## Thursday, December 22, 2016

### Advanced Math Solutions – Vector Calculator, Simple Vector Arithmetic

Vectors are used to represent anything that has a direction and magnitude, length. The most popular example of a vector is velocity. Given a car’s velocity of 50 miles per hour going north from the origin, we can draw a vector. In this blog post, we will focus on the simpler aspects of vectors. We won’t talk about how to graph vectors.

Position vectors are vectors that give the position of a point from the origin. The vector is denoted as \vec{v}=<a_1,\:a_2,\:a_3> and starts at point A=(0, 0, 0) and ends at point B=(a_1,\:a_2,\:a_3).

This brings us to how to find a vector given an initial and final point. Given two points A=(a_1,\:a_2,\:a_3) and B=(b_1,\:b_2,\:b_3), the vector \vec{AB}, which goes from point A to B, is \vec{v}=<b_1-a_1,\:b_2-a_2,\:b_3-a_3>.

Given two vectors \vec{a}=<a_1,\:a_2,\:a_3> and  \vec{b}=<b_1,\:b_2,\:b_3> , \vec{a}+\vec{b}=<a_1+b_1,\:a_2+b_2,\:a_3+b_3>

Subtracting vectors is just as simple

Given two vectors \vec{a}=<a_1,\:a_2,\:a_3> and  \vec{b}=<b_1,\:b_2,\:b_3> , \vec{a}-\vec{b}=<a_1-b_1,\:a_2-b_2,\:a_3-b_3>

Scalar multiplication is used to lengthen or shorten a vector
Given two vectors \vec{a}=<a_1,\:a_2,\:a_3> and any number c, c\vec{a}=<ca_1,\:ca_2\:ca_3>

Every vector has a magnitude and a direction. The direction is where its arrow is pointed and the magnitude is the length of the vector. If the magnitude of a vector is 1, then we call that vector a unit vector.

Magnitude is denoted as |\vec{a}| or ||\vec{a}||.
We will use ||\vec{a}||, so we don't get confused with absolute values.
||\vec{a}||=\sqrt{a_x^2+a_y^2}

A unit vector \hat{u\:}, is a vector with length 1.
\hat{u\:}=\frac{u}{||u||}

Here’s an example of finding the unit vector of a vector (click here):

Here are some properties to memorize about basic vector arithmetic:

The topics we covered in this blog are simple. I recommend practicing a few examples and memorizing the formulas, and you should be good to go. We are going to cover some of the heavier vector topics in next blog.

Until next time,

Leah

## Tuesday, December 13, 2016

### High School Math Solutions – Inequalities Calculator, Exponential Inequalities

Last post, we talked about how to solve logarithmic inequalities. This post, we will learn how to solve exponential inequalities. The method of solving exponential inequalities is very similar to solving logarithmic inequalities. In these problems, the variable is in the exponent and our goal is to isolate the variable, which means getting it out of the exponent. We will see some these problems in the forms:

e^x>a

a^x<b

Let’s see how to solve these problems step by step.

1. Use algebraic manipulation to move everything that is not in the exponential expression to one side
2. Isolate the variable by getting rid of the exponential expression

Example:
e^x>a
e^x>e^{\ln(⁡a)}
x>\ln(a)

Example:
a^x<b
a^x<a^{\log_a(b)}
x<\log_a⁡(b)

3. Solve the inequality

These problems are a little simpler than solving logarithmic inequalities. Let’s see how to solve an example step by step.

3^{x+1}+1<2

Step 1: Use algebraic manipulation to move everything that is not in the exponential expression to one side

3^{2x+1}+1<2

3^{2x+1}<1

Step 2: Isolate the variable by getting rid of the exponential expression

3^{2x+1}<1

3^{2x+1}<3^{\log_3(1)}

2x+1<\log_3⁡(1)

Step 3: Solve the inequality

2x+1<\log_3(1)

2x+1<0

x<\frac{-1}{2}

That was pretty simple. Let’s see some more examples.

Solving exponential inequalities are simpler than solving logarithmic inequalities. However, it can still get a little tricky when solving these inequalities with more parts. For more help and practice on solving exponential inequalities, visit Symbolab’s practice.

