While systems of two linear equations with two unknowns can be solved using algebra, it is also possible to solve them by graphing each equation. In the examples below, you will see how to find the solution to a system of equations from a graph, how to determine if there are no solutions, and how to determine if there are infinitely many solutions.

Systems with one solution

Consider the following system of equations.

\begin{array}{l} x+2y=8\\ \\x-2y=-4\end{array}

If you solve this system using the addition method, you will find that x = 2 and y = 3. But, what does this look like on a graph?

solution-to-system-of-equations

Notice that the two lines intersect at exactly one point, and that this point is (2, 3). For any system with one solution, the graphs will intersect at one point and this point will represent the solution to the system of equations.

Systems with no solutions

The following system of equations has no solution.

\begin{array}{l} x-y=-2\\ \\x-y=1\end{array}

How would you know? One way would be to try and solve the system, and see that you get an untrue statement. Another would be to realize that the first equation is saying that x–y is one number, while the second equation is saying that x–y has a different value. These can’t both be true. The difference must have a single value.

You could also tell by looking at the graph of the system of equations.

no-solution-to-system-of-equations

The two lines are parallel, meaning that they will never intersect. For any system of equations, if there is no solution the the system, the two graphs will not intersect at any point. For linear equations, this will result in a graph of two parallel lines.

Infinitely many solutions.

Let’s look now at a system of equations with infinitely many solutions. While it will not always be so obvious, you can tell that this system has infinitely many solutions because the second equation is just a multiple of the first.

\begin{array}{l} x+y=-2\\ \\2x+2y=-4\end{array}

If you were to graph these two equations, you would get the following result.

infinitely-many-solutions

Even though the system of equations includes two linear equations, you end up with a single line. This will always be the case when there are infinitely many solutions. The solution set is actually all points along the line.

Can there be two or three or four solutions?

You may have noticed that we covered only three cases: one solution, no solutions, and infinitely many solutions. When working with linear equations, these are the only possibilities.

This is because the graphs of the equations are lines. A straight line can only intersect another straight line at one point, at no points, or at all points (they are just the same line). If two lines intersected at, say exactly two points, then the lines would have to bend and would no longer be lines.

If you are working with nonlinear equations, however, things are quite different! For example, in the graph below, the circle and the line intersect at exactly two points.

graph-of-system-of-equations-ex2

These two points would represent solutions to the system of equations. But again, this is only because both graphs are not from a linear equation.

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Matrices are used in a variety of different math settings from algebra and linear algebra to finite math. Of course, to be able to work with matrices, you need to understand the notation used and simple (but important) ideas like the size of a matrix.

Elements of a matrix

A matrix is a way of organizing numbers into rows and columns. It may represent a system of equations, a real-life situation, or simply be a matrix of interest all on its own. Matrices are most commonly labeled with capital letters such as A, B, or C. Below, you can see a matrix we will refer to as “matrix A“.

example-of-a-matrix

The numbers within the matrix, are referred to as elements. One way to talk about a specific element is to use a lowercase letter and label it with the row and column of the element.

elements-of-a-matrix

Note that you could also say “5 is the (1,2) entry” and “8 is the (2,3) entry”.

The size of a matrix

The common theme with matrices is “think rows-columns”, and this holds even when discussing the size or dimension of a matrix. If a matrix has 4 rows and 6 columns, we say it is a 4 x 6 matrix (read: four by six).

size-of-a-matrix

Summary

As you study matrices, remember the following ideas:

  • Elements are referred to by their location in terms of the row, then column.
  • The size of a matrix is: (number of rows) x (number of columns). For a 2 x 3 matrix, you would say “the size of the matrix is 2 by 3”.

Continue studying matrices

Now that you have reviewed important notation and ideas like the size of a matrix, you are ready to study how to add, subtract, and multiply matrices.

Operations with matrices:

As you will see below, in order to add or subtract polynomials, you really just need to pay attention to signs and combine like terms. This is especially important with subtraction, as you will be distributing a negative sign.

Table of contents:

  1. Adding polynomials
  2. Subtracting polynomials

Adding polynomials

There are only really two steps required for adding two polynomials. For example, consider adding x^2 + 4x + 1 and 4x^2 - 2. In the picture, like terms are underlined the same number of times.

how-to-add-polynomials

For practice, let’s apply these steps to a couple more problems.

