[Learn Math by Example] Today I solve a math problem: solving and graphing inequalities + using set builder and interval notations. Please scroll down to see my whole course plan.
I am teaching math through examples. If you have any questions, please leave a comment.
Here is my course plan: ----------------------- Algebra 1 -----------------------
- Solving absolute value equations - Writing absolute value equations from its solutions - Solving and graphing inequalities on a number line + using set builder and interval notations - Solving + graphing (on a number line) ‘and’ & ‘or’ inequalities (write answers in set builder and interval notations) - Relations and functions relationship (plus identifying functions) - Properties of lines that slant up, lines that slant down, horizontal lines, and vertical lines (domain, range, end behavior, intercepts, max/min points, positive and negative regions, symmetry) - Graphing and finding the properties of linear equations in standard form - Finding the slope (also mention the slopes of horizontal and vertical lines) - Writing linear equations in point-slope form from given criteria - Graphing inequalities through shading - Relationship of slopes of parallel and intersecting lines - Solving systems of equations with substitution and elimination - Simplifying monomials through multiplying and dividing - Classifying polynomials by terms and degree and identifying the leading coefficient - Zero and negative exponents - Order of magnitude - Identifying vertical and horizontal shifts and dilations (k, h, a) in equations (graph transformations) - Graphing and finding the properties of exponential functions (a+b)(a-b)=a^2-b^2, (a-b)^2=a^2-2ab+b^2, (a+b)^2=a^2+2ab+b^2 theorems - FOIL method - Factoring and solving expressions - Projectile problems - Graphing and finding the properties (including the axis of symmetry) of quadratic functions - Identifying the different forms of quadratic equations (vertex form, intercept form, and standard form) - Completing the square - Factoring using the quadratic formula - Using the discriminant (b^2-4ac) to find the number of real solutions of a quadratic equation - Graphing and finding the properties of square root functions - Simplifying the exponent of a radicand variable - Using conjugates to simplify radical expressions that contain a sum/difference of square roots in the denominator - Arithmetic vs geometric sequence - Identifying irrational vs rational numbers
【Algebra 1】 Lesson 3: Solving and Graphing Inequalities + Using Set Builder and Interval Notations
Hi, I'm Marlene and today we will find the solutions of the inequality 2x-3 is less than
or equal to 4x plus 5 that are negative integers.
So for the first step you want to solve this inequality.
To do that you do 2x-3 less than or equal to 4x plus 5 and we're stating the problem
and then I can move the 2x over to the right side so it's negative 3 less than minus 2x
plus 5.
I move the 5 over so I mean I could put the 2x in front then.
Remember to if you're moving sides just remember to flip the side.
I can move the 5 over there so it would be 2x is greater than or equal to negative 8.
So then you can do x is greater than or equal to negative 4.
So if I want to graph this I could do graph for example and then set 0, negative 1, negative
2, 3, negative 4.
You could take a part of the graph from anywhere.
You can for example you could do it from negative 2 to 5.
For this I'll just do 0 to negative 4 and then so I plot it since it's greater than
or equal to I set a black dot instead of an open dot like this.
And then since it's greater than or equal to negative 4 you want to go to the right
of it.
And now there are two ways to write the answer to an inequality.
So you could do interval notation.
This is more of a range of solutions.
If a solution is included in the answer you would use a bracket like that.
And if it's not included or if it's at infinity or negative infinity you would use a parenthesis.
So for example in interval notation all real solutions is like this and you would just
do negative infinity to infinity.
And since you technically can't reach infinity or negative infinity these values would not
be included in the solution so you would use parenthesis.
Then you have set builder notation.
And set builder notation is writing the actual final inequality.
So what it is is you use one of these curly brackets and you would use x.
This line means such that sign here.
And then for example here you would do, for example the inequality was x is less than
1.
And here it would be x such that x is less than 1.
For both of these you could do compound inequalities which means more than one inequality.
You could express it using these things which I will cover more in another lesson.
So in this case first we want to do interval notation.
So first you know it goes from negative 4 and to this way it will keep on going.
It goes infinitely.
So you know one side is negative 4.
But since it is greater than or equal to then negative 4 is included in the solution.
So you want to do a square bracket and then negative 4 comma and it goes to infinity.
So infinity and then parenthesis.
You always want to put the smaller value here on the left and the larger value on the right.
Now for set builder the final inequality is x is greater than or equal to negative 4.
So that would be the set builder.
You just do x such that x is greater than or equal to negative 4 and you do a parenthesis.
So we're not finished yet because it says you want to find solutions that are negative
integers.
And negative integers are values like negative 1, negative 2, negative 3.
They're all less than 0 but they're not fractions.
So what I can do is so you have the values.
So to find that just look at this and say like oh where is this less than 0?
I'm not a fraction.
So look at the graph 0 is here.
So anything from here between here every little line in this area is a negative integer solution.
So negative 4 is included in the solution.
So negative 4 then negative 3 and then negative 2 and then negative 1.
So these are all the values of this inequality that are negative integers.
And now for the next question it says solve for a if 2x minus 1 is less than or equal
to negative 1 and x is less than or equal to negative 1.
So first what you want to do in this first inequality is to isolate the x.
So if I restate the problem 2x minus a less than or equal to negative 1 and then 2x less
than or equal to negative 1 plus a.
And you could do x and then divide it both sides by 2 so like this.
So here in this inequality it says x is less than or equal to negative 1 and here x is
less than or equal to negative 1 plus a over 2.
So here the value of this is going to equal value of negative 1 plus a over 2 is going
to equal negative 1.
So I set that equal to each other equals negative 1.
So negative 1 plus a equals negative 2 and then a you move it over so it is a equals
negative 1 and that's how you solve for a.
So for the next question it says if a is less than 0 find the range of the solutions of
x when the absolute value of a times x is less than a.
So what you want to do first is simplify this so that x is isolated less than a.
I can do x less than a over the absolute value of a.
So here you are given the solution and you know that a is less than 0 and you also know
that the absolute value of a equals negative a since it is negative so then absolute value
is always positive.
So here it would be like saying for example if I take a to be negative 5 then the absolute
value of a would equal 5 and then negative 5 over 5 is negative 1 because you are dividing
the number by itself and then putting a negative in front of it.
Okay so then since you know that a divided by the absolute value of a is negative 1 because
it would be the same for any numbers for example negative 17 over 17 would also be negative 1.
Okay so then x would be less than negative 1 and that's how you solve it.
And then if I wanted to graph it I could do example 0, 1, 2, negative 1, negative 2 and
then since it's not included in the solution I do an open circle and since it's less than
I go that way.
And now to practice interval notation and set builder notation.
The interval notation for this would be negative infinity comma negative 1 but negative 1 would
have a parenthesis because it's not included in the solution and the set builder would
just be x such that x is less than negative 1.
