Finite Geometric Series

Happy Wednesday math friends! In today’s post, we are going to go over what the phrase, “finding the sum of a finite geometric series” means and then use the finite geometric series formula to solve an example one step at a time! A finite geometric series happens when we add together a finite amount of terms from a geometric sequence together. We will go into more detail below and as always if you have any questions be sure to check out the video and comment with any questions below.

*If you need a refresher on geometric sequences (otherwise known as a geometric progression) before tackling these types of questions, don’t hesitate to check out this post! And, if you’re looking for an infinite geometric series, as opposed to a finite one, also be sure to check that out.

What does it mean to find the “Sum of the Finite Geometric Sequence”?

We already know what a geometric sequence is: a sequence of numbers that form a pattern when the same number is multiplied or divided to each subsequent term. In the example below, we can see that each number in the sequence is being multiplied by the common ratio, the number 2.

Geometric Sequence Example:

But what happens if we wanted to sum the first 20 terms of our geometric sequence together? Adding 2+4+8+16+……a20 . Notice I used the notation a20 to represent our unknown 20th term of the finite sequence.

Example:

How would we calculate that?  Instead of finding the first 20 terms of our sequence and adding them all together, thankfully there is a better way, which is where our Finite Geometric Series formula comes in handy!

Finite Geometric Series Formula

Why is it called “finite”? Adding the first 20 terms of our geometric sequence are considered to be “finite” because the first 20 terms have a definite ending as opposed to a sequence that is infinite and goes on forever. Adding together an infinite geometric series comes with a different formula.

Finite Geometric Series Formula:

a1=The first term of our sequence. In this case, we can see that the first term will be the number 2 in the example above. Therefore, we can say a1=2.

r= The common ratio is the number that is multiplied or divided by each consecutive term within the sequence. In the example above, each number is multiplied by 2, therefore we can say, r=2.

n= The total number of terms we are trying to sum together. In the example mentioned above, we are trying to sum 20 terms in total, so in this case n=20.

Finite Geometric Series Formula

Now that we have this finite geometric series equation to work with, let’s take another look at our question and apply our finite geometric series formula to answer the solution:

Finite Geometric Series Example:

Step 1: First let’s write out the finite geometric series formula and identify what each part represents/what numbers need to be filled in.

a1=2 The first term of our sequence. a1=2.

r= 2 The common ratio is the number that is multiplied or divided by each consecutive term within the sequence.

n= 20 The total number of terms we are trying to sum together.

Finite Geometric Series Formula

Step 2: Now let’s plug in our numbers into the finite geometric series formula and calculate and solve with the given values.

Finite Geometric Series Formula

We have found our solution! Remember the number here, 2,097,150 represents the sum of the first 20 numbers of the geometric sequence given to us 2+4+8+16+……a20=2,097,150.

Summation Notation – Finite Geometric Sequence

You may come across the finite geometric series formula in a different format, known as summation notation or sigma notation. Writing in summation notation, we are actually writing the sum of the geometric sequence in a different way, but will still come to the same solution as we did using the formula. Check it out below:

Now that we know what each part of this summation notation looks like, let’s actually identify what each part of this equation means:

Now let us go back and try to solve our original question, finding the sum of the first 20 terms of the sequence 1, 4,8, 16, …. all the way to thew 20th term:

Notice we get the same exact solution as we did in the previous example, mission accomplished!

Note! Does the above summation notation totally freak you out? Fear not! Learn more about how summation notation works here,what it means, and don’t be intimidated by these math symbols anymore!

Think you are ready to try questions on your own? Check out similar practice questions and find the sum of each finite series below!

Practice Questions:

1) Find the sum of the first 15 terms of the following geometric sequence:

4, 12, 36, 108, ….

2) Find the finite sum of the first 12 terms of the following sequence and round to the nearest tenth:

128, 64, 32, 16, ….

3) Find the sum of the first 18 successive term of the following geometric sequence and round to the nearest tenth:

400, 100, 25, 6.25, ….

4) Find the sum of the first 12 consecutive terms of the following geometric sequence:

3, 6, 12, 24, ….

Solutions:

1) 28,697,812

2) 255.9

3) 533.3

4) 12,285

Related Posts:

Looking to learn more about sequences? You’ve come to the right place! Check out these sequence resources and posts below. Personally, I recommend looking at the finite geometric sequence or the geometric infinite series posts next!