Example

Find the sum: -3x^4 + 5x^3 - 7 and -8x^4 - 10x^3 + 3x  + 2

Remember here that “find the sum” is another way of saying “add”. So, we will first set up the addition and then combine like terms. Just as in the picture, like terms are underlined.

(-3x^4 + 5x^3 - 7) + (-8x^4 - 10x^3 + 3x  + 2)

= -\underline{3x^4} + \underline{\underline{5x^3}} - \underline{\underline{\underline{7}}} -\underline{8x^4} -\underline{\underline{ 10x^3}} + 3x  + \underline{\underline{\underline{2}}}

= \boxed{-11x^4 -5x^3 + 3x -5}

In some problems, you may find that the set up is already done. In these cases, you need only worry about combining like terms.

Example

Simplify: (14y^3 + 2y^2) + (3y^3 + 7y^2)

Although the instructions say simplify, you can see that this is actually a polynomial addition problem. Therefore, we can find the answer with the following steps.

(14y^3 + 2y^2) + (3y^3 + 7y^2)

=\underline{14y^3} + \underline{\underline{2y^2}} + \underline{3y^3} + \underline{\underline{7y^2}}

=\boxed{17y^3 + 9y^2}

Subtracting polynomials

When subtracting two polynomials, there is one extra step we must apply before combining like terms: distribute the negative. As an example, consider (x^2 + 5) - (2x^2 -1).

how-to-subtract-polynomials

Notice that when you distribute the negative, you distribute it to every term in the parentheses. Pay close attention to this in the next couple of examples.

Example

Find the difference: (-4x^3 + x^2 - 1) - (-2x^3 + 3x^2 + 5)

Just as “find the sum” means to add, “find the difference” means to subtract. Therefore, we will set up the subtraction, distribute the negative, and then collect like terms to find the answer.

(-4x^3 + x^2 - 1) - (-2x^3 + 3x^2 + 5)

= -\underline{4x^3} + \underline{\underline{x^2}} - 1 + \underline{2x^3} - \underline{\underline{3x^2}} - 5

= \boxed{-2x^3 -2x^2 -6}

As you can see, the main idea is to keep track of the signs. Also note, that just as with addition, instructions may just say “simplify”. Even so, you will still distribute the negative before combining like terms.

Example

Simplify: (12x^5 -x) - (18x^5 + 4x^2 + x + 1)

=\underline{12x^5} -\underline{\underline{x}} - \underline{18x^5} - 4x^2 - \underline{\underline{x}} - 1

=\boxed{-6x^5 - 4x^2 - 2x - 1}

Summary

Now that you have studied how to add and subtract polynomials, let’s review the overall steps.

  1. Set up the addition or the subtraction.
  2. Distribute the negative (if subtracting).
  3. Combine like terms / simplify.

You will find later that these skills are important when working with rational expressions, long division of polynomials, and even when working with functions!

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When working with exponents, there are two new words that come up: exponent and base. As an example, in x^2, 2 is the exponent while x is the base and in  4^{12}, 12 is the exponent while 4 is the base.

exponent-and-base

Table of contents:

  1. Whole number exponents are about multiplication
  2. Negative exponents
  3. Fractions as exponents
  4. Exponents on the calculator

Whole number exponents are about multiplication

As long as the exponent is a positive whole number, you can think of it as telling you how many times you should multiply the base by itself.

  2^3=2\cdot2\cdot2=8
  x^3=x\cdot x\cdot x
  5^2=5\cdot5=25

As you can see,   4^{12} means “4 multiplied by itself 12 times” and that is a really big number. Many values that you calculate from exponents will be quite large.

There are a couple of things to note:

  • Anything with an exponent of zero is defined to be 1. So,   4^{0}=1,   100^{0}=1.
  • Ok right above, when I said anything, I meant “almost anything”. 0^0 is an indeterminate form. For everything else however, the rule above holds.
  • An exponent of 1 is the same as just writing the number by itself:   4^{1}=4.

Negative Exponents

Negative exponents are treated a bit differently. By definition, if we have a positive number n and a nonzero b, then the following rule holds:


b^{-n} = \dfrac{1}{b^n}

Using this rule, if you see a value with a negative exponent, then it can be rewritten as 1 “over” that term with a positive exponent. Here are a few examples of how this works.

