• Writing and Plotting Inequalities

    Posted by Mark Rinaldi on 4/29/2019

    Equations are a comparison between expressions that are equivalent, meaning they have the same value.  That is why we have an equal sign between them.  Sometimes, when you are comparing expressions you see that they are not equal so we can't use an equal sign.  These types of comparisons are called inequalities.  There are four basic symbols used for inequalities.

    • greater than  >
    • less than  <
    • greater than or equal to  ≥
    • less than or equal to  ≤

    Many students have been taught to think of the symbol as a mouth that is opening up to eat the bigger number.  They also have a hard time remembering which one is called "greater than" and which one is called "less than".  I always tell them to think of an easy example such as 5 > 2.  The only way to read that out loud that makes sense is if you say "5 is greater than 2".  So that is the "greater than" symbol.

    To plot the solution to an inequality on a number line is pretty straight forward.

    Let's look at two examples.  First is x > 2.  We take a our number line and put an open circle at 2.  We use an open circle because x can't equal 2 so 2 isn't actually included as a solution.  The open circle tells us that our graph starts there but isn't included.  We then take a number on either side of 2 and plug it into the inequality.  If we choose 4 for x, we see that 4 > 2 is a true statement.  So our arrow is drawn in that direction.  If we had chosen 1, 1 > 2 not a true statement.  The graph would look like this:

    number line

    Now let's try w ≤ -1.  We place a solid dot on -1 because it is "less than or equal to".  -1 is equal to -1 so it is a solution.  We then test numbers on both sides to see which way we should draw our arrow.

    number line

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  • Balancing Equations

    Posted by Mark Rinaldi on 4/24/2019

    We have been working on balance equations in class.  We've been working on the proper format of showing our work and showing our answer.  Balancing equations is all about doing the same thing to both sides of the equation to make sure it stays balanced.  Let's look at a simple example using a one-step equation:

    m + 13 = 45

    We find the variable and see what is being done to it.  In this case the variable is "m" and we are adding 13 to it.  To get rid of a +13, we do the opposite which would be to subtract 13.

    m + 13 - 13 = 45 - 13

    The +13 and the -13 would equal 0 and just leave us with m.  On the right side of the equation, we just do the math.  We then right our answer in the form of a simple equation showing what the variable equals.

    m = 32


    Khan Academy Links:


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  • Combining Like Terms

    Posted by Mark Rinaldi on 3/25/2019

    We can use the properties of operations discussed in one of the previous blog entries to combine like terms to make an expression simpler to use.  Like terms are terms that have the same variable and the same exponents.


    3k and 4k

    4a2 and 2a2

    9wxy and 11wxy

    We can take an expression and use the properties to move around terms and combine the like terms so that our final expression has fewer terms in it.  Lets look at an example.

    9(2a + 3b) + 4(5a - 2b)

    In this expression, we can not combine the like terms yet because they are currently grouped in separate parentheses.  Since we can't combine 2a and 3b to remove the parentheses, we would use the distributive property to multiply each item in the group separately.

    9 * 2a + 9 * 3b + 4 * 5a - 4 * 2b

    18a + 27b + 20a - 8b

    We can then rearrange the terms with the commutative property.

    18a + 20a + 27b - 8b

    We can then group the like terms with the associative property.

    (18a + 20a) + (27b - 8b)

    And then we can combine like terms to get the expression in simplest form.

    38a + 19b

    Therefore, 9(2a + 3b) + 4(5a - 2b) is equivalent to 38a + 19b.

    Khan Academy = Finding Equivalent Expressions

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  • Evaluating Algebraic Expressions

    Posted by Mark Rinaldi on 3/13/2019

    We've been working on evaluating algebraic expressions.  To evaluate means to assign a value to something.  We have been taking our algebraic expressions and plugging in values for the variables and then following the order of operations to find the value of the expression.  Let's look at a simple example:

    7w + 5h - 8

    In this expression, we have two variables with three terms.  To evaluate the expression, we need to be given some values to plug in for w and h.  In this case, let's say that w = 4 and h = 9.  We would then rewrite the expression replacing the variables with the given values.

    7 x 4 + 5 x 9 - 8

    We then solve it using order of operations.

    28 + 45 - 8

    73 - 8


    This section of Khan Academy has both review videos and practice problems.

