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## SAT

### Course: SAT > Unit 10

Lesson 1: Heart of algebra- Solving linear equations and linear inequalities — Basic example
- Solving linear equations and linear inequalities — Harder example
- Interpreting linear functions — Basic example
- Interpreting linear functions — Harder example
- Linear equation word problems — Basic example
- Linear equation word problems — Harder example
- Linear inequality word problems — Basic example
- Linear inequality word problems — Harder example
- Graphing linear equations — Basic example
- Graphing linear equations — Harder example
- Linear function word problems — Basic example
- Linear function word problems — Harder example
- Systems of linear inequalities word problems — Basic example
- Systems of linear inequalities word problems — Harder example
- Solving systems of linear equations — Basic example
- Solving systems of linear equations — Harder example
- Systems of linear equations word problems — Basic example
- Systems of linear equations word problems — Harder example

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# Graphing linear equations — Basic example

Watch Sal work through a basic Graphing linear equations problem.

## Want to join the conversation?

- i want to know where can i get a video on statistics

i really need it as i lack in it

please respond!(29 votes)- In the math tab, there is a section called "Probability and statistics".(10 votes)

- I am confused why and how did sal know it was the first graph?(14 votes)
- Sal knew that the graph had to start at $1000 because that is what the question said. That eliminated two of the possibilities that started at 0 and at $500. Two down, two to go in about 10 seconds of your valuable time.

We can choose between the other two by calculating a couple of points and comparing to the graphs that remain. The words of the problem say that the bond increases at $75 per year. Because the vertical scale is so large, we need a bunch of years to show a good enough increase to be able to read the difference.

At 75 per year, this would increase $300 in 4 years (75 x 4 = 300). Add that to $1000 to get $1300 at the 4th year. The third graph example grows 900 in 4 years, ($1900), while the first graph is right on track at under $1500 in the 4th year. It looks really close to $1300.

Choose graph one and`happy dance`

on to the next question.(20 votes)

- i should probably concentrate alot more(13 votes)
- Yes, you must because it's math so yeah.(2 votes)

- When graphing an inequality do we shade the region with a solution or without a solution?(7 votes)
- If the inequality is „ or … then we draw a solid line. If the inequality is < or > then we draw a dotted line. After drawing the line, we need to shade the unwanted region. Rewrite the inequality 2 x – 3 y ≥ 6 as y ≤ x – 2. Since the inequality is ≤ , the wanted region is below the line. We shade below the line.

When graphing inequalities, you shade all areas that x and/or y can be. If the number is x, you shade left and right. If x is anywhere from -11 to ∞, then shade the area to the right of -11. If it is from -∞ to 5, shade the areas to the left of 5(1 vote)

- yall from 8 years ago yall like grown ups(6 votes)
- In1:09, Sal explains that "you want one of them to be 180". Why would that be? Why wouldn't it be 300? Is it because we're dealing mainly with time in the word problem?(4 votes)
- because every year, they increase by 75$ only(3 votes)

- considering i have my psat on the 14th, I need help!(5 votes)
- Me need help on me assignment!

pls?(3 votes) - The points A(2, 3) and B(m, 11) are 10 units part. Which equation could describe the line that contains points A and B?(2 votes)

## Video transcript

- [Instructor] A line is graphed
in the xy-plane as shown. Which of the following
equations represents the line? They give us a bunch of equations here and so there is several
ways we can tackle it. When we look at this, I could see there's two interesting points here, there's the point when x is, let me just write this and actually I'm gonna write a little lower so we can look at it the same time that I look at the equation choices. So we see that when x is a zero, y is one. So that is the y intercept we could say. And then when we could see
when x is six, y is zero. When x is six, y is zero,
so a very kind of basic way of approaching this is see well, when x is a zero, y
needs to be equal to one. When x is zero we get
six y is equal to one, well then y is gonna
be equal to one sixth, rule that one out. When x is equal to zero, y
needs to be equal to one. If x is zero then six y equals six. Yeah, y is going to be equal to one. Now when y is zero, x
needs to be equal to six. So if y is 0, this goes
away and x is equal to six. So we're done, this is our choice. Now there's other ways
that we could do it. We could write it first
in slope intercept form and then convert to this
form right over here. So let's do it that way as well. We could say that the
equation of this line is gonna be y, if I write it
in y equals mx plus b form where m is the slope and
b is the y intercept. We already know that b is equal to one. So we already know that's
one, and what's the slope? Well slope is our change
in y for given change in x and we see when our change
in x is positive six, when our change in x is positive six, our change in y is negative one, so our slope is we decrease in y by one when we increase in x
by six is negative 1/6. So the equation of the line
y is equal to negative 1/6 x, this is the slope plus one. And then we could convert to the forms that we have here, so lets
see, we could add 1/6 x to both sides and you're gonna get one over six x plus y is equal to one, and that's not quite what we have here. All the coefficients on x are just one. So we can multiply both
sides of this times six, and we would get x plus
six y is equal to six. Which is exactly the
choice that we picked.