3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}
10^{-1} = \dfrac{1}{10^1} = \dfrac{1}{10}
(-2)^{-4} = \dfrac{1}{(-2)^4} = \dfrac{1}{16}

As you can see, it is simply about following the formula when dealing with negative exponents. After this, you can simply calculate the value using the rules above.

Fractions as exponents

In algebra and more advanced math, it is very common to see an exponent of one-half or some other fraction. This is one way to represent terms involving roots, or radical terms. By definition (if n is nonzero):


b^{\frac{m}{n}}= \sqrt[n]{b^m}

Working with these types of terms (radical terms) is a more advanced topic and will be the subject of a different article, but it is important that you be able to rewrite such terms using the definition as below.

3^{\frac{1}{2}}=\sqrt{3^1}=\sqrt{3}
4^{\frac{2}{3}}=\sqrt[3]{4^2} = \sqrt[3]{16}

As you study terms involving exponents, you will find that you need to combine rules quite often. For example, the rule above can be combined with the rule for negative exponents to simplify even more complex expressions.

Using a Calculator

Finding the value of terms with exponents on scientific or graphing calculators is quite easy. The usual key you will need will use a carat symbol ^ to represent that the next number is an exponent. So, 3^4 can be represented as 3^4 in most calculators. On the TI83 or TI84:

exponent-on-calculator

screenshot-exponent-calculator

You can also always use the carat symbol on google (just type it into google! they now do automatic calculations for you) or wolfram alpha.

It doesn’t matter what function you are looking at, the y intercept is the point at which the graph of that function crosses the y-axis. That means the point can always be written (0, c) for some number c.


The y-intercept is at y = -1.

The y-intercept is at y = -1. As a point, we would write (0, -1)



Finding y Intercepts

Along the y-axis, the x value of any point is zero. This means that to find the y intercept of any function, we can just let x = 0 and solve for y. For example, consider the function y=\frac{1}{3}x+4. If we let x = 0, then y=\frac{1}{3}(0)+4 = 4. As a point, this would be the point (0, 4).

Similarly, the function in the graph at the top of the page is, y=x^2-1. When x = 0, y = 0^2 - 1 = -1., meaning the y intercept is -1, the same as on the graph. If this is still confusing, in the video to the right, I go through a couple more examples.

A Closer Look at the y Intercept

Now that we have seen how to find them, there are two interesting questions that can come up:

  1. Can a function have more than one y intercept?
  2. Can a function have no y intercept?

In answering these, remember that a function can only have one output (y-value) for each input (x-value). If a function has more than one y intercept, that means that there are two y’s for the input x = 0. This isn’t possible, so no, it is not possible for a function to have more than one y-intercept.

What about no y intercept? Well, take a look at the graph of y=\frac{1}{x} below. It never crosses the y-axis, so it has no y-intercept.


This function has no y-intercept.

This function has no y-intercept.


What happened? If you let x=0, you get y = \frac{1}{0} which is undefined. This is just one example, but there are other similar functions, so yes, it is certainly possible for a function to not have a y-intercept.

Using a Graphing Calculator

Graphing calculators can be used to find x and y intercepts pretty quickly. A while back, I made a short video that can show you how this works. You can find it on the mathbootcamps youtube page.

What About x-Intercepts?

Well, you can take a look at my discussion on x intercepts right over here!

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Given the graph of any function like y = x – 2, the x intercept is simply the point where the graph crosses the x-axis. There might be just one such point or many. In either case, the idea stays the same. Another way to think about this is, the x intercept is the point on the graph where y = 0. This can be used to easily find the x intercept for any graph. In this guide, we will talk about how to find x intercepts using a graph, an equation, and a graphing calculator.

Finding the x-intercept or intercepts using a Graph

As mentioned above, functions may have one, zero, or even many x intercepts. These can be found by looking at where the graph of a function crosses the x-axis. This point (or these points) is graphed for each of the functions below.


Graph of y = x - 2
The x intercept is 2

The x intercept is 2



Graph of y = x^2 + 1
The x intercepts are -1 and 1.

The x intercepts are -1 and 1.



Graph of y = x^4-8x^3-49x^2+260x+300
The x intercepts are -6, -1, 5, and 10.

The x intercepts are -6, -1, 5, and 10.


