High School Math Solutions – Inequalities Calculator, Absolute Value Inequalities Part II

Last post we talked about absolute value inequalities with one absolute value expression. In this post, we will learn how to solve absolute value inequalities with two absolute value expressions. To do this, we will use some concepts from a previous post on rational functions. This can get a little tricky and confusing, so take time to read everything carefully.

I’ll show you the steps to solving these inequalities and then I will go through one example step by step.

Steps:

1. For each absolute value expression, figure out their negative and positive ranges.
2. Combine ranges if needed
3. For each range, evaluate the absolute value inequality
4. Combine the ranges

Here’s our first example (click here):

\frac{|3x+2|}{|x-1|} >2

Step 1: For each absolute value expression, figure out their negative and positive ranges

	3x+2	x-1
Positive (≥0)	3x+2≥0 Range:x\ge-\frac{2}{3} Positive absolute value expression:|3x+2|=3x+2	x-1≥0 Range:x≥1 Positive absolute value expression:|x-1|=x-1
Negative (<0)	3x+2<0 Range:x\le-\frac{2}{3} Positive absolute value expression:|3x+2|=-(3x+2)	x-1<0 Range:x<1 Positive absolute value expression:|x-1|=-(x-1)

Step 2: Combine ranges if needed

Let’s see our ranges: x≥-\frac{2}{3},\quad x≥1,\quad x<-\frac{2}{3},\quad x<1
Now let’s combine any ranges if possible: x<-\frac{2}{3},\quad -\frac{2}{3}≤x<1, \quad x≥1

Step 3: For each range, evaluate the absolute value inequality

For x<-\frac{2}{3}:

\frac{-(3x+2)}{-(x-1)}>2	We use the negative absolute value expressions because for both expressions the range is less than their negative range.
\frac{3x+2}{x-1}>2
\frac{x+4}{x-1}>0	Now, we see that this is just a rational inequality. So we solve the rational inequality
x<-4\:or\:x>1	The steps aren’t shown to get the answer. You can refer back to the rational inequalities blog if you forget how to solve it

For -\frac{2}{3}≤x<1:

\frac{3x+2}{-(x-1)}>2	We use the negative absolute value expression for the |x-1| because the range is less than its range.
\frac{5x}{1-x}>0
0<x<1

For x≥1:

\frac{3x+2}{x-1}>2
\frac{x+4}{x-1}>2
x<-4\:or\:x>1

Step 4: Combine the ranges

(x<-\frac{2}{3}\:and\:x<-4\:or\:x>1)\:or\:(-\frac{2}{3}≤x<1\:and\:0<x<1)\:or\:(x≥1\:and\:x<-4\:or\:x>1)

Answer: x<-4 or x>1 or 0<x<1


Hopefully, that wasn’t too complicated. Now let’s see one more example.

Last example (click here):

As you can see, this topic can be very confusing and tricky. The more you practice, the more natural it will become. For more practice examples check out Symbolab’s practice.

Until next time,
Leah

High School Math Solutions – Inequalities Calculator, Absolute Value Inequalities Part I

Last post, we learned how to solve rational inequalities. In this post, we will learn how to solve absolute value inequalities. There are two ways to solve these types of inequalities depending on what the inequality sign is. Let’s see how to solve them . . .

Case I (< or ≤):
Given |x|<a, we read this as “x is less than a units for 0.” On a number that looks like:

The red line is the solution for x. Remember because it is “less than,” a is not included in the points (this is why the circle is not filled in).

So we rewrite

The solution is in this form: -a<x<a.
The same goes for ≤. Just substitute ≤ for < and fill in the circles in the number line.

Case II(> or ≥):
Given |x|>a, we read this as “ x is more than a units from 0.” On a number line, that looks like:

The red line is the solution for x. Remember because it is “greater than.” A is not included in the points.

So we rewrite

The solution is 2 inequalities, NOT one. It is in the form: x<-a or x>a.
The same goes for ≥. Just substitute ≥ for > and fill in the circles in the number line.


NOTE: Remember to carry out any algebraic manipulations outside of the absolute value before continuing, for example:

2|x|-2>4
|x|>3

ANOTHER NOTE: Make sure a is positive or else there is no solution.


Let’s see some examples . . .


First example (click here):

Notice that the algebraic manipulations to be done are inside the absolute value, so that is done last.


Last example (click here):

That wasn’t too complicated. Next post, we will learn how to solve absolute value inequalities with two absolute value expressions in it. Make sure you practice because next post the problems will become trickier.

Until next time,

Leah

High School Math Solutions – Inequalities Calculator, Rational Inequalities

Last post, we talked about solving quadratic inequalities. In this post, we will talk about rational inequalities. Let’s recall rational functions are an algebraic fraction that contain polynomials in the numerator and denominator. Solving rational inequalities is a little different than solving quadratic inequalities. Both share the concept of a table and testing values.

Let’s see how to solve rational inequalities.

