Hi, I am Dr. Shah, I was the National Lecture Competition winner in 1989, and I am the Maths master at Mathscool. Now, ready for a new way in maths? Again, we got rational equation, we got fractions, and again we want to put brackets around any sums, any plus or minus signs inside the fractions. Multiplying by the denominators, to get rid of the fractions, so, my first denominator is this divided by x, so times x. So, that divided by x, and times x, cancel each other out, here I am just left with 2. When you are multiplying a fraction, you are actually only multiplying the top of the fraction, so on the top of the fraction, I get x over (x+2) and here again, I am multiplying the top by x and so here, x on top of the fraction. You can see I haven't got rid of all the fractions here, so I am going to have it multiplied again with this denominator, trying to get rid of that fraction, so times (x+2). Here, the divided by (x+2) and the times (x+2) cancel each other out, so I am left with -x there and here again, when you are multiplying a fraction, you are multiplying the top of the fraction, again x(x+2) over 3. I still have not got rid of all the fractions, I have still got a divide by 3 here, so I have to multiply every term by 3, so times 3 here, times 3 here and times 3 here. 2 times 3 would give me 6, 3 times -x gives me -3x, and here the divide by 3 and times 3 cancel each other out, giving me x(x+2). So I had to get rid of all of the fractions. A lot harder this time because I have lot of the fractions in the original equation, now I carry on to the next thing, multiplying out the brackets. So I just copy my equation that I reached so far up to the top here, and I am going to multiply out the brackets, so, that it gives me 6x, and 6 times 2 is 12, -3x here and I am multiplying these brackets here, which are going to give me x² + 2x. So I multiplied out the brackets and the brackets are gone. Now, I am going to simplify it. When you serve an equation, while you are working through it, simplify it along. So I have got 6x and -3x here, I'm going to simplify it, 6x-3x is 3x. Now, I am going to collect all the terms to one side. So I am going to add -3x on both sides, so that 3x disappears there, and here I am just left with 12, I have got 2x-3x which is going to give me -x, and I am going to take all the terms on to one side, so I am going to add -12 on both sides. I just wipe these out to make it here more space, and the 12 and the -12 cancel out, and so I am left here with 0, then here x²-x-12. And again I have got here a quadratic, so to solve that quadratic, I am going to factorize it, so I am going have an x at the front in each of these factors, to give me x², to get -12, may be I could try -4 times 3, say -4 times +3 would give me -12, and let us check whether it works. , x² +3x-4x, 3x minus 4x is -x, which is correct, and then -12, and that is fine. When you have quadratic equal to zero, and factorize, you know either the first factor has to be zero, or the second factor has to be zero. So if x-4 = 0, x would have to be 4, and if x+3 = 0, then x would have to be -3. So those are my answers, and that is how a rational equation has to be solved. .

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