Geometric Sequence

Recursive Formula

Arithmetic Sequence

Finite Arithmetic Series

Infinite Geometric Series

Golden Ratio in the Real World

Fibonacci Sequence

Still, got questions? No problem! Don’t hesitate to comment with any questions below or check out the video above. Thanks so much for stopping by and happy calculating! 🙂

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How to Construct a 45 Degree Angle with a Compass

Greeting math friends and welcome to another wonderful week of MathSux! In today’s post we are going to break down how to construct a 45 degree angle with a compass. We will take this step by step starting with a simple straight edge, then we will create a 90 degree angle, and finally we will use an angle bisector to bisect that 90 degree angle to get two 45 degree angles. If you have any questions please don’t hesitate to check out the video and step-by-step GIF below. Thanks so much for stopping by and happy calculating! 🙂

How to Construct a 45 degree Angle with a Compass

How to Construct a 45 Degree Angle with a Compass:

Step 1: Using a straightedge or a ruler, draw a straight line, labeling each end point A and B.

Step 2: Next, using a compass, place the point of the compass on the edge of point A and draw a circle.

Step 3: Keeping the same length of the compass, take the point of the compass to the point where the circle and line AB intersect. Then swing the compass to make a new arc on the circle above line AB.

Step 4: Keeping that same length of the compass, go to the new intersection we just made and mark another arc along the circle.

Step 5: Now, take a new length of the compass (any will do), and bring it to one of the intersections we made on the circle.  Then create a new arc above the circle by swinging the compass.

Step 6: Keep the same length of the compass, bring the compass to the other intersection we made on our circle to create a new arc above the circle.

Step 7: Mark a point where these two lines intersect and using a straight edge, connect this intersection to point A. Notice this forms a 90 degree angle (or a right angle).

Step 8: Now to bisect our newly made 90 degree angle, we are going to focus on the pink highlighted points where the original circle intersects with line AB and our newly made line.

Step 9: Using a compass (any length), we are going to take the compass point to one of these hi-lighted points and draw another arc.

Step 10: Keeping that same length of the compass, go to the other highlighted point and make another arc as well.

Step 11: Now with a straight edge, draw a line from point A to the new intersection of arcs we just made.

Step 12: Notice we split or 90 degree angle in half and now have two equal 45 degree angles?!

Still got questions? No problem! Don’t hesitate to comment with any questions below or check out the video above. Thanks for stopping by and happy calculating! 🙂

Constructions and Related Posts:

Looking to construct more than just a square inside a circle? Check out these related posts and step-by-step tutorials on geometry constructions below!

Construct an Equilateral Triangle

Perpendicular Line Segment through a Point

Angle Bisector

Construct a Square Inscribed in a Circle

Altitudes of a Triangle (Acute, Obtuse, Right)

How to Construct a Parallel Line

Bisect a Line Segment

Best Geometry Tools!

Looking to get the best construction tools? Any compass and straight-edge will do the trick, but personally, I prefer to use my favorite mini math toolbox from Staedler. Stadler has a geometry math set that comes with a mini ruler, compass, protractor, and eraser in a nice travel-sized pack that is perfect for students on the go and for keeping everything organized….did I mention it’s only $7.99 on Amazon?! This is the same set I use for every construction video in this post. Check out the link below and let me know what you think!

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Looking for more constructions? Check out how to construct a square inscribed in a circle and an equilateral triangle by clicking on their respective links! And if you’re looking for even more geometry constructions, check out the link here!

Multiplying Radicals Expression

Hi everyone and welcome to MathSux! In today’s post we are going to explore the rules for multiplying radicals, mainly focusing on multiplying binomials expressions that contain radicals. We will go over several types of examples in this post starting with the basics and working our way up to expanding radical binomial expressions and simplifying. If you need a quick review on how to expand binomials using the distributive property/box method, check out this post here. Also, be sure to check out the video and practice questions below. Thanks so much for stopping by and happy calculating! 🙂

What Are the Rules for Multiplying Radicals?

  1. Multiply the coefficients.
  2. Multiply the term or terms under the radical.
  3. Simplify the solution whenever possible.
Multiplying Radicals

Check out more completed Examples of multiplying radicals that involve some simplifying below:

Multiplying Radicals

The above shows a few simple cases of multiplying radicals but let us take a look at how to multiply radicals that are placed within a binomial within these next Examples.

Step 1: To solve this, we are going to use the distributive property to multiply each term by each term, sometimes known as FOIL (for more on the FOIL method, check out this post, here.)

Multiplying Radicals

Step 2: Before we say we are done and have a solution, notice that we can simplify this even more by combining like terms, 12+2=14, we can also combine the like radical terms below because they have the same value of 2 under the radical.