Until next time,

Leah

## Tuesday, November 29, 2016

### High School Math Solutions – Inequalities Calculator, Logarithmic Inequalities

Last post, we talked about radical inequalities. In this post, we will talk about how to solve logarithmic inequalities. We’ll see logarithmic inequalities in forms such as \log_b(f(x))<a or \ln(⁡f(x))<a. In order to solve these inequalities, the goal will be to isolate the variable, just as in any inequality, and we will do this by getting rid of the log function. Let’s dive in and see how to solve logarithmic inequalities.

Steps to solve logarithmic inequalities:
1. Use algebraic manipulation to move anything that is not in the logarithmic expression to one side
2. Combine logarithmic expressions
3. Isolate the variable by getting rid of the logarithmic expression
Ex:    \log_b⁡(f(x))<a

b^(\log_b⁡(f(x))) <b^a

f(x)<b^a

Ex: \ln(f(x))<a

\ln⁡(f(x))<\ln⁡(e^a )

f(x)<e^a
4. Solve inequality
5. Get the range for the expression in the original log function
Ex:  \log_b⁡(f(x))
Range: f(x)>0
6. Combine ranges

Let’s do an example step by step now.

\log_4⁡(x+3)-\log_4⁡(x+2)\ge\frac{3}{2}

Step 1: Use algebraic manipulation to move anything that is not in the logarithmic expression to one side

There’s nothing to move, so we can skip this step.

Step 2: Combine logarithmic expressions

\log_4⁡(x+3)-\log_4⁡(x+2)\ge\frac{3}{2}

\log_4⁡(\frac{x+3}{x+2})≥\frac{3}{2}

Step 3: Isolate the variable by getting rid of the logarithmic expression

\log_4⁡(\frac{x+3}{x+2})≥\frac{3}{2}

4^(\log_4⁡(\frac{x+3}{x+2})) ≥4^(\frac{3}{2})

\frac{x+3}{x+2}≥8

Step 4: Solve inequality

\frac{x+3}{x+2}≥8

We can see that this is now a rational inequality. We won’t solve this step by step; I will show the answer after solving this inequality. If you are struggling with solving this inequality, visit the blog post on rational inequalities.

-2<x≤\frac{-13}{7}

Step 5: Get the range for the expression in the original log function

\log_4⁡(x+3)-\log_4⁡(x+2)\ge\frac{3}{2}

x+3>0              x+2>0

x>-3              x>-2

Step 6: Combine ranges

-2<x≤\frac{-13}{7},    x>-2,    x>-3

-2<x≤\frac{-13}{7}

That wasn’t too bad! Let’s see some more examples.

Solving logarithmic inequalities is not too difficult. Just remember to get the ranges inside the logarithmic expressions and to double check your work. For more help and practice on this topic visit Symbolab’s  practice.

Until next time,

Leah

## Tuesday, November 22, 2016

### High School Math Solutions – Inequalities Calculator, Radical Inequalities

Last post, we went over how to solve absolute value inequalities. For today’s post, we will talk about how to solve radical inequalities. Solving radical inequalities is easier than solving absolute value inequalities and require fewer steps. Let’s see the steps on how to solve these inequalities.

Steps:
1. Isolate the square root
2. Check that the inequality is true (i.e. not less than 0)
3. Find the real region for the square root, (i.e. see when the expression inside square root igreater than or equal to 0)
4. Simplify and compute the inequality
5. Combine the ranges

Let’s see how to do one example step by step.

\sqrt{5+x}-1<3

Step 1: Isolate the square root

\sqrt{5+x}<4

Step 2: Check that the inequality is true

Yes, this inequality is true the radical is not less than 0.

Step 3: Find the real region for the square root

5+x≥0

x≥-5

Step 4: Simplify and compute the inequality

\sqrt{5+x}<4

(\sqrt{5+x})^2<4^2

5+x<16

x<11
Step 5: Combine the ranges

x≥-5 and x<11

-5≤x<11

That wasn’t too difficult. Let’s see some more examples.

Solving radical inequalities isn’t too difficult, however, they require practice. For more practice examples check out Symbolab’s practice.

Until next time,

Leah

## Thursday, November 3, 2016

### High School Math Solutions – Polynomials Calculator, Dividing Polynomials (Long Division)

Last post, we talked dividing polynomials using factoring and splitting up the fraction. In this post, we will talk about another method for dividing polynomials, long division. Long division with polynomials is similar to the basic numerical long division, except we are dividing variables. This is where it gets tricky. I will talk about the steps to dividing polynomials using long division to help make the process easier and go into detail.

