Applied Linear Equations: Consecutive Numbers - Problem 1
So, one of the tricks of consecutive number problem, is they don't always have to be directly next to each other. So sometimes you'll have consecutive numbers 3, 4, 5, 6, sometimes you have a little bit more restrictions on what actually is going on.
For this particular problem, that we're going to look at, now we're actually dealing with two consecutive odd integers, so by that 3, 5, 11, 13 something like that, where you're actually skipping a little bit in between.
So, looking at this problem, if the smaller of two consecutive odd integers is doubled, the result is nine more than the larger of the two numbers. Find the two integers. So, let's take a look at this, so as I was saying, we're just dealing with two consecutive odd integers, so, basically example 3, 5, 11, 13 something like that. The trick is, from going from one to the other and so, instead of just adding one to get to the next number, we're actually adding 2 to go to the following odd number, so here we're actually dealing with x and x plus 2.
By this language we don't actually know if they're even or odd because you could be going from 2 to 4 or from 1 to 3, but hopefully when we actually work it out we end up with a odd number, if we end up with an even number, we did something terribly wrong.
So for this problem the smaller of two consecutive odd numbers is doubled, you double the smaller one x is the smaller one so this is just going to be 2x. The result we said the answer is nine more than the larger, so this number is going to be nine more than this so this is going to be equal x plus 2 plus 9.
So this number is nine more than this number, we could just as easily have subtracted 9 from this number, either way would have worked. From here pretty straight forward an equation 2x is equals to x plus 11, subtract the x over, x is equal to 11.
Again making sure we answer the question properly as for the two integers so if we found x is 1, this represented our smaller number so our larger number is 13, so we know that 11 and 13 are the answers to this equation.