What are all the values of 𝑥 for which it is true that 𝑥 plus three over 𝑥 minus one is greater than or equal to three?
We write down the inequality again. And we can note that if we multiply both sides by 𝑥 minus one, then we’ll get a linear inequality. We’ll have 𝑥 plus three on the left-hand side and three times 𝑥 minus one on the right-hand side. But what inequality sign will we have between them?
Remember, if we multiply both sides of an inequality by a positive quantity, we keep the same sign. But if we multiplied by a negative quantity, then we have to flip the sign, reversing its direction. So the question is, is the quantity that we’re multiplying by 𝑥 minus one positive or negative?
Well, of course this depends on the value of 𝑥. If 𝑥 is greater than one, then the quantity that we’re multiplying by is positive. And so we’ll have the same sign in the inequality, the greater than or equal to sign. However, if 𝑥 is less than one, then the quantity that we’re multiplying by is negative. And so upon the multiplying by 𝑥 minus one, the sign of the inequality will flip from a greater than or equal to to a less than or equal to.
There is of course another possibility. 𝑥 could be equal to one. But when 𝑥 is equal to one, then the left-hand side of the original inequality is undefined because we’re dividing by 𝑥 minus one, which would be zero. And so clearly 𝑥 is equal to one is not a value of 𝑥 for which the inequality holds.
We have two cases remaining: the case when 𝑥 is greater than one and the case when 𝑥 is less than one. Both of these cases have linear inequalities, which we can solve in the normal way. Starting with the case 𝑥 is greater than one, we distribute on the right-hand side, subtract 𝑥 from both sides, add three to both sides, and divide by two, noting that as two is positive, we don’t have to flip the sign of the inequality.
And writing 𝑥 on the left-hand side as is convention, three is greater than or equal to 𝑥 becomes 𝑥 is less than or equal to three. We might be tempted to think that this is our answer. But we should remember that we made the assumption that 𝑥 is greater than one to solve this inequality.
Combining these two conditions, we get that 𝑥 is between one and three, where the end point one is not included because we have a less than sign. But the end point three is included. We now move to the case 𝑥 is less than one. And it’s exactly the same process to solve this. We distribute, subtract 𝑥 on both sides, add three to both sides, divide both sides by two without flipping the sign because two is positive, and finally rewriting three is less than or equal to 𝑥 in a more conventional way as 𝑥 is greater than or equal to three.
Again we need to combine these two constraints that we have on 𝑥. 𝑥 must be less than one and also greater than or equal to three. There are no real numbers which satisfy both constraints simultaneously. A number can’t be simultaneously less than one and also greater than or equal to three. And so we get no solutions from this case.
The values of 𝑥 for which it is true that 𝑥 plus three over 𝑥 minus one is greater than or equal to three are the values of 𝑥 which are greater than one and less than or equal to three. Had we not been careful about whether the quantity we’re multiplying by was positive or negative, we might have naively just kept the same sign rather than dividing it into cases.
In that case, we would have got the incorrect answer 𝑥 is less than or equal to three because we hadn’t realized that we’d make the additional assumption that 𝑥 was greater than one. And it’s easy to see that 𝑥 is less than or equal to three can’t be the right answer because zero is less than or equal to three. But zero does not satisfy our initial inequality.
Taking the left-hand side of the inequality and replacing 𝑥 by zero, we get zero plus three over zero minus one, which is negative three, which is not greater than or equal to the right-hand side of the inequality, three. But you can check therefore what we’re saying is the right answer. One is less than 𝑥 is less than or equal to three. Values inside this interval satisfy the original inequality. And equally importantly, values outside this interval do not satisfy the original inequality.