Steps for polynomial long division:

1.  Organize each polynomial by higher order
• We want to make sure that each polynomial is written in order of the variable with the highest exponent to the variable with the lowest exponent
• you can skip this step if they are already in high order
2.  Set up in long division form
• The denominator becomes the divisor and the numerator becomes the dividend

3.  Write 0 as the coefficient for missing terms in the dividend
• Since we’ve put in order the terms based on their exponent, we can see which terms are missing (i.e. x^4+x^2 we can see we are missing an x^3 term so we will add that in to its proper spot and make the coefficient 0)
• This will help you with step 6, so you don’t subtract the wrong terms
• You can skip this step if there are no missing terms
4.   Divide the first term of the dividend (numerator) by the first term of the divisor (denominator)
• This is allows us to see what we need to multiply the divisor by to get rid of the first term of the dividend
5.   Multiply the divisor by that term
•  Write the term down on top of line where the term that is getting eliminated is
6.  Subtract this from the dividend
• This gives you a new polynomial to work with
7.  Repeat steps 4-6 until you get a remainder
• When you repeat step 4, move onto the newest first term from step 6
8.  Put the remainder over the divisor to create a fraction and add it to the new polynomial

This may seem a bit confusing, so we will go through two examples step by step to understand better how to solve these problems.

\frac{(x^4+6x^2+2)}{(x^2+5)}

1.  Organize each polynomial by high order
We can skip this step because the polynomials are already in high order
2.  Set up in long division form
3.  Write 0 as the coefficient for missing terms in the dividend

4.  Divide the first term of the dividend (numerator) by the first term of the divisor (denominator)
\frac{x^4}{x^2}= x^2

5.  Multiply the divisor by that term
x^2∙(x^2+5)=x^4+5x^2

6.  Subtract this from the dividend

7.  Repeat step 4-6 until you get a remainder
\frac{x^2}{x^2} =1

8.  Put the remainder over the divisor to create a fraction and add it to the new polynomial
x^2+1+\frac{(-3)}{(x^2+5)}

\frac{(2x^2-18+5x)}{(x+4)}

1.  Organize each polynomial by high order
\frac{(2x^2+5x-18)}{(x+4)}

2.  Set up in long division form

3.  Write 0 as the coefficient for missing terms in the dividend
We can skip this step because there are no missing terms.

4.  Divide the first term of the dividend (numerator) by the first term of the divisor (denominator)
\frac{2x^2}{x}=2x

5.  Multiply the divisor by that term
2x(x+4)=2x^2+8x

6.  Subtract this from the dividend

7.  Repeat steps 4-6 until you get the remainder
\frac{(-3x)}{x}=-3

8.  Put the remainder over the divisor to create a fraction and add it to the new polynomial
2x-3+\frac{(-6)}{(x+4)}

Dividing polynomials using long division is very tricky. It is so easy to skip an exponent, have an algebraic error, and forget a step. This is why practicing this type of problem is so important. The only way to get better at it is to keep practicing it. Check out Symbolab’s Practice for practice problems and quizzes.

Until next time,

Leah.

## Wednesday, October 26, 2016

### High School Math Solutions – Polynomials Calculator, Dividing Polynomials

In the last post, we talked about how to multiply polynomials. In this post, we will talk about to divide polynomials.  When we divide polynomials, we can write these problems in the form of a fraction. This allows us to reduce the fraction and get our answer. Although this method doesn’t work every time we divide polynomials, when it does, it is a quick, simple method.

Once written in the form of a fraction, there are two ways to reduce the fraction. One method is to split the fraction up, so there is one term per fraction, having the denominator be the same. The other method is to factor the numerator and denominator and cancel out terms that are the same in the numerator and denominator.

Let’s see some examples to better understand how to solve these problems.

We will use the splitting method first for our first example.

