Question 1

Trying to figure out where I can find Sal giving a video on these types of equations. x+1 over 3 + x+2 over 7 = 2. What is this problem called? a linear equation or rational? Need to learn how to do these. thanks for your help

Accepted Answer

Your example looks like a rational equation, Sal has 3 videos on this topic (they are called solving rational equations). I solved your equation and you have to find a common denominator first, which is 21. Then you multiply 7(x+1) + 3(x+2), all that over 21, which equals 2. After a few steps, you`ll have 10x +13=42, then 10x=29, and finally, x=10/29. I hope this helps, but like I said, Sal teaches how to solve those types of equations, just watch "solving rational equations" and you`ll be fine!

Question 2

I'm having a lot of trouble with inequalities, because I don't really understand the entire idea of an inequality... What is it? How do you solve it? The tutorial in my mathbook was almost useless... I don't really know what it is! How do you solve it? What kind of answer are you supposed to give? Is there a video where Sal explains this?

Accepted Answer

Inequalities describe a relationship between two values that are not equal.
a < b states that the value of a is less than the value of b, and a > b states that the value of a is greater than the value of b.
Keep in mind that a "value" being greater or less than another value refers to its position on the number line: those with lesser values are "more negative," or further left on the number line, while those with greater values are "more positive," or further to the right on the number line.
When you are solving algebraic equations with inequalities, you treat them almost like equations. You may add or subtract on both sides without any difference.
When you multiply or divide, however, you must consider whether the operation you are performing will change the nature of the problem. Multiplying or dividing by a negative will change the signs of both sides, and thus change the relative positions of the numbers on the number line, effectively mirroring them about zero. So, if you have 2 > 1, and multiply by a -1, you get -2 > -1, which is not true, since -1 is more positive. We are forced to switch the sign, and make it: -2 < -1, in order to ensure it remains a true statement.
So -x < 1 will become x > -1, and -5x > -10 will become x < 2.
In the above inequalities, there is only one variable: x. We are only concerned with the x-values. If you choose to graph the left and right sides as separate equations, in order to find the intersection points, we are looking for when one function is above or below the other. Just keep in mind that the y-values are unimportant to the answer. The only reason we would need them is to see the relationship between the two graphs.
If you want any more practice or instruction on inequalities, here is the link to the Algebra I section: https://www.khanacademy.org/math/algebra/linear_inequalities

Question 3

I need help with a question such as 4/(5-3x)<0 . I can't seem to apply this knowledge to that inequality.

Accepted Answer

Isolate x, so the first step would be to realize that as long as  (5-3x) is a negative then 4/(5-3x) will be less than 0. So you know that 5-3x < 0 (4 can be removed). From here you can move the 3x to the other side so 5 < 3x. Divide both sides by 3 to get 5/3 < x. This can be reversed to put x on the left side and so the answer would be that x > 5/3. Vote me up if this helped :)

Question 4

At 8:50, how can (x+2) be considered positive when -1 still falls in the range of x>-2?

Accepted Answer

SG,
When x=-1 x+2 = -1+2 = +1 so even though x is not positive, (x+2) is positive.

I hope that is of some help.

Question 5

how do you solve a linear equation using the "substitution" method?

Accepted Answer

Substitution, otherwise known as "plugging in," is a method that is used in a system of equations with two or more variables. Once we know the value of one variable, we can substitute in order to find the value of another variable.

For instance, if 2x+3y=27, and we know x=6, then we would write 2(6)+3y=27, thus 3y=27-12=15, so y=5.

Substitution is also sometimes used in an equation with one variable, simply to check your answer. If we have solved 6a=18 and we found a=3, then we can write 6(3)=18.

Question 6

what does this sign > mean

Accepted Answer

it means greater than e.g 2>1 it means 2 is bigger than 1.

Question 7

1/(x-2) >= 3/(x+1)  i have (-2x+7)/(x-2)(x+1) but the answer is x-2/7 over denominator,,,, help

Accepted Answer

1/(x-2) >= 3/(x+1), x =/= 2, x=/= -1
1/(x-2) - 3/(x+1) >= 0
( 1(x+1) - 3(x-2) ) / ( (x-2)(x+1) ) >= 0
( x + 1 -3x + 6 ) / ( x^2 + x - 2 ) >= 0
(-2x + 7) / (x^2 + x - 2) >= 0

Since we're asked when the expression is not negative, let's make a sign chart.

Zeroes of -2x + 7:
-2x + 7 = 0
-2x = -7
x = 2/7

Zeroes of x^2 + x - 2:
Since x^2 + x - 2 was expanded from (x-2)(x+1), we can easily see that the zeroes are x = 2 and x = -1. Also remember that there are the forbidden values for x.

-2x + 7 is a line with negative slope, so on the left side of its zero it's positive and on the right side it's negative.

x^2 + x - 2 is a parabola opening up, so between its zeroes it's negative, otherwise positive.

Let's call -2x + 7 = f(x) and x^2 + x - 2 = g(x) so (-2x + 7) / (x^2 + x - 2) = f(x)/g(x) and I don't have to write that horrendously long expression.

Here's the sign chart:

     f(x)  +   #  +   |   -  #  -
     g(x)  +   #  -   |   -  #  +
          --- -1 --- 2/7 --- 2 --->
f(x)/g(x)  +   #  -   |   +  #  -

         <-----)      [------)

The first row is the sign of -2x + 7 in each interval, the second row is the sign of x^2 + x - 2 in each interval, and the third row is the sign of the products of the signs above, being the sign of (-2x + 7) / (x^2 + x - 2).

On the last line I marked the intervals when (-2x + 7) / (x^2 + x - 2) is not negative (that means we have to include all boundaries we can. We can't include -1 or 2 because they're forbidden but we have to include 2/7.

So the answer is x < -1 or 2/7 <= x < 2.

Question 8

At around 00:58: Why is a>0 and b>0 the consequence of a/b>0?

Accepted Answer

A/B can only be greater than 0 (in other words, positive) only if A and B are both positive (A>0 and A>0) or A and B are both negative (A<0 and B<0). Any other combination will result in a negative fraction (A/B<0 instead of A/B>0).

Question 9

I'm working  on a problem that states: What rate of interest compounded annually is required to double an investment in 5 years. I got part of the answer as r=5th root2-1, but I have no clue how to enter this into a ti-84 to get the approximation of this 5th root. Please help soon finals!

Accepted Answer

Look for a button that says x^y (that is x with the exponent of y).  That is the power button.  Use that and put in the reciprocal of the order of the root.  For a fifth root that would be 1/5 or 0.2.  A root is really a reciprocal of the power.  So a square root is really the same as raising to the power of one half and a fifth root is the same as raising to the power of 1/5 or 0.2
So, if your answer is -1 + fifth root of 2, key in this:
(2 [x^y] 0.2) - 1 =
The answer is = 0.148698355... or 14.87%

Course: Algebra (all content) > Unit 13

Rational inequalities: one side is zero

Want to join the conversation?

Video transcript