MrE's Algebra Blog

Algebra: Chapter 9, Lesson 6, page 421.

Graphing Systems of Linear Inequalities

We continue Chapter 9-5 techniques and solve 2 inequalities.

For the first equation, we can use any technique we learned from Chapter 7. Usually, the slope-intercept or ${\displaystyle x}$ and ${\displaystyle y}$ intercepts can be used to quickly define the line. Check a simple point like (0, 0) to see if that part of the ½ plane is true. If so, then shade that area.

Do the same for the other inequality and shade the appropriate ½ plane. The IMPORTANT PART is WHERE THE 2 INEQUALITIES OVERLAP THEIR SHADING, IS THE SOLUTION TO BOTH INEQUALITIES.

Remember, boundary lines of the form or ${\displaystyle >}$ are DASHED. The line is NOT part of the solution. Lines of the form ${\displaystyle \le }$ or${\displaystyle \ge }$ are solid because their line ARE part of the solution.

Again, the textbook is pretty good here but here are some more examples from purplemath.com too!

Two of tonight’s homework problems solved by MrE are here! Just click it.

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Algebra 1a: Chapter 3 Review

Mixed Practice #6, page 623.

Make sure that your notes are up-to-date.

Bring your problems worked out and your questions for the Chapter 3 review. These type of questions will appear on the test. Make sure that your notes are up-to-date. Practice makes perfect

Remember how to do the whickity-whack divide thingy.

Remember too, the steps to SOLVING 2-STEP EQUATIONS:

1. Multiply both sides to clear fractions or decimals, if necessary.
2. Collect like terms on each side, if necessary.
3. Use the addition property to move the variable to one side and all other terms to the other side of the equation.
4. Collect like terms again, if necessary
5. Add or subtract to isolate the variable and finally
6. Use the multiplication or division or reciprocal properties to solve for the variable.

Algebra: Chapter 9, Lesson 5, page 417.

Inequalities in 2 variables

Given an inequality, treat it as an equality and using the ${\displaystyle x}$ and ${\displaystyle y}$ intercepts, find the solution to the equality. Plot it on your graph paper.

• If the inequality is just a , or ${\displaystyle >}$ problem, then the boundary line (the line you draw connecting the dots) will itself be dotted or dashed. This mean that the points on the line are NOT part of the solution.
• If the inequality has a ${\displaystyle \le }$ or ${\displaystyle \ge }$, then the line will be solid, signifying that the line is part of the solution.

There are 2 ½ planes on the graph, one side of the boundary line that belongs to the solution set (this side will be shaded as part of the solution) and the other side of the line that does not satisfy the inequality.

Now to figure out what ½ plane to shade, pick a point [I like to pick ${\displaystyle (0,0)}$ or ${\displaystyle (1,1)}$] and try those ${\displaystyle (x,y)}$ values in the inequality.

• If the point chosen makes the inequality TRUE, then shade that part of the plane.
• If the point chosen does not satisfy the inequality, then shade the OPPOSITE side ½ plane.

The textbook is actually pretty good in this area, see pages 417-419 for good examples. Purplemath.com has these examples as well.

Two of tonight’s homework problems solved by MrE are here! Just click it

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Algebra 1a: Chapter 3 Review

Skills Practice 8 and 9, pages 16 and 17 AND Mixed Practice #6, page 623.

Make sure that your notes are up-to-date.

Bring your problems worked out and your questions for the Chapter 3 review. These type of questions will appear on the test. Make sure that your notes are up-to-date. Practice makes perfect

Remember how to do the whickity-whack divide thingy.

Remember too, the steps to SOLVING 2-STEP EQUATIONS:

1. Multiply both sