[Learn Math by Example] Today I solve a math problem: solving absolute value equations. 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 1: How to Solve Absolute Value Equations?
Hi, I'm Marlene, and today I will solve absolute value equations.
Okay, so the definition of absolute value is the distance of a number from zero.
And one thing to remember is that distance is always positive or equal to zero.
So, for example, absolute value is denoted as this.
And then, for example, x as to represent number.
And another line.
These two vertical lines denote absolute value.
So, for example, if I have x equals negative 5.
So it would be these two lines, negative 5.
So the distance of negative 5 from 0 is 5.
So the absolute value of negative 5 would be 5.
But the absolute value of 0, the opposite of 0, and 0 itself are all equal to 0.
And that's something important to remember.
Also, for example, absolute value of x cannot equal negative 2.
Because absolute value can never be negative.
For the first question, we are solving for x.
And the absolute value of x minus 2 equals 1.
So, for example, if you have absolute value of x equals 1.
Then x can equal 1 or x can equal negative 1.
So the same thing happens for this equation here.
So x minus 2 could equal 1 or x minus 2 could equal negative 1.
And then you would solve it like an equation.
And x equals 1.
Okay, so here x is equal to 1 and 3.
For the next question, you want to solve for both x and y.
So here there is this value plus this value equals 0.
And since there's the absolute value symbol on both sides of the plus sign.
You know that both sides can't be negative values.
And they somehow both add up to 0.
So they both have to equal 0.
Okay, so you know that you just have to set the absolute value of x minus 2y as 0.
Because 1 half times 0 is 0.
So, x minus 2y equals 0.
And here you can set this one as y minus 1 half as 0.
So y equals 1 half.
And here you can plug in y.
So x minus 2 times 1 half equals 0.
So x minus 1 equals 0.
x equals 1.
So here you have x equals 1 and y equals 1 half.
Okay, so for the next question, it states if a and b are opposites.
And the absolute value of a minus b equals 6.
Find all possible answers of the absolute value of b minus 1.
So here since a and b are opposites, then a could equal negative b.
Or negative a could equal b.
Okay, here.
And you will find that opposites, like for example, negative 3 and 3,
their absolute values, 3, they are both equal to 3.
So opposites have the same absolute values.
So here the a and b will have the same absolute values.
So you want to find which two of the same positive numbers added together equals 6 and 3 and 3.
But the thing is, one's positive and one's negative.
And you don't know which one's negative and which one's positive.
So you want to set up two scenarios.
One is a equals 3, b equals negative 3.
Other one is a equals negative 3 and b equals 3.
To prove that both of them have equal 6 when the absolute values are 8.
To prove that both scenarios work for answering that absolute value of a minus b equals 6,
I'll just solve it.
So a is 3 minus negative 3 equals 6.
So this would be absolute value of 6, which equals 6.
So this is correct.
And if I take it here, it would be negative 3 minus 3 equals 6.
And it's absolute value of negative 6, which is also 6.
So that's also correct.
So now here, in these two scenarios, you have two different values of b.
So you can plug them into this absolute value of b minus 1.
So one of them is negative 3 minus 1.
And so absolute value of negative 4.
So it would equal 4.
Or it can be b equals 3.
So it would be absolute value of 3 minus 1.
And that would be absolute value of 2.
And that would be 2.
So all the possible answers of the absolute value of b minus 1 is 4 and 2.
And for the last question, it's used the smallest possible answer in the equation
absolute value of 2x minus 1 plus 2 to solve for x.
So you know that the smallest possible value of an absolute value is 0.
So for example, if I said 2x minus 1 equals 0,
then you can write this equation as 0 plus 2 equals 2.
And that would be the smallest possible answer.
So then I would write it as the absolute value of 2x minus 1 plus 2 equals 2.
Now I subtract 2 from both sides.
So the absolute value of 2x minus 1 equals 0.
So 2x minus 1 equals 0.
Since the absolute value of 0 is 0 and the opposite of 0 is 0.
So there's only one answer for the absolute value of 0.
So now you're going to do 2x equals 1 and x equals 1 half.
