TenMarks teaches you how to solve fractional equations and write the solution in simplest form
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How to Solve Fractional Equations In this video lesson, let’s learn about solving fractional equations. We’re given two different equations to solve. Let’s do the first one first. The equation is x+1 3/4=3 1/4. We know this is fractional equation because it has at least one fraction. This one actually has two but a fractional equation has at least one fraction. In order to solve this, the first thing we’ve got to do is take these mixed fractions and convert them to improper fractions which make it easy to work with. 1 ¾, we know we can convert it by taking the whole number which is one, multiplying it by the denominator which 4, adding the numerator which is 3 and having the whole thing be over the denominator which is 4. This becomes 4×1=4+3=7/4. Similarly 3 ¼, I take the whole number which is 3, multiply it by the denominator which 4, add the numerator which is 12+1=13/4. What we’re given, I can write the equation as x+ instead of 1 ¾ we can write 7/4=13/4. Now that we have this equation, we can see this is an addition equation. On the left hand side, I've got 7/4 and what I need to do is make sure that on the left hand side, I’m only left with a variable which is x. The way to do this is to subtract 7/4 from both sides. Let’s subtract 7/4 from the left hand side and from the right hand side. If we subtract 7/4 from the left hand side and the right hand side, on the left hand side we’ll be left with x+7/4-7/4, so the left side is left with x. The right hand side is 13/4-7/4 which I know how to subtract because these are like fractions. Since both the denominators are the same, I simply have to subtract 13-7 which is 6/4. So, x=6/4 is the answer to the question but since 6/4 is not in its simplest form, I can divide this by 2, both the numerator and the denominator. Two is the Greatest Common factor by the way, 6÷2=3. 4÷2=2, so x=3/2 is the answer I was looking for. That’s the solution to the equation. Let’s use the same principle to do the second one. The second question or the second problem gives us, x-6 2/3 = 11. Again step one; let’s convert this to an improper fraction. 6 2/3 becomes 6×3+2/3=11 or x-6×3 is 18+2, so 20/3=11. Since I have 20/3 on the left, I need to add 20/3 to both sides. When I add this, left side gets x, right side gets 11+20/3. Instead of 11, I can put 11/1.Since these are unlike fractions; I need to take an LCM. The LCM of 1 and 3 is 3. In this scenario, I have to multiply 11 with 3, so I get a 33 plus in this scenario, I copy 20, so x becomes 33+20/3 which is 53/3. Since this is still an improper fraction, I can convert this to a proper fraction or a mixed fraction if I wanted. 53/3 is the same as times 17 is 51 2/3. I can write the final answer as x= either 53/3 or x=17 2/3, both would be correct. Quickly recapping what we’ve learned. When we are looking at a fractional equation which means that there’s at least one fraction, the way we solve it is we take a mixed fraction, convert it to regular fractions or improper fractions. Once we have that and we have an equation like x+7/4 = 13/4, since there is an addition of 7/4 on the left hand side, I subtract both left and right by 7/4 which gives x= 6/4 which I can write in simplest form as x=3/2. Similarly, we did another problem where we converted the mixed fraction into a regular fraction which we got x-20/3=11. 11 is the same as 11/1 which means x=53/3 which when converted to mixed fraction gives us 17 2/3.
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