Here the steps:

1. Move everything to one side of the inequality sign
2. Simplify the rational function 
3. Find the zeros from the numerator and undefined points in the denominator
4. Derive intervals
5. Find the sign of the rational function on each interval
6. Select the proper inequality

Let’s go through our first example step by step to understand the concept better.


Here’s our first example (click here):

\frac{x}{x-3}<4

Step 1: Move everything to one side of the inequality sign

\frac{x}{x-3}<4
\frac{x}{x-3}-4<0
\frac{-3x+12}{x-3}<0

Make sure you combine everything into one rational function.

Step 2: Simplify the rational function

\frac{-3x+12}{x-3}<0
\frac{3(-x+4)}{x-3}<0

It’s already simplified. Nothing to cancel out.

Step 3: Find the zeros from the numerator and undefined points in the denominator

-3x+12=0                                                        x-3=0
x=4                                                                     x=3

          Zero                                                                     Undefined point

Step 4: Derive intervals

Step 5: Find the sign of the rational function on each interval

Table Header	x\lt3	3\ltx\lt4	x\gt4
\frac{-3x+12}{x-3}	\frac{-3(0)+12}{(0)-3}=-4	\frac{-3(3.5)+12}{(3.5)-3}=3	\frac{-3(5)+12}{(5)-3}=-\frac{3}{2}
Sign	\quad\quad\quad\quad-	\quad\quad\quad\quad+	\quad\quad\quad\quad-

We changed the format of the intervals to inequalities. We pick a number in the interval, plug the number in the rational function and see what sign the answer is (negative or positive).

Step 6: Select the proper inequality

\frac{-3x+12}{x-3}<0
x<3\quad\:and\quad\:x>4

We refer back to the original inequality and see which inequality satisfies the original inequality. We are looking for a number that produces a negative number. We refer back to the table and see that x<3 and x>4 satisfy this.

Alright, that was a mouthful. Let’s see some more examples now…


Second example (click here):

Last example (click here):

Make sure you double check your work for calculation errors because that is where it’s very easy to make a mistake. For more practice, check out Symbolab’s practice.

Until next time,

Leah

High School Math Solutions – Inequalities Calculator, Quadratic Inequalities

We’ve learned how to solve linear inequalities. Now, it’s time to learn how to solve quadratic inequalities. Solving quadratic inequalities is a little harder than solving linear inequalities. Let’s see how to solve them.

There are a couple ways to solve quadratic inequalities depending on the inequality. I’ll focus on explaining the more complicated version.

We’re given the quadratic inequality: x^2+2x-8\le0

Here are the steps to solving it:

1. Move everything to one side of the inequality sign
2. Set the inequality sign to an equal sign and solve for x
3. Create three intervals
4. Pick a number in each inequality and see if it satisfies the original inequality
5. Select the proper inequality

Now we will go through this example step by step to understand a little better how to solve it.

Step 1: Make sure you start with 0 on one side

x^2+2x-8\le0

Looks good!

Step 2: Set the inequality sign to an equal sign and solve for x

x^2+2x-8=0
(x-2)(x+4)=0
x=2\:and\:x=-4

Step 3: Create three intervals

We are able to pick three intervals from looking at the number and seeing where the function crosses the x-axis (i.e. where the function is equal to 0).


Step 4: Pick a number in each inequality and see if it satisfies the original inequality

x<-4               -4<x<2               x>2

Table Header	x<-4	-4<x<2	x>2
x^2+2x-8	(-5)^2+2(-5)-8=7	(0)^2+2(0)-8=-8	(3)^2+2(3)-8=7
Sign	\quad\quad\quad\quad+	\quad\quad\quad\quad-	\quad\quad\quad\quad+

We’ve turned the intervals into inequalities. Then, we picked a number in each inequality to see if it satisfied the original inequality, x^2+2x-8\le0.

Step 5: Select the proper inequality

-4<x<2
-4\lex\le2

We’ve selected -4<x<2 because it satisfies the original inequality because the quadratic is negative when x is between -4 and 2. However, don’t forget that the original equality contains the ≤ symbol so that means x can equal 0 too. We change the inequality signs because we know -4 and 2 are the zeros of the quadratic.


Let’s see some more examples…

Second example (click here):

Last example… This one is simple (click here):

Hopefully, that wasn’t too hard! Solving quadratic inequalities sometimes require patience to write everything out. For more help check out Symbolab’s practice.

Until next time,

Leah

High School Math Solutions – Inequalities Calculator, Compound Inequalities

In the previous post, we talked about solving linear inequalities. In today’s post we will focus on compound inequalities, which are just a little more complicated than linear inequalities.

A compound inequality is an equation of two or more inequalities joined together by “and” or “or”. Sometimes you see compound inequalities joined by “and” written like -3<2x-1<5. Compound inequalities have three parts: the left, the middle, and then right. Just like compound inequalities, our goal is to isolate the variable. However, we want the variable to be isolated in the middle part.