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  • Algebraic Expressions

    Posted by Mark Rinaldi on 3/11/2019

    We have started to look at the "dreaded" subject of Algebra.  Many students hear the word algebra and automatically associate it with math that is more difficult.  What they don't realize is that they have been using algebra since their earliest days in elementary school.  When the teacher wrote ____ + 7 = 10 on the board and students had to solve for the blank, that was basic algebra.  When they did input/output machines, that was basic algebra.  The only difference as you get higher up, is that the math itself gets more complex, and now instead of blanks, boxes, stars, etc. we begin to use letters in the problems.

    We are beginning our unit on Algebra by going over some of the basic terms.

    • An expression is like an incomplete sentence.  There is no equal sign involved.  An example of a simple expression is 6 + 5.  Notice there is no equal sign so we do not solve it at all.
    • An algebraic expression expands on the previous definition by including one or more variables in the expression.  Now it would look something like 8m-23 or 5(x+3).
    • variable is a letter or symbol that represents some unknown number.  As its name suggests, the value of the variable can change.
    • coefficient is a number that is multiplied by a variable.
    • constant is a specific number by itself whose value does not change.
    • Within an expression, there are terms.  The terms in an expression are the parts that are added or subtracted together.

    Also, an important thing to notice in algebra is the formatting for certain operations.  For multiplication between a number and a variable or between multiple variables, you don't usually include the multiplication sign.  Instead of writing 7 * r or h * k, we would just write it as 7r or hk.  For division between variables or a variable and a number, we usually write it in fraction form.  Instead of g÷4 we would write g/4.

    Let's look at an example of an algebraic expression and name the parts.


    In this expression we can start off by seeing that there are 2 terms:  3w and 8.  Our variable is w with a coefficient of 3.  8 would be a constant.

    Khan Academy links:
    Expression terms, factors, and coefficients video
    Identifying parts of expressions practice problems

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  • Order of Operations

    Posted by Mark Rinaldi on 2/28/2019

    We are reviewing the Order of Operations to make sure the students have an understanding of how it works.  It is important background knowledge I need to make sure they have before we can move on to our next unit.  Most people remember the steps of the Order of Operations using the acronym, PEMDAS.  This is a little misleading as it makes many people believe that it is a six-step process.  It is actually only four steps.

    Order of Operations:

    1. Solve everything inside parentheses.
    2. Solve any exponents.
    3. Solve all multiplication and division going left to right across the expression.
    4. Solve all addition and subtraction going left to right across the expression.

    In Step 1, when you are solving any operations within the parentheses, you still have to follow the proper order.  It is very important that students learn how to write out their work in a step by step method solving one step at a time.  Let's look at an example. (The ^ symbol is a way to show a number raised to a power.  5^2 is five to the second power or five squared which is 25).

    (2 + 4 x 3)^2 ÷ 7

    (2+12)^2 ÷ 7

    14^2 ÷ 7

    196 ÷ 7


    Notice in each line, I only solved one operation.  This is how the work should be written out.

    I would check out some Khan Academy videos and practice to review some examples:  Order of Operations

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  • Percent Word Problems

    Posted by Mark Rinaldi on 2/5/2019

    When looking at percent word problems, there are two main "outlines" we can use to set up our problem.  We can set it up using equivalent ratios or we can set up it as a number sentence.

    Let's look at a sample problem:

    Heather and her family are going to the grand opening of a new amusement park. There is a special price on tickets this weekend. Tickets cost $56 each. This is 70% of the cost of a regular price ticket. What is the cost of the regular price ticket?

    Now let's look at the two different ways we've been using in class to approach this type of problem.

    The Number Sentence:

    All percent problems can be boiled down to this simple number sentence:

    part = % x whole

    They will always give you two of the three parts somewhere in the word problem.  You can read the world problem and plug in the values that they give you to find the one that is missing.  In this problem, the percent is obvious - 70%.  Then we have the $56.  We have to figure out if this is the part or the whole.  Since it says that $56 is 70% of the cost of a regular ticket price, then $56 is part of the price.

    $56 is 70% of the cost.

    56 = .70 x ______

    Since we are missing part of a multiplication problem, we can reverse it into division.

    56 ÷ .70 = ____

    If we solve that problem, we get $80, so that is the cost of a regular ticket.