How to Find x Intercepts

When the graph of a function is crossing the x-axis, what does the point look like as an ordered pair? For instance, is the x-intercept is said to be 5, what point is this? Well, since there is no “height” to that point, we would write it as (5, 0). In fact, any x-intercept would have a 0 as the y-value.

We can use this to find the x-intercept of intercepts of any function if we have the equation. The general rule is let y = 0 and solve for x. How easy this is depends entirely on the equation!

As an example, let’s use the first equation graphed above:  y = x - 2. If we let y = 0, then we get the equation 0 = x - 2 This equation can be solved by using the usual rules for solving linear equations: we just add 2 to both sides and find the x intercept is x = 2.

Things are a little more complicated with the second example. If  y = 0 in the equation y = x^2 - 1, then we get the quadratic equation 0 = x^2 - 1. Adding 1 to both sides, we find that x^2 = 1 which has solutions 1 and -1 by the square root rule.

Of course, the idea remains the same even when we get to the last (much more complicated) equation y = x^4-8x^3-49x^2+260x+300. Setting y to zero and solving for x would give us all of the x-intercepts we see on the graph. However, this is a bit more algebra than we need for this article so we will save that equation for another day. Instead, if you need a few more examples, I have gone through some different types of functions in this video from the mathbootcamps youtube!



Finding x-intercepts on a Graphing Calculator

This is a bit more involved, but on this old youtube video I go through how to find both x and y intercepts using a TI 84.

What About y Intercepts?

Good question! I go over y-intercepts in this article.

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Matrix multiplication is just one of those things that is not intuitive – at least not at first. You have just had so many years of multiplication meaning one thing and then you have this entirely new definition to work with! It certainly takes some getting used to. (and if you continue to study advanced math – get used to that idea of “getting used to” things)

The easiest way to think of it is “rows hit columns and fill up rows”. This process is repeated until you run out of rows in the first matrix. This is seen in the following animation (no sound).

You can read more about how to multiply matrices here, including examples and important properties: Multiplying matrices.

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While there are plenty of methods to get a good hand drawn graph of a linear equation, many plot more points than necessary and (if you aren’t careful) misrepresent the location of two important points: the x and y intercepts.

Using the intercepts to plot your line takes care of both of these problems and more importantly: its fast! In this video, I will use two examples to show you exactly how it’s done.

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At this point, I haven’t even written out any equation or expression and you already know what I’m talking about. There are problems that come up in algebra where it seems like you should be able to cancel a variable out and move on. Yet, in these very situations your professor keeps telling you it can’t happen!

It may not always seem like it, but there really is a pattern and a simple rule of when you can and can’t cancel a term or a variable. Let’s focus on rational expressions specifically. Those are expressions (like the one below) where both the numerator and the denominator are polynomials.


\dfrac{x+5}{x^2+1}

When can you cancel terms in a rational expression?

You can cancel any matching factors that occur in both the numerator and the denominator. Let’s use numbers to understand this a little better:


\dfrac{4}{14}


In the fraction above, 2 is a factor of 4 since 2 x 2 is 4. Similarly, 2 is a factor of 14 since 2 x 7 = 14. Based on my rule, I should be able to cancel the 2’s and get a fraction that is still the same thing! Let’s see:



\dfrac{4}{14}=\dfrac{2(2)}{2(7)}=\dfrac{2}{7}

If you check on a calculator you will find that 4 divided by 14 is the same as 2 divided by 7 – so this rule seems to be working here. What is really going on?

Why does this work?

Every number other than zero, when divided by itself is 1. Since a fraction simply represents division (in one sense), \dfrac{2}{2} = 1 and we really just have 1\times\dfrac{2}{7}.

Trying it with Variables

Now that everything makes some sense with numbers, let’s try variables. For us, the variables represent numbers so they will behave in the same way. Take for instance the expression


  \dfrac{x^2-x}{x^3}, x\neq 0

The numerator has a couple of factors which we can find by factoring out an x (doesn’t that phrase make a lot more sense now?!): x^2-x = x(x-1). This shows us that x and x-1 are factors of the numerator. Looking at the denominator, we see x is also a factor since x^3=x(x^2). Remember, when both the numerator and the denominator share a factor, you can cancel that factor since it is really just 1.

