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# Tutor profile: Stacey W.

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Stacey W.
Middle & High School Mathematics Teacher & Tutor for 5 years
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## Questions

### Subject:Linear Algebra

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Question:

How do I graph a line in slope-intercept form that has a negative slope?

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Stacey W.

This can be tricky because we automatically think that up and right on a graph are positive directions and down and left are negative directions, which is accurate information. So when students go to graph negative slope, they think they should count down for the "rise" of the slope and count to the left for the "run" of the slope. When you do this, however, it creates a positive line. Let's look at why this happens. We know that slope is a ratio of the change in y to the change in x written as a fraction, and is more commonly known as rise over run. Because slope is written as a fraction, we must treat it as a fraction. If the slope is negative, which piece of the fraction becomes negative? The top? The bottom? Or both? To best understand this, we have to also remember something else about a fraction, which is that a fraction is the same as a division problem. If I know the slope is negative and I make the rise (numerator) negative and the run (denominator) negative, then I have a division problem with a negative divided by a negative which gives me a what? A positive, correct. But if my slope is supposed to be negative, then I cannot make both the top and bottom negative. What if I make the rise (numerator) negative? Then I have a negative divided by a positive, which equals a negative, so it works to make just the rise negative. What about making the run (denominator) negative? Then I have a positive divided by a negative, which gives a negative, so it also works to make just the run negative. So what we have determined is that we can make the rise negative OR the run negative, but we cannot make them both negative because that actually gives us a positive slope. Therefore when you are graphing a line with negative slope and are counting your rise over run to find additional points to plot, you have to keep this in mind. If you make your rise negative you would count down, but then you must have a positive run and count to the right. Or you can make you rise positive and count up, but then you must make your run negative and count back to the left. However, if you count your rise down and your run to the left then you have used to negatives which is going to result in a positive slope. Try it!

### Subject:Pre-Algebra

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Question:

How do I solve a two-step equation?

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Stacey W.

### Subject:Algebra

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Question:

How do I use substitution to solve a system of equations?

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Stacey W.

First of all, let's think about what the word substitution means. When I think of substitution, I think about a basketball game when they substitute players into the game for a certain position. When a coach wants to put someone else into the game, they pull a person from the court and "substitute" a person from the bench to go onto the court. Keep in mind that in order to substitute one player for the other, they must play the same position. We wouldn't substitute a point guard for a center. Using substitution to solve a system of equation is very similar to this process. We have two equations that each contain two different variables (let's use x and y). In order to determine the value of one of the variable (x), we need to pull out the other variable (y) in one of the equations and substitute in a value of y that only contains numerical terms and/or x terms, so that we only have the one variable x and are able to solve for the x-value. Once we know the x-value, then we can again substitute using this value for x, which enables us to solve for the y variable. Let's look at an actual example. We have the equations y = 5x - 7 and -3x - 2y = -12. The first equation tells me that y is equal to 5x - 7, so where ever there is a y, I can substitute it for 5x - 7. So as they do in basketball, in the second equation, I can pull out the y and substitute in the 5x - 7 in order to create an equation with only the one x variable (notice I pulled out the y variable and substituted it with a value of y, but I cannot substitute this value in for the x just like I substitute point guard for point guard and not point guard for center). I can now solve this new equation with my substitute value of y: -3x - 2(5x - 7) = -12 to determine what the x is equal to. When you so so, you should get x = 2. You can now use this value as a substitute for x in order to solve one of the equations for the y-value (y = 5(2) - 7). You should find that y = 3. Therefore, the solution to your system is (2, 3).

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