Multiplying Radicals

Now, let’s take a look at another type of Example, where we need to first expand binomials that contain radicals.

Step 1: Notice in with this example, we are going to need to expand this binomial and then use the distributive property (or FOIL) to multiply each term by each term.

Multiplying Radicals

Step 2: In this case, notice that we can simplify our answer even more by combining like terms.

Multiplying Radicals

Try the following practice questions on your own to truly master the topic!

Practice Questions:

Solutions:

Still got questions? No problem! Don’t hesitate to comment with any questions below or check out the video above. Thanks for stopping by and happy calculating! 🙂

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Focus and Directrix of a Parabola: Algebra 2

Happy Wednesday math friends! In today’s post are going to go over what the focus and directrix of a parabola are and how to find them. We will also take a look at how to find the equation of a parabola when given the focus and directrix. You may already be familiar with parabolas and quadratic equations, but with this post we will dive deeper into the definition of a parabola and understand it on a whole new level! Don’t forget to check out the video and practice questions, below. Thanks for stopping by and happy calculating! 🙂

Parabola? Vertex? Focus? Directrix?  What does it all mean?!

Focus and Directrix of a Parabola:

Parabola: A set of all points in a plane that are equidistant from a given point (the focus) and a given line (the directrix). The equation of a parabola in relation to the focus and the directrix is:

Vertex: The maximum or minimum point on a parabola.  This coordinate point always lands on the parabola itself. Takes the coordinate form (a,(.5(b+k))) from the parabola equation above.

Focus: A coordinate point that is “inside” the parabola that has the same distance from the vertex as it does the distance between the vertex and directrix. Usually denotes as (a,b) within the parabola equation above.

Directrix: A horizontal line denoted as, “y= ” or “k=  “, that is the same distance from the vertex when compared to the distance between the vertex and the focus.

The above definitions explain that every point on a parabola is equidistant from the vertex to the focus as it is to the directrix.  In the picture below, we can see that the vertex is 2 units away from the focus and 2 units away from the directrix. They are equidistant:

Focus and Directrix of a Parabola:

This phenomenon, works not just the vertex, but for each and every point found on a parabola!

We can see below the equidistance between several points on the parabola that compare the distance between the vertex to the directrix and the distance between the vertex to the focus. Notice that they are all the same!

Focus and Directrix of a Parabola:

Ready to test our new knowledge with the following Example?

Step 1: First, let’s sketch out our parabola with the given information.  We know that the vertex is at point (2,-1) a point that lies on the parabola itself, so let us map that out.  We also know that the directrix is k=3 and is represented as a line as y=-3.

Focus and Directrix of a Parabola:

Step 2: Next, let us measure the distance between the directrix and the vertex.  The distance between the vertex and directrix is 2 units. That means we must also measure the 2 units on the opposite side of the vertex to find the value of the focus. This leads us to the point (2,1).

Focus and Directrix of a Parabola:

Now for part two of our question, how to find the equation of a parabola now that we have the value of the focus and the given directrix.  For this, we will need to use the funky looking equation for a parabola mentioned earlier in this post.

Step 1: First we need to gather all of our information, the formula for the equation of a parabola , the given directrix, k=-3 and the focus we found in the previous example (2,1) which corresponds to the formula as a=2 and b=1.

Step 2: Now, let’s plug everything into our formula where a=2, b=1, and k=-3, to find the equation to our parabola:

Practice Questions:

Solutions:

Still got questions? No problem! Don’t hesitate to comment with any questions below. Thanks for stopping by and happy calculating! 🙂

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Check out similar posts for Algebra 2/Trig. here!

Looking for more on Quadratic Equations and Functions? Check out the following Related posts!

Factoring Review

Factor by Grouping

Completing the Square

4 Ways to Factor a Trinomial

Is it a Function?

Imaginary and Complex Numbers

Quadratic Equations with 2 Imaginary Solutions

What is the Discriminant?

How to find the Area of a Parallelogram: Geometry

Hey math peeps! In today’s post, we are going to go over how to find the area of a parallelogram. There is an easy formula to remember, A=bh, but we are going to look at why this formula works in the first place and then solve a few examples. Just a quick warning: The following examples do use special triangles and if you are need of a review, check out the posts here for 45 45 90 and 30 60 90 special triangles. Also, don’t forget to watch the video and try the practice problems below. Thanks so much for stopping by and happy calculating!

Area of a Parallelogram Formula:

How to find the Area of a Parallelogram

Why does the Formula for Area of a Parallelogram work?