-3<2x-1<5\quad\quad\quad\quad\to\quad\quad\quad-1<x<3

How do we get that answer? The key is to break the compound inequalities.

Here are the steps for solving compound inequalities:

1. Break the inequality into two parts
2. Solve each linear inequality by isolating the variable
3. Combine the inequalities 

Simple enough? Let’s break down an example step by step to understand the concept.

Here’s our problem (click here):

-17<3+10x\le33

1. Break the inequality into two parts

            -17<3+10x                                                    3+10x\le33

Based on the definition of a compound inequality, we know that this compound is made of two inequalities. We break the compound inequality down into its two inequalities.

2. Solve each linear inequality by isolating the variable

              -17<3+10x                                                    3+10x\le33
                -20<10x                                                        10x\le30

                 -2<x                                                              x\le3

Using algebra, we can isolate x and get two simplified inequalities.


3. Combine the inequalities
                 -2<x                                                           x\le3
                                                 -2<x\le3

Make sure after combining the inequalities that the simplified compound inequality makes logical sense.


Let’s see two more examples . . .

Second example (click here)

Last example (click here)

Solving compound inequalities is just little harder than solving linear inequalities. Remember to keep your work nice and neat and practice and you will do great! For more practice on compound inequalities, check out Symbolab’s practice.

Until next time,

Leah

High School Math Solutions – Inequalities Calculator, Linear Inequalities

Solving linear inequalities is pretty simple. A linear inequality is an inequality which involves a linear function. Think of a linear inequality as a linear equation. You solve a linear inequality just like you would a linear equation, by isolating the variable using algebraic manipulations.

One thing to note: if you divide or multiply a linear inequality by a negative number you must switch the inequality sign. That means if you have -2y\le4 and you divide the linear inequality by -2 to isolate y, then the sign switches to ≥. The inequality becomes y\ge2.

Let’s see some examples.

First example (click here):

By using simple algebra, x was isolated and the answer was calculated. You can see that the sign was switched because the inequality was multiplied by -2.


Second example (click here):

Last example (click here):

Solving linear inequalities is pretty simple if you’ve solved linear equations before. As long as you remember to switch the inequality sign when multiplying or dividing the inequality by a negative number, then you are good to go! If you need more help or practice on this topic, check out Symbolab’s inequality practice.

Until next time,

Leah

High School Math Solutions – Systems of Equations Calculator, Nonlinear

In a previous post, we learned about how to solve a system of linear equations. In this post, we will learn how to solve a system of nonlinear equations. A system of nonlinear equations is a system in which one or more variables has an order greater than 1, and/or there is a product of variables in one of the equations. There are two ways to solve systems of nonlinear equations, substitution and elimination.

Substitution

Very similar to the method of substitution for systems of linear equations

1. Solve for one variable in one of the equations
2. Substitute the value of that variable in another equation
3. Solve for variable in that equation
4. Solve for other variables using known variable value
5. Double check to make sure the ordered pairs work by plugging them back into an equation

Elimination

Also very similar to the method of substitution for systems of linear equations

1. Multiply by a constant
2. Add/subtract to get rid of a variable
3. Solve for the variable
4. Solve for other variables using known variable value
5. Double check to make sure the ordered pairs work by plugging them back into an equation

It is a lot easier to visualize these two methods than to just read about them. I will go through two examples step by step, one for each method.

Here’s an example using substitution (click here):

3-x^2=y
x+1-y=0

1. Solve for y in the second equation

y=x+1

2. Substitute x+1 for y in the first equation

3-x^2=x+1

3. Solve for x

x^2+x-2=0
(x+2)(x-1)=0
x=1\:or\:x=-2

4. Solve for y by substituting in the values for x

y=x+1
y=(1)+1 \:\:\:\: y=(-2)+1
y=2 \:\:\:\:\:\: y=-1

5. Double check

(1,2) \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: (-2,-1)
3-(1)^2=2 \:\:\:\:\:\:\:\:\:\:\: 3-(-2)^2=-1
2=2 \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-1=-1

Here’s an example using elimination (click here):

x^2+y=5
x^2+y^2=7

1. We don’t need to multiply by a constant because x^2 already has the same constant of 1 in both equations.

2. Subtract to get rid of x^2

\:\:\:x^2+y=5
-\:x^2+y^2=7
-------
y-y^2=-2

3. Solve for y

y^2-y-2=0
(y-2)(y+1)=0
y=2\:or\:y=-1

4. Solve for x

y=2 \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: y=-1
x^2+(2)=5 \:\:\:\:\:\:\:\:\:\: x^2+(-1)=5
x^2=3 \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: x^2=6
x=\pm\sqrt{3} \:\:\:\:\:\:\:\:\:\:\:\:\:\: x=\pm\sqrt{6}

5. Double check

Last example (click here):

You have to do a lot of algebraic manipulation when solving systems of equations, especially with nonlinear equations. Practice an array of different types of systems of nonlinear equations to become very familiar with solving these problems.

Until next time,

Leah