hello and welcome to esip Nayan for today's video we are going to discuss how to solve rational equations the learning objectives are the following first distinguish irrational equations then solve rational equations We Begin by defining what rational equation is rational equation is an equation that contains one or more rational expressions recall that irrational expression is simply a ratio of two polynomials some examples of rational equations are given here below the slide first we have 1 over x equals 1 over 5 minus X another example with b x plus 4 over x equals negative five and last example we have x squared over X plus one equal to 1 over X plus one so notice that in all of the examples given here there is a presence of rational expressions and that is how we Define rational equations and some examples moving on let's discuss how to solve rational equations before we proceed to actual examples let me share to you the three steps that we will be following in solving rational equations first we will be solving for the value of x then check if the solution is a true solution to the equation lastly decide which Solutions are considered now for Step 1 we will be utilizing our knowledge and skills in solving algebraic expressions to be able to solve the value of x given the rational equation for number two this is the checking part we will see if the value of x that we were able to obtain from Step 1 is a true solution to the equation why do we need to do this what is true solution does that mean there's a false solution the answer is the presence of the term extraneous solution so extreme solution is actually a false solution to the rational equation when does a solution becomes false it happens when the salt value of x does not satisfy the equation later on given the examples we will be explaining when the solution is true or false then lastly we will be deciding which Solutions are considered and we will only be considering true Solutions through Solutions on the other hand are the values of X that when checked satisfies the given equation let's now have the first example we have X over 5 plus 1 over 4 equals x over 2. the first step in solving rational equations is to solve for the value of x we will be using our algebraic computational skills to be able to isolate X and therefore solve for its value notice that on the left hand side of the equation we have here two fractions that we need to combine first before we can solve for the value of x now to be able to combine X over 5 plus 1 over 4 we need to write them as one fraction by getting their LCD the LCD of 5 and 4 is equal to 20. now we rewrite the left hand side of the equation in such a way that the LCD becomes 20 or both denominators rather becomes 20. so we have now 20 divided by 5 that is 4 times x so we are going to have here for X added to 20 divided by 4 is 5 times 1 we get the five so basically we can combine them as one fraction by writing it as 4X plus 5 we simply combine the numerator then we copy the denominator as one so the left hand side of the equation can also be written as 4X plus 5 over 20. now let us rewrite the equation this way the left hand side becomes now 4X plus 5 over 20 whereas the right hand side stays as is we will simply copy X over 2 and this is now the new equation we will work with moving on we'll now solve for the value of x so what we can do here is to cross multiply so we will multiply for X plus 5 by 2 so that becomes 2 times 4X plus 5 and it will be equated to the product of 20 times x so that is 20 x we need to further simplify so that we can isolate X so what are we going to do next we just need to distribute multiplying 2 to the binomial inside the parenthesis so that becomes 2 times 4X we have 8X plus 2 times 5 is 10. equated to the value of 20 x now since our plan is to isolate X we need to combine all x's on one side of the equation that leaves us 10 only on the left hand side of the equation equated to 20x from positive 8X it becomes negative 8X and so simplifying further we get 10 equals 12 x and to finally isolate X we divide both sides by 12. so this will be canceled and therefore the value of x will now become 10 over 12 or that is equal to 5 over 6 and this is now the value of x and we are done with step one now that we already have solved for the value of x which is 5 over 6 we shall now proceed to step 2 which is the checking part we will simply check if the solved solution of x is indeed a true solution to the equation so how do we do that we will simply substitute the value of x to any of this form of the equations and then we will see if the equation will be satisfied so basically we will simply let X be equal to 5 over 6 given this option or this one so the original form is X over 5 plus 1 over 4 equals x over 2 and the derivative form combining the fractions on on the left hand side we have 4X plus 5 over 20 equals x over 2. so you can choose any of the following when substituting now for this case I'll be using form number two so I'll let X be equal to 5 over 6 give event for X plus 5 over 20 equals x over 2 and see if the equation will be satisfied now for the checking part it's quite simple since we are just simply going to substitute the value so that becomes 4 multiplied to the value of x which is 5 over 6 added to 5 all over 20 equals on the right hand side of the equation we will still be substituting the value of x after this we are now going to simplify now we have 4 times 5 that gives us 20 added to 5 divided by 20 then on the right hand side of the equation we are dealing with a complex fraction so that becomes 5 over 6 multiplied to 1 over 2. so simplifying this part we have 20 over 6 plus 5 over 1 we can get the LCD the LCD is 6 6 divided by 6 is 1 multiplied to 20 that's 20. add that to 6 divided by 1 is 6 multiplied to 5 that gives us Thirty and this will still be all over 20 which is the denominator on the left hand side of the equation meanwhile on the right hand side we will simplify this factors we have 5 over 12. moving on we can further simplify the numerator on the left hand side of the equation by having 50 over 6 but it will still be divided to 20 and that is equal to 5 over 12. now we are checking if the left hand side is equal to the right hand side of the equation the left hand side can also be Rewritten as 50 over 6 multiplied to 1 over 20 and we are checking if it is equal to 5 over 12 so we can further simplify the left hand side by making this five and two and so that will give us a product of 5 over 12 which is basically equal to the right hand side of the equation which is also 5 over 12 and so that will lead us to the last step which is to decide if the solution can be considered as a true solution when do we consider a solution as a true solution it happens if when you are done checking the equation will be satisfied and in this case since the left hand side and the right hand side equation are the same then we can say that x equals 5 over 6 is indeed a true solution to the equation and we're done with example number one let's now have the second example y plus 3 all over over y minus 1 equals 4 over y minus 1. so at this point we are going to proceed immediately with the step one which is to solve for the value of x now unlike on the previous example the left hand side and the right hand side of the equation only contains one fraction so this is already good to go to simplify this easily what we can do is to multiply both sides of the equation to Y minus one notice that the denominators for both sides of the equations are the same and so what we can do is to multiply y minus 1 to both sides so that they can easily be canceled so that will give us y minus 1 multiplied to Y plus 3 over y minus 1 equal to 4 over y minus 1 multiplied to Y minus 1. after this the denominators will now be canceled then this is an easy part already what's left the are the following doing we have y plus 3 on the left hand side of the equation and that is equated to four to isolate why this becomes 4 3 is being transferred to the right hand side so that becomes negative three and so the value of y is equal to one notice that the variable used here is not anymore X but y but it can be any variable in general so this is now the value of y step two simply check if the solution is a true solution to the equation how do we do that again we just need to do the checking so let's check if indeed Y is a true Solution by substituting it to the original equation so we will let all y's be equal to one given the equation so that becomes 1 plus 3 all over 1 minus one we are checking if it will be equal to 4 over 1 minus 1. so the left hand side can be simplified as four over zero while the right hand side becomes four over zero notice that the left hand side and the right hand side are equal but there is a problem here any number that is having a denominator of zero will give us an undefined value so if the value is undefined definitely we cannot get any value in reality so although the left hand side and the right hand side are equal if it becomes undefined then this is already a false solution therefore y equals one is a false solution or what we call the extraneous solution so basically since this is an extraneous solution we do not have actually a solution for the rational equation