Did you notice that the formula for area of a parallelogram above, base times height, is the same as the area formula for a rectangle?  Why?

If we cut off the triangle that naturally forms along the dotted line of our parallelogram, rotated it, and placed it on the other side of our parallelogram, it would naturally fit like a puzzle piece and create a rectangle! Check it out below:

How to find the Area of a Parallelogram

Now that we know where this formula comes from, let’s see it in action in the examples below:

Example #1:

How to find the Area of a Parallelogram

Step 1: Write out the formula:

Step 2:  Fill in the formula with values found on our parallelogram, b=12 inches h=4 inches, and multiply them together to get 48 inches squared.

How to find the Area of a Parallelogram

That was a simple example, but lets try a harder one that involves special triangles.

Example #2:

How to find the Area of a Parallelogram

Step 1: Write out the formula:

Step 2:  Label the values found on our parallelogram, b=10 ft and notice that we are going to need to find the value of the height.

Step 3: In order to find the value of the height, we need to remember our special triangles! We are not given the value of the height, but we are given some value of the triangle that is formed by the dotted line.  Let us take a closer look and expand this triangle:

Step 4: We can add in the missing 45º degree value so that our triangle now sums to 180º.

Step 5: Remember 45 45 90 special triangles(If you need a review click the link). Because that is exactly what we are going to need to find the value of the height! Below is our triangle on the left, and on the right is the 45 45 90 triangle ratios we need to know to find the value of the height.

Based on the above ratios, we can figure out that the height value is the same value as the base of the triangle, 2.

Step 6: If we place our triangle back into the original parallelogram, we can plug in our value for the height, h=2, into our formula to find the area:

How to find the Area of a Parallelogram

When you’re ready, check out the practice questions below!

Practice Questions:

Find the area of each parallelogram:

Solutions:

Still got questions? No problem! Don’t hesitate to comment with any questions or check out the video above. Happy calculating! 🙂

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Quadratic Equations with Two Imaginary Solutions

Hi everyone and welcome to MathSux! In today’s post we are going to be solving quadratic equations by using the quadratic formula. You may have used the quadratic formula before, but this time we are working with quadratic equations with two imaginary solutions. All this means is that there are negative numbers under the radical that have to be converted into imaginary numbers. If you need a review on imaginary numbers or the quadratic formula before reading this post, check out these links! Thanks so much for stopping by and happy calculating! 🙂

What is the Quadratic Formula?

The Quadratic formula is a formula we use to find the x-values of a quadratic equation. When we find the x-value of a quadratic equation, we are actually finding its x-values on the coordinate plane. Check out the formula below:

where, a, b, and c are coefficients based on the quadratic equation in standard form:

What does it mean to have “Imaginary Roots”?

When we solve for the x-values of a quadratic equation, we are always looking for where the equation “hits” the x-axis. But when we have imaginary numbers as roots, the quadratic equation in question, never actually hit the x-axis. Ever. This creates a sort of “floating” quadratic equation with complex numbers as roots. See what it can look like below:

Quadratic Equations with Two Imaginary Solutions

Ready for an Example?  Let us see how to use the quadratic formula specifically, quadratic equations with two imaginary solutions:

Quadratic Equations with Two Imaginary Solutions
Quadratic Equations with Two Imaginary Solutions
Quadratic Equations with Two Imaginary Solutions

Think you are ready to try practice questions on your own? Check out the ones below!

Practice Questions:

Quadratic Equations with Two Imaginary Solutions

Solutions:

Still got questions? No problem! Don’t hesitate to comment with any questions below or check out the video above. Thanks for stopping by and happy calculating! 🙂

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Factor By Grouping Examples: Algebra

Hey there math peeps and welcome to MathSux! In today’s post we are going to cover factor by grouping examples, a surprisingly cool and easy factoring method used to factor quadratic equations when “a” is greater than one. It can also be used to factor four term polynomials. We are going to look at an example of each below. If you have any questions, please don’t hesitate to check out the video and try the practice problems at the end of this post. Thanks for stopping by and happy calculating! 🙂

What is Factor by Grouping?

Factor by Grouping is a factoring method that groups common factors of an algebraic expression together.  Many times, we use factoring to find the x-values of a quadratic equation when the coefficient “a” is greater than 1.

When should we use Factor by Grouping?

1) If the first coefficient in a quadratic equation, a, is greater than 1:

Factor By Grouping examples

2) When there is a polynomial with 4 terms:

Factor by Grouping Examples:

Ready for an Example?  Let us look at how to factor a quadratic equation when a is greater than one.