(32x^4-56x^2)\div 8x
1. Rewrite the problem in the form of a fraction
\frac{32x^4-56x^2}{8x}
2. Split the fraction
\frac{32x^4}{8x}-\frac{56x^2}{8x}
3. Reduce the fractions
\frac{8\cdot 4x^4}{8x}-\frac{8\cdot7x^2}{8x}
4x^3-7x

Next example:
We will use our same problem above, but use factoring to solve the problem.
(32x^4-56x^2)\div 8x
1. Rewrite the problem in the form of a fraction
\frac{32x^4-56x^2}{8x}
2. Factor the numerator and denominator
\frac{8x^2(4x^2-7)}{8x}
3. Cancel out common factors

4. Simplify
x(4x^2-7)
4x^3-7x

(2x^3+11x^2+5x)\div(2x^2+x)
1. Rewrite the problem in the form of a fraction
\frac{2x^3+11x^2+5x}{2x^2+x}
2. Factor the numerator and denominator
\frac{(2x+1)(x+5)x}{(2x+1)x}
3. Cancel out common factors
4. Simplify
x+5

As you can see, factoring is a big part of dividing polynomials. Before attempting these problems, make sure you’ve mastered factoring. Dividing polynomials by these two methods is pretty simple. However, these methods may not work for certain division problems. Next blog post, we will talk about another method to use, when these methods don’t work. For more practice and help on this topic, checkout Symbolab’s Practice.

Until next time,

Leah

### Middle School Math Solutions – Polynomials Calculator, Factoring Quadratics

Just like numbers have factors (2×3=6), expressions have factors ((x+2)(x+3)=x^2+5x+6). Factoring is the process of finding factors of a number or expression, where we find what multiplies together to make the expression or number.  In today’s blog post, we will talk about how to factor simple expressions and quadratics.

Factoring simple expressions –
Given a simple expression, ax+b, pull out the greatest common factor from the expression. Pretty simple!
Factor\:2x+6
2x+2\cdot3
2(x+3)
Quadratics have the form: (ax)^2+bx+c, where a, b, and c are numbers.
Here are the steps for factoring quadratics:
1. Find u and v such that u∙v=a∙c and u+v=b
This means u and v are factors of a∙c that when added together equal b
2. Rewrite the expression as (ax^2+ux)+(vx+c)
3. Factor out what you can from each parentheses
4. Factor out a common term
5. Check by multiplying the factors together (FOIL)

Factor\:x^2-5x+6
1. Find u and v such that u∙v=a∙c and u+v=b
a∙c=1∙6=6
6 can be written as the product of 1 and 6, -1 and -6, 3 and 2, or of -3 and -2. We need to pick the factors of 6 that equal -5.
-3+(-2)=-5
-3∙-2=6
u=-3,\:v=-2
2. Rewrite the expression as (ax^2+ux)+(vx+c)
(x^2+(-3x))+(-2x+6)
(x^2-3x)+(-2x+6)
3. Factor out what you can from each parentheses
(x^2-3x)+(-2x+6)
x(x-3)+2(-x+3)
x(x-3)-2(x-3)
We factored out a -1 from (-x+3) on the last step because we want the expressions inside the parentheses to be the same for step 4.
4. Factor out a common term
x(x-3)-2(x-3)
(x-3)(x-2)
5 .Check by multiplying the factors together
(x-3)(x-2)=x^2-2x-3x+6=x^2-5x+6

Factor\:2x^2+x-6
1. Find u and v such that u∙v=a∙c and u+v=b
a∙c=2∙-6=-12
Factors of -12:    -12 and 1, 12 and -1, -6 and 2, 6 and -2, -4 and 3, 4 and -3
4-3=1
4∙-3=-12
u=-4,\:v=3
2. Rewrite the expression as (ax^2+ux)+(vx+c)
(2x^2-4x)+(3x-6)
3. Factor out what you can from each parentheses
(2x^2-4x)+(3x-6)
2x(x-2)+3(x-2)
4. Factor out a common term
(x-2)(2x+3)
5. Check by multiplying the factors together
(x-2)(2x+3)=2x^2+3x-4x-6=(2x)^2-x-6

The best advice for factoring quadratics is to practice factoring as many quadratics as you can. The more you practice factoring, the faster you will get. Soon, you’ll be able to skip the steps and factor it all in your head within seconds. For more help or practice on the topic, check out  Symbolab’s Practice.

Until next time,

Leah

### Symbolab Study Groups…groups that work

stud•y group
noun
plural noun: study groups
a group of people who meet to study a particular subject and then report their findings or recommendations.

Study groups are great; more brainpower, boost motivation, support system.  But let’s be honest, it’s not always the most effective way to learn.  Sessions can turn into social events, schedule doesn’t work for everyone, we’re not always prepared for the sessions…

Symbolab Groups is a game changer.  There’s no better way to connect, share notes, and work through difficult problems together.  Symbolab Groups is a stress free study environment.  You get all the benefits of a study group minus the distractions.

