot to keep track of in this problem. However, before we approach this problem, we always want to make sure that our denominators are in its simplest form. Currently, all of the given denominators are in the form of factor trinomial. So let's go ahead and take a look at the first denominator. We have M squared plus nine M plus 20. And this is going to factor to give us the quantities M plus four times M plus five. In the second denominator, we have M squared plus M minus 12. And this is going to factor to give us M plus four times M minus three. And for the final denominator, we have M squared plus two M minus 15. This is going to factor to give us the quantity M plus five times M minus three. So we can go ahead and replace all the current denominators with its new factored forms. And now what we want to go ahead and do is we want to go ahead and figure out what is the lowest common denominator between the three rational expressions in order to simplify the equation. Well, in the first fraction, we have M plus four times M plus five, in the second fraction, we have M plus four times M minus three. And in the third fraction, we have M plus five times M minus three. What we want is a combination of the different multiples from the three different denominators. And that means that the lowest common denominator for the three, given rational functions is going to be M plus four times M plus five times M minus three, which again is a product of the different combinations of factors in each of the denominators. So what we want to go ahead and do is we want to multiply each of these quantities by the lowest common denominator in order to eliminate the denominators. So just to write that out, we're going to take six M over M plus four times M plus five. And we're going to multiply that by the quantity M plus four times M plus five times M minus three. And we're going to do the same thing to the other quantities. We're multiplying the second fraction by the L CD, which is again, is going to be M plus four times M plus five times M minus three. Then this is going to be equal to four M over M plus five times M minus three, multiplied by the L CD, which is going to be M plus four times M plus five times M minus three. And so what's going to happen is that in the first fraction, when we multiply the quantity by the L CD, we can go ahead and cancel out the factors of M plus five and M plus four from the denominator. And the same rule applies to the other two fractions M plus four times M minus three is going to cancel out from the denominator and M plus five times M minus three is going to cancel out from the denominator. And so what we are left with is the sum of products. We are left with six M times the quantity M minus three plus three times the quantity M plus five equal to four M times the quantity M plus four. Now, we want to go ahead and simplify the given equation. We can do that by first distributing six M into the quantity M minus three and we can distribute the quantity three into M plus five. We can also distribute four M into the quantity M plus four. And doing so is going to give us six M squared minus 18 M plus three M plus 15, equal to four M squared plus 16 M. Now, what we want to go ahead and do is we want to go ahead and combine like terms on the left hand side of the equation. So the combinable light terms here are going to be negative 18 M and three M. And that is going to leave us with six M squared minus M plus 15, equal to four M squared plus 16 M. Now, in order to solve for M, we need to set the equation equal to zero. And in order to do that, we're going to have to take each of these terms and move them to the left hand side of the equation. In order to do that, we can subtract four M squared from both sides of the equation. And we can subtract 16 M from both sides of the equation. And that is going to leave us with the equation two M squared minus 31 M plus 15 is equal to zero. And the left hand side is in the form of a factor trinomial through the trial and error method, we can factor this to give this the quantities to M minus one multiplied by M minus 15. And that is going to equal to zero. Now, finally, what we have here are two factors of M. And using our zero property, we can go ahead and solve for M by sending both of the factors equal to zero. Doing so is going to give us two individual equations. We're going to have two M minus one equal to zero. And we're going to have M minus 15 equal to zero. On the right hand side, we're going to add 15 to both sides of this equation. And that is going to give us M is equal to positive 15, which is going to be one of our solutions. And on the left hand side, if we add one to both sides of this equation, we get two M is equal to one. And finally, if we divide by two, we either we are left with the quantity M is equal to 1/2, which is going to be the second solution to our given equation. So the solutions for M that we are left with is M is equal to 15 and M is equal to one half. With that being said, the answer to this problem is going to be C. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

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In Exercises 1–34, solve each rational equation. If an equation has no solution, so state.3y/(y²+5y+6) + 2/(y²+y−2) = 5y/(y²+2y−3)

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