Factor By Grouping examples
Factor By Grouping examples

Now, let’s take a look at another type of example, that can be solved with the help of factor by grouping!

Notice that this question is actually easier to solve than the last! The polynomial above, is already split into 4 terms, therefore, we can jump ahead, skipping the product/sum steps we did in the previous example!

Factor By Grouping examples

Ready to try practice questions on your own? Check them out below to master Factor by Grouping!

Practice Questions:

Solutions:

Still got questions? No problem! Don’t hesitate to comment with any questions below. Thanks for stopping by and happy calculating! 🙂

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Looking to learn about different Factoring methods and functions? Check out the related posts below!

GCF, Product/Sum, Difference of Two Squares, Quadratic Formula

Quadratic Formula, Product/Sum, Completing the Square, Graph

Completing the Square

The Discriminant

Is it a Function?

Quadratic Equations with 2 Imaginary Solutions

Free High School Math Resources

Free High School Math Resources

Greetings math friends! Today’s post is for the New York state teachers out there in need of a lesson boost. In this post, we’ll go over what MathSux has to offer for free high school math resources including videos, lessons, practice questions, etc. Remember everything you see here is 100% free and designed to make your life (and your students’ life) easier.

Signing up with MathSux will get you access to FREE:

1) Math Videos

2) Math Lessons

3) Practice Questions

4) NYS Regents Review

Everything designed here aligns with the NYS Common Core Standards for Algebra, Geometry, Algebra 2/Trig. and Statistics.

I am a NYS math teacher that creates free math videos, lessons, and practice questions every week, right here, for you! On the YouTube channel, you’ll also find NYS Common Core Regents questions reviewed one question at a time.

Featured on Google Classrooms around the world, MathSux.org is a great resource especially now, in the time of COVID and zooming and schooling from home. I hope you stick around and find these resources helpful.

And if you’re looking to get the latest MathSux.org videos and emails straight to your inbox, don’t forget to sign up on the right hand-side of the website. Thanks so much for stopping by and happy calculating! 🙂

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Bisect a Line Segment

Hi everyone and welcome to MathSux! In this post we are going to be constructing a perpendicular bisector by using a compass and stright edge. A perpendicular bisector, (also known as a segment bisector), is a line that cuts a line segment in half and creates four 90º angles. It is a super fast and super simple construction! We’ll also go over the Perpendicular Bisector Theorem we can infer from our original construction. If you’re looking for more geometric constructions, don’t forget to check more out here. Thanks so much for stopping by and happy calculating! 🙂

What is the Perpendicular Bisector of a line segment?

  • A perpendicular bisector slices our line segment AB (or any line segment) in half at its midpoint, creating two equal halves.
  • This will also create a 90º angle ( or right angle) about the line.
Constructing a Perpendicular Bisector through line segment AB

What is happening in this GIF?

Step 1: Notice we are given line segment AB. First, we are going to measure out a little more than halfway across the line segment AB by using a compass.

Step 2: Next we are going to place the point of the compass on point A and swing above and below line segment AB to create a half circle.

Step 3: Keeping the same distance on our compass, we are then going to place the point of the compass onto Point B (the opposite side) and repeat the same step we did on point A, drawing an arc in the shape of a semi-circle.

Step 4: Notice the intersection point above and below line segment AB!? Now, we are going to connect these two points by drawing a line with a ruler or straightedge.

Step 5: Yay! We now have a segment bisector! This cuts line segment AB right at its midpoint, while also dividing line segment AB into two equal halves and creating a 90º angle around our two intersecting lines.

Perpendicular Bisector Theorem:

The Perpendicular Bisector Theorem explains that any point along the perpendicular bisector line we just create is equidistant to each end point of the original line segment (in this case line segment AB).

Therefore, if we were to draw points C,D, and E along the perpendicular bisector, then draw imaginary lines stemming from these points to each end point, we’d get something like the image below:

AC = CB

AD = DB

AE = EB

Line Segment Theorem

Notice that with each of our points on the perpendicular line above, we can now state that the following is true:

AC = CB

AD = DB

AE = EB

The fact that we can state the above is true is reason for the Perpendicular Bisector Theorem!

Constructions and Related Posts:

Looking to construct more than just a perpendicular segment bisector? Check out these related posts and step by step tutorials on geometry constructions below!