Symbolab helps you stay focused.  No need to worry about missing out on group meetings, or not taking notes.  Stuck on a problem?  Just ask, your friends can help you instantly. Share problems, exercises and graphs, start a discussion, answer questions.

Symbolab Groups are there for you, always.

Cheers,
Michal

## Wednesday, September 28, 2016

### Middle School Math Solutions – Polynomials Calculator, Multiplying Polynomials

Multiplying polynomials can be tricky because you have to pay attention to every term, not to mention it can be very messy. There are a few ways of multiplying polynomials, depending on how many terms are in each polynomial. In this post, we will focus on how to multiply two term polynomials and how to multiply two or more term polynomials.

Multiply two term polynomials

When multiplying polynomials with two terms, you use the FOIL method. The FOIL method only works for multiplying two term polynomials. FOIL stands for first, outer, inner, last. This lets you know the order of how to distribute and multiply the terms. Let’s see how it works.

After FOILing, multiply the terms, group like terms, and add like terms if there are any.
Here is another helpful identity to use when multiplying two term polynomials:
(a+b)(a-b)=a^2-b^2
Multiplying these polynomials is pretty simple because if you memorize these identities then you just plug in the values and have an answer.

Multiplying multiple term polynomials

You cannot use the FOIL method to multiply these polynomials. Instead, you have to multiply each term in one polynomial by each term in the other. You can do this by multiplying each term of one polynomial by the other polynomial. This can be tricky because it is easy to miss one term. When we do examples of this, it will become easier to understand how to solve them.

When multiplying polynomials, you may come across multiplying variables with exponents by variables with exponents. In this case, we use this exponent rule:
x^n\cdot x^m=x^(n+m)
For this rule, the base or variable must be the same. When multiplying variables with exponents, you add the exponents together.

Let’s see some examples to understand how to multiply polynomials.
(2x-1)(5x-6)
We will use the FOIL method to solve this.
1.   Use FOIL identity
(2x-1)(5x-6)
2x\cdot 5x+2x\cdot -6+(-1)\cdot 5x+(-1)\cdot -6
2.   Multiply terms
10x^2-12x-5x+6
3.   Group like terms
10x^2-12x-5x+6
(Luckily, everything was already grouped together)

10x^2-17x+6

(2x^2+6)(2x^2-6)
Here, we can use another one of the identities for multiplying two term polynomials.
1.   Use (a+b)(a-b)=a^2-b^2
(2x^2+6)(2x^2-6)
(2x^2 )^2-6^2
2.   Simplify
4x^4-6^2
4x^4-36
(x^2+2x-1)(2x^2-3x+6)
1.   Multiply each term in one polynomial by the other polynomial
x^2 (2x^2-3x+6)+2x(2x^2-3x+6)-1(2x^2-3x+6)
2.   Distribute and multiply
2x^2\cdot x^2-3x\cdot x^2+6\cdot x^2+2x^2\cdot 2x-3x\cdot 2x+6\cdot 2x+2x^2\cdot -1-3x\cdot -1+6\cdot -1
2x^4-3x^3+6x^2+4x^3-6x^2+12x-2x^2+3x-6
3.   Group like terms
2x^4-3x^3+6x^2+4x^3-6x^2+12x-2x^2+3x-6
2x^4-3x^3+4x^3+6x^2-6x^2-2x^2+12x+3x-6
2x^4+x^3+6x^2-6x^2-2x^2+12x+3x-6
2x^4+x^3-2x^2+12x+3x-6
2x^4+x^3-2x^2+15x-6

Multiplying polynomials looks intimidating, but as long as you keep your work neat and double check your work, it should be pretty easy. Practice will be one of the biggest things that will help you. The more you practice, the easier multiplying polynomials will be because you will get the hang and flow of how to multiply them. Check out Symbolab’s Practice for more help and practice.

Until next time,

Leah

## Tuesday, September 20, 2016

### Middle School Math Solutions – Polynomials Calculator, Subtracting Polynomials

In the previous post, we talked about how to add polynomials. In this post, we will talk about subtracting polynomials. The key to subtracting polynomials is to make sure that you distribute the minus sign to the expression in the parentheses. Imagine that instead of – there is a -1. This should help you visualize why you are distributing the minus sign. Once the minus sign is distributed, the minus sign will turn into a plus sign. Then, you’ll be able to add the polynomials together. Let’s see the steps for subtracting polynomials.

