# Sample Math eQuestions

**Example 1: **

**Question:**I am preparing for my exam, and I’m having a hard time solving equations. I’m working on this equation: 2(x-3)=5x. Here’s my solution:

I got x=-1, but the answer key says the correct answer is -2. Can you help me figure out what the problem is?

**Response:**I am happy to help you with this question. To solve this equation, you need to expand the left hand side using distributive law. You made a good start by multiplying the x by 2. But you have made a small mistake here: distributive law says that you need to multiply*both*of the numbers inside the brackets by the number outside the brackets. You need to multiply 2 by both x and -3. You’ve done the next steps correctly: you have brought similar terms together and then divided by the coefficient of x to get your answer. If you remember to multiply 2 by x and by -3, you will get -2 as the answer.

Also, check the following website https://www.chem.tamu.edu/class/fyp/mathrev/mr-algeb.html. It has a couple of examples of how distributive law works, and then it has a couple of practice questions down at the bottom. Try them out!

**Example 2: **

**Question:**I have my midterm for Math 100 on Thursday, and I’m stuck on a limit question. This is the limit I want to calculate: lim(x^2-1)/|x-1| when x approaches 1. My answer is 2, but, in fact, this problem does not have a limit.**Response:**Thank you for submitting this question! I think the tricky part of this limit is that you have to deal with the absolute value: |x-1|. Remember that the absolute value of x is the distance of x from zero, so x could be a positive number or a negative number. For example, 3 and -3 are both three units from 0, so they both have the absolute value of 3. With |x|, we do not know whether it is x units to the left of 0 or x units to the right of zero, so we have to consider both possibilities. In other words, we have to think about what happens when we approach 0 from the left-hand side and from the right-hand side of the number line.In your specific question, the absolute value is |x-1|, so you need to take the limit when x approaches 1 from the left-hand side and when x approaches 1 from the right-hand side. From the right-hand side, x-1 is positive, so you can simply remove the absolute value bars: (x-1). However, when x approaches 1 from the left-hand side, x-1 is negative, so you need an extra negative sign to make it positive: -(x-1). Now you have the information you need to find the limit from both sides. By definition, if the limit from left-hand side and from the right-hand side are equal, then the limit exists. Otherwise the limit does not exist.

For some examples of limits with absolute values, check out http://math.stackexchange.com/questions/222927/absolute-value-and-limit-reasoning.

**Example 3: **

**Question:**I’m working on some questions about probability, and one question talks about how students did on a midterm exam. The scores were normally distributed, and the mean was 65%. The standard deviation was 11.5%. A student named Jim scored 72%, and I need to know what percentage of students got a score between the mean score and Jim’s score. I know I’m supposed to use the normal distribution table, and I think I have to use Jim’s score as the z-score, but I’m not sure how to convert it. Can you please explain?**Response:**Sure! I’d be glad to explain. Start by looking at this graph: https://www.mathsisfun.com/data/standard-normal-distribution-table.html. On this graph, zero represents the mean, which, in your example, is 65%. If we want to compare a particular student’s score to the rest of the population, we need to know how far that score was from the mean. We measure that difference in standard deviations, and we represent that difference on the x axis. In your example, a student who scored 76.5% would be 11.5% above the mean. That’s exactly the standard deviation of the distribution, so we would say that the student had a z-score of 1, and we would put that student on the graph at x=1. A student who scored 88% would have a z-score of 2 because the student is 23% (two times the standard deviation) above the mean.

But Jim’s z-score won’t be exactly 1 or 2: he’s in between 0 and 1 on the graph. So, first, you need to figure out how far his score is from the mean: just take his score and then subtract the mean. Then, you need to divide by the standard deviation, and you’ll end up with his z-score. We can represent that process with this formula:

Then you can compare Jim’s z-score to the normal distribution table. The link I’ve given you has a fun, easy-to-use graph at the top. Just click on the graph to set your z-score, and it will tell you the percentage of students in the area between the mean and Jim’s z-score. (If you have a problem where you need to know the percentage of students above Jim’s score or below Jim’s score, you can also click the button on the graph to say that you want the percentage that is less than z or greater than z.) You can also use the table down below the graph. For some examples of how to use this table, try reading the example questions below the table on the Math Is Fun site. I hope this response helps!