Construct an Equilateral Triangle

Perpendicular Line through a Point

Angle Bisector

Construct a 45º angle

Altitudes of a Triangle (Acute, Obtuse, Right)

Construct a Square inscribed in a Circle

Best Geometry Tools!

Looking to get the best construction tools? Any compass and straight edge will do the trick, but personally, I prefer to use my favorite mini math toolbox from Staedler. Stadler has a geometry math set that comes with a mini ruler, compass, protractor, and eraser in a nice travel-sized pack that is perfect for students on the go and for keeping everything organized….did I mention it’s only $7.99 on Amazon?! This is the same set I use for every construction video! Check out the link below and let me know what you think!

Still got questions? No problem! Don’t hesitate to comment with any questions below or check out the video above. Thanks for stopping by and happy calculating! 🙂

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Want to see how to construct a square inscribed in a circle? Or maybe you want to construct an equilateral triangle? Click on each link to view each construction! And if you’re looking for even more geometry constructions, check out the link here! And if you’re looking for a construction you don’t yet see, please be sure to suggest it in the comments below!

How to Solve Inequalities with 2 Variables: Algebra

Hi everyone and happy Wednesday! Today we are going to look at how to solve inequalities with 2 variables. You may hear this in your class as “Simultaneous Inequalities” or “Systems of Inequalities,” all of these mean the same exact thing! The key to answering these types of questions, is to know how to graph inequalities and to know that the solution is always found where the two shaded regions overlap each other on the graph. We’re going to go over an example one step at time, then there will be practice questions at the end of this post that you can try on your own. Happy calculating! 🙂

How to Solve Inequalities with 2 Variables:

Just to review, when graphing linear inequalities, remember, we always want to treat the inequality as an equation of a line in  form….with a few exceptions:

1)Depending on what type of inequality sign we are graphing, we will use either a dotted line and an open circle (< and >) or a solid line and a closed circle (> or <) and  to correctly represent the solution.

2) Shading is another important feature of graphing inequalities.  Depending on the inequality sign we will need to either shade above the x-axis ( > or > ) or below the x-axis ( < or < ) to correctly represent the solution.

3) Solution: To find the solution of a system of inequalities, we are always going to look for where the shaded regions of both inequalities overlap.

How to Solve Inequalities with 2 Variables

Now that we know the rules, of graphing simultaneous inequalities, let’s take a look at an Example!

How to Solve Inequalities with 2 Variables

Step 1: First, let’s take our first inequality, and get it into y=mx+b form. To do this, we need to move .5x to the other side of the inequality by subtracting it from both sides. Once we do that, we can identify the slope and the y-intercept.

Step 2: Before graphing, let’s now identify what type of inequality we have here.  Since we are working with a < sign, we will need to use a dotted line and open circles when graphing.

Step 3: Now that we have identified all the information we need to, let’s graph the first inequality below:

Step 4: Now it is time for us to shade our graph, since this is an inequality, we need to show all of our potential solutions with shading.  Since we have a less than sign, <, we will be shading below the x-axis.  Notice all the negative y-values below are included to the left of our line.  This is where we will shade.

Step 5: Next, let’s start graphing our second inequality! We do this by taking the second equation, and getting it into y=mx+b form. To do this, we need to move 2x to the other side of the inequality by adding it to both sides. Then we can simplify the inequality even further by dividing out a 2.

Step 6: Before graphing, let’s now identify what type of inequality we have here.  Since we are working with a > sign, we will need to use a solid line and closed circles when creating our graph.

Step 7: Now that we have identified all the information we need to, let’s graph the second inequality below:

Step 8: Now it is time for us to shade our graph.  Since we have a greater than or equal to sign, >, we will be shading above the x-axis.  Notice all the positive y-values above are included to the left of our line.  This is where we will shade.

How to Solve Inequalities with 2 Variables

Where is the solution?!

Step 9: The solution is found where the two shaded regions overlap. In this case, we can see that the two shaded regions overlap in the purple section of this graph.

How to Solve Inequalities with 2 Variables

Step 10: Check!  Now we can finally check our work.  To do that, we can choose any point within our overlapping purple shaded region, if the coordinate point we choose holds true when plugged into both of our inequalities then our graph is correct!

Let’s take the point (-4,-1) and plug it into both original inequalities where x=-4 and y=-1.

Practice Questions:

Solutions:

How to Solve Inequalities with 2 Variables
How to Solve Inequalities with 2 Variables

Still got questions? No problem! Don’t hesitate to comment with any questions below. Thanks for stopping by and happy calculating! 🙂

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Looking to review graphing linear inequalities Check out this post on here!