Middle School Math Solutions – Expand Calculator, Distributive Law

The distributive law helps with multiplication problems by breaking down large numbers into smaller numbers. In this blog post, we will talk about the distributive law and how to use it.

The law says that multiplying a number by a group of numbers added together is the same as multiplying each separately. What does this mean? Here we can see it in a formula:

You can see that the multiplication of a has been distributed among the sum of b and c.

Let’s see some examples of how to use the distributive law.

First example (click here):

                                                                Expand 4(2x+5)

1. Define the values for a, b, c

                                                               a=4,   b=2x,   c=5

Refer to the formula to see what these values are.

2. Plug these values into the distributive law formula

                                                               4(2x+5)=4∙2x+4∙5

3. Simplify your answer

                                                                4∙2x+4∙5=8x+20

8x+20 is our answer.

Next example (click here):

                                                                Expand 5(10-9p)

1. Define the values for a, b, c

                                                              a=5,   b=10,   c=-9p

2. Plug these values into the distributive law formula

                                                            5(10-9p)=5∙10+5∙(-9p)

3. Simplify your answer

                                                             5∙10+5∙(-9p)=50-45p

50-45p is the answer.

In this last example, we will see another application of the distributive law.

Last example (click here):

                                                                     5(106)

Just by looking at this problem, this might be difficult to calculate quickly. So let’s use the distributive law.

1. Make 106 the sum of two numbers

                                                                 106=100+6

2. Plug this in for 106

                                                                  5(100+6)

3. Define values a, b, c

                                                            a=5,   b=100,   c=6

4. Plug these into the distributive law formula

                                                           5(100+6)=5∙100+5∙6

5. Simplify your answer

                                                         5∙100+5∙6=500+30=530

530 is the answer.

As you can see, the distributive law is very handy. Therefore, I highly suggest you memorize this formula and become familiar with how to use it. You can do so by doing as many examples as you can. For practice examples and more help, check out Symbolab’s Practice

Until next time,

Leah

High School Solutions – Functions Calculator, Inverse

Last blog posts, we focused on how to find the domain and range of functions. In this blog post, we will discuss inverses of functions and how to find the inverse of a function.

An inverse of a function f(x), denoted f^(-1)(x), is a function that reverses or undoes f(x). What does that mean? This means that the domain or inputs of a function is the range or output of the function’s inverse and the range or outputs of a function is the domain or inputs of the function’s inverse. If f(x)=y, then f^(-1)(y)=x. Let’s see some pictures to better understand.

                                                         
It is important to note that some functions have more than one inverse. For example, quadratic equations have two inverses because the negative and positive value for an input goes to the same y value (For x^2+4, when x = 1 and x = -1 , we get y = 5). When you find the inverse of quadratics, you’ll notice you get two inverses, one is a postive square root and the other is a negative square root. We will see an example of this later in the post.

Now, that you’ve got the concept of what the inverse of a function is, we will see the steps on how to find the inverse of a function.

Steps to find the inverse of a function:

1. Replace y for f(x)
2. Solve for x
3. Substitute y = x
4. If you want to check the function, then f(f^(-1) (x))=x  and f^(-1) (f(x))=x

The steps are pretty simple to remember and follow. Now, let’s see some examples.

First example (click here):

                                                 Find the inverse of f(x)=3x+5

1. Replace y for f(x)

                                                                 y=3x+5

2. Solve for x

                                                                 y=3x+5

                                                                 y-5=3x

                                                               \frac{y-5}{3}=x

3. Substitute y = x

                                                       \frac{x-5}{3}=y=f^(-1) (x)

4. Check to make sure it is correct (f(f^(-1)(x))=x  and f^(-1)(f(x))=x)

                                              f^(-1)(\f(x))=\frac{(3x+5)-5}{3}=\frac{3x}{3}=x

                                             f(f^(-1)(x))=3(\frac{x-5}{3})+5=x-5+5=x

Next example (click here):

                                                     Find the inverse of f(x)=\sqrt{x+3}

1. Replace y for f(x)

                                                                     y=\sqrt{x+3}

2. Solve for x

                                                                     y=\sqrt{x+3}

                                                                     y^2=x+3

                                                                     y^2-3=x

3. Substitute y = x

                                                             x^2-3=y=f^(-1)(x)

4. Check to make sure it is correct (f(f^(-1)(x))=x  and f^(-1)(f(x))=x)

                                               f^(-1)(\f(x))=(\sqrt{x+3})^2-3=x+3-3=x

                                              f(f^(-1)(x))=\sqrt{(x^2-3)+3}=\sqrt{x^2}=x

Last example (click here):

                                                   Find the inverse of f(x)=2x^2-2

1. Replace y for f(x)

                                                              y=2x^2-2                      

2. Solve for x

                                                                 y+2=2x^2

                                                               \frac{y+2}{2}=x^2

                                                              ±\sqrt{\frac{y+2}{2}}=x

3. Substitue y = x

                                                      ±\sqrt{\frac{x+2}{2}}=y=f^(-1)(x)

4. Check to make sure it is correct (f(f^(-1)(x))=x  and f^(-1)(f(x))=x)

You can do this step on your own. We will skip it to save some time.

As you can see, finding the inverse of a function is pretty simple. It is easy to make algebraic answers, so make sure you check your answer to see if it is correct. With practice, you’ll be able to master this topic easily. For more help or practice on this topic, check out Symbolab’s Practice.

Until next time,

Leah