  \dfrac{x^2-x}{x^3} = \dfrac{x(x^2-1)}{x(x^2)}= \dfrac{x^2-1}{x^2}

The only time you can’t cancel terms in the numerator and denominator is when they are both NOT factors.. That’s why you can’t cancel x in \dfrac{x-5}{x}. The x is not a factor of the numerator; its just a term being added. Cancelling the x here would be like cancelling the 5 in \dfrac{5+1}{5} and saying that is 1. If you check it in a calculator it certainly isn’t!

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A quadratic equation is any equation that can be written as ax^2+bx+c=0, for some numbers a, b, and c (a is nonzero). There are many ways to solve equations like this, including graphing, factoring, and using technology. One other method is using the quadratic formula.

the-quadratic-formula

Applying the quadratic formula

Looking at the formula, you can see that a, b, and c are the numbers straight from your equation. Applying this formula is really just about determining the values of a, b, and c and then simplifying the results. Let’s take a look at a couple of examples.

Example

Solve: x^2-x=6

Before we do anything else, we need to make sure that all the terms are on one side of the equation. As you can see above, the formula is based on the idea that we have 0 on one side.

Subtract 6 from both sides to get:

x^2-x-6=0

Now that we have it in this form, we can see that:

a=1 , b= -1 , c=-6

Why are b and c negative? The formula is based off the form ax^2+bx+c=0 where all the numerical values are being added and we can rewrite x^2-x-6=0 as x^2 + (-x) + (-6) = 0.

To keep it simple, just remember to carry the sign into the formula.

Once you have the values of a, b, and c, the final step is to substitute them into the formula and simplify.

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-6)}}{2(1)}

=\dfrac{1\pm\sqrt{1+24}}{2}

=\dfrac{1\pm\sqrt{25}}{2}.

At this stage, the plus or minus symbol (\pm) tells you that there are actually two different solutions:

\dfrac{1+\sqrt{25}}{2}=\dfrac{1+5}{2}=\dfrac{6}{2}=3

and

\dfrac{1- \sqrt{25}}{2}=\dfrac{1-5}{2}=\dfrac{-4}{2}=-2

Therefore the final answer would be:

x=3\;,-2

This particular quadratic equation could have been solved using factoring instead, and so it ended up simplifying really nicely. Often, there will be a bit more work – as you can see in the next example.

Example

Solve: 2x^2+2x-7=0

This time we already have all the terms on the same side. So, we just need to determine the values of a, b, and c.

a=2 , b=2 , c=-7

Applying the formula:

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-2\pm\sqrt{(2)^2-4(2)(-7)}}{2(2)}

=\dfrac{-2\pm\sqrt{4+56}}{4}

=\dfrac{-2\pm\sqrt{60}}{4}

=\dfrac{-2\pm 2\sqrt{15}}{4}

Notice that 2 is a FACTOR of both the numerator and denominator, so it can be cancelled.

=\dfrac{-1\pm\sqrt{15}}{2}

This answer can not be simplifed anymore, though you could approximate the answer with decimals. Therefore the final answer is:

x=\dfrac{-1+\sqrt{15}}{2},\dfrac{-1-\sqrt{15}}{2}

Complex Solutions

When using the quadratic formula, it is possible to find complex solutions – that is, solutions that are not real numbers but instead are based on the imaginary unit, i. Recall the following definition:

i = \sqrt{-1}

If a negative square root comes, your equation has complex solutions which can be written in terms of i.

Example

Solve: x^2-2x+5=0

Just as in the previous example, we already have all the terms on one side. So, we will just determine the values of a, b, and c and then apply the formula.

a=1 , b=-2 , c=5

Applying the formula:

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(1)(5)}}{2(1)}

=\dfrac{2\pm\sqrt{4-20}}{2}

=\dfrac{2\pm\sqrt{-16}}{2}

A negative value under the square root means that there are no real solutions to this equation. However, there are complex solutions. Using the definition of i, we can write:

\sqrt{-16} = 4i

This allows us to keep simplifying.

=\dfrac{2\pm 4i}{2}

=1 \pm 2i

The solutions to this quadratic equation are:

x=1+2i , 1 - 2i

There are three cases with any quadratic equation: one real solution, two real solutions, or no real solutions (complex solutions). Each case tells us not only about the equation, but also about its graph – something we will study more in a later article!

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