Steps:
1. Distribute the negative sign
• Include this step only if there is a minus sign in front of a polynomial in parentheses
2. Remove the parentheses
• Include this step if there are polynomials in parentheses
3. Group like terms
• Put together and order like terms (terms with the same variables and the same exponent)
• Add the coefficients of the like terms

Let’s see some examples to better understand distributing the minus sign.

(x^2+2x-1)-(2x^2-3x+6)
1.  Distribute the negative sign
(x^2+2x-1)-(2x^2-3x+6)
(x^2+2x-1)-1(2x^2-3x+6)
(x^2+2x-1)+(-2x^2+3x-6)
When we imagine that there is a -1 instead of – outside the parentheses, it is easy to see and remind ourselves that we have to distribute the negative. Distributing the negative sign allows that minus sign to turn into a plus sign.

2.   Remove the parentheses
x^2+2x-1+-2x^2+3x-6
x^2+2x-1-2x^2+3x-6
3.  Group like terms
x^2+2x-1-2x^2+3x-6
x^2-2x^2+2x+3x-1-6
x^2-2x^2+2x+3x-1-6
-x^2+5x-7

(2x^3+2x-1)-(2x^2-5x-6)
1.  Distribute the negative sign
(2x^3+2x-1)-(2x^2-5x-6)
(2x^3+2x-1)-1(2x^2-5x-6)
(2x^3+2x-1)+(-2x^2+5x+6)
2.  Remove the parentheses
2x^3+2x-1+-2x^2+5x+6
2x^3+2x-1-2x^2+5x+6
3.  Group like terms
2x^3+2x-1-2x^2+5x+6
2x^3-2x^2+2x+5x-1+6
2x^3-2x^2+7x-1+6
2x^3-2x^2+7x+5

(4x^3-x^2+x-2)-(-x^2+3)
1.  Distribute the negative sign
(4x^3-x^2+x-2)-(-x^2+3)
(4x^3-x^2+x-2)-1(-x^2+3)
(4x^3-x^2+x-2)+(x^2-3)
2.  Remove the parentheses
4x^3-x^2+x-2+x^2-3
3.  Group like terms
4x^3-x^2+x-2+x^2-3
4x^3-x^2+x^2+x-2-3
4x^3+x-2-3
4x^3+x-5
Subtracting polynomials takes some practice to get the hang of distributing and remembering to distribute. Once you’ve mastered adding polynomials, subtracting them should be simple. For more practice, check out Symbolab’s Practice.

Until next time,

Leah

## Monday, August 15, 2016

### Middle School Math Solutions – Polynomials Calculator, Adding Polynomials

A polynomial is an expression of two or more algebraic terms, often having different exponents. Adding polynomials is pretty simple. The only tricky part is that there are different terms. The key is to add the like terms. Like terms are terms that have the same variable with the same exponent, even if they have different coefficients. There are three steps to adding polynomials:
1. Remove parentheses - only include this step if the polynomials are in parentheses
2. Group like terms - put together and order like terms (terms with the same variable and same exponent)
3. Add like terms - add the coefficients of the like terms

Let’s check out some examples step by step.

(x^2+2x-1)+(2x^2-3x+6)
1. Remove parentheses
x^2+2x-1+2x^2-3x+6
2. Group like terms
x^2+2x^2+2x-3x-1+6
When I group like terms, I like to group and put the terms in order from the largest exponent to the smallest exponent.

3x^2+2x-3x-1+6
3x^2-x-1+6
3x^2-x+5

(2x^3+4x+7)+(2x^2+5x-6)
1. Remove parentheses
2x^3+4x+7+2x^2+5x-6
2. Group like terms
2x^3+2x^2+4x+5x+7-6
2x^3+2x^2+9x+1

4x+(4x-2)+(x^2-3)
1. Remove parentheses
4x+4x-2+x^2-3
2. Group like terms
x^2+4x+4x-2-3
x^2+8x-2-3
x^2+8x-5

Adding polynomials is pretty simple and requires just a few steps. The trickiest part about adding polynomials is making sure that you pay attention the exponents. It is easy to overlook and confuse a 2 for a 3. For more practice on this topic, check out Symbolab’s Practice.

Until next time,

Leah

## Thursday, June 9, 2016

### 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

\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

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

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.

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 . . .

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

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.

\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

\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}

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
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…