    Equivalent Ratios:

    We can also try to take the information and plug them into the same equivalent ratios that work for any percent problem:

    percent ratio

    Again, looking at the problem, we have 70% and $56 for the part.

    70/100 = 56/?

    We can solve this using equivalent rations by reducing the percent and then working our way back up.

    70/100 = 7/10

    7 x 8 = 56 so 10 x 8 = 80.

    70/100 = 56/80 so the whole ticket price is $80.

    If students can remember one or both of those two basic setups, it will help make finding the solution to the problems that much easier.

    You can check out this section on Khan Academy to see some sample problems explained.

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  • Converting between Percent, Decimal, and Fraction

    Posted by Mark Rinaldi on 1/23/2019

    We have been working with converting between percent, decimal, and fraction.  There are six basic conversions students need to be able to do.

    Decimal to Percent - Move the decimal point over two places to the right.  

    0.49 = 49%, 0.07 = 7%

    Percent to Decimal - Move the decimal point over two places to the left.  

    84% = 0.84, 62.7% = 0.627, 129% = 1.29

    Decimal to Fraction - Read the decimal and rewrite it as a fraction read the same way.  Then simplify if possible.  0.047 is read as forty-seven thousandths.  Rewrite it as 47/1000 so it is still read as forty-seven thousandths.

    Fraction to Decimal - There are two options.  One is to try and make an equivalent fraction with the denominator being a decimal place value (10ths, 100ths, 100ths, etc.).  You can then turn it into its synonymous decimal, like in Decimal to Fraction above.  Also, you can remember that a fraction is just another way to write a division problem.  Turn it into a division problem and solve.  You may have to round it off if there is a repeating decimal.  5/8 = 5 ÷ 8 = 0.625

    Fraction to Percent - One way is to set the fraction up equivalent to another fraction over 100.  If it can't easily be turned into something over 100, you can turn it into a decimal first and then turn the decimal into a percent.

    Percent to Fraction - Change the percent into a fraction over 100.  73% = 73/100.  Simplify if possible.

    This section of Khan Academy has some videos and practice problems you may find helpful.

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  • Rates and Unit Rates

    Posted by Mark Rinaldi on 12/18/2018

    We are taking our work with ratios and applying it to rates.  A rate is a ratio where the two quantities being compared are measured in different units.  A simple example of a rate is speed.  Speed is measures as distance over time.  The most common rate of speed seen in everyday life is miles per hour.  Notice the two values are measured in different units: miles and hours.

    An important type of rate we are working with is the unit rate.  A unit rate is a rate in which the second quantity in the comparison is always one unit.  For example, speed is almost always given as a unit rate.  We don't see signs on the highway saying the speed limit is 195 miles per 3 hours.  It is always something per 1 hour.

    Many of the examples we are working with in class are related to finding unit rates.  For example, if you are told that one bottle of juice costs $3.84 for a 16 ounce bottle and another bottle costs $4.50 for 25 ounces, you would want to look at the unit price to see which is the better deal.

    $3.84 / 16 ounces = ? / 1 ounce

    We would divide 16 by 16 to get to 1 so we divide $3.84 by 16 to get to $0.24 so the first bottle is $0.24 per ounce.  

    $4.50 25 ounce = ? / 1 ounce

    We would divide 25 by 25 to get to 1 so we divide #4.50 by 25 to get to $0.18 so the second bottle is $0.18 per ounce.  The second bottle is a better value.

    Khan Academy links:
    The Rates section has some videos about unit rates and some practice problems.

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  • Ratio Word Problems

    Posted by Mark Rinaldi on 12/17/2018

    We have been working on solving word problems using ratios.  I've been trying to get them to set up their problems the same way every time.  Let's look at an example:

    The ratio of boys to girls in the school is 8 to 11.  If there are 336 boys in the building, how many girls are there?

    The first thing we do is write the ratio using the words of what we are comparing.  In this case, our starting ratio would be:



    Then we write the starting ratio:

    boys/girls = 8/11

    Then we write the new information they give us to solve the problem:

    boys/girls = 8/11 = 336/?

    Now we have equivalent ratios and we can find the multiplicative relationship between the ratios.  We need to figure out 8 times what equals 336.  So we divide 336 by 8 and get 42.  The multiplicatve relationship between the ratios is x42.  So then we multiply 11 by 42 to find the number of girls.

    11 x 42 = 462.

    There are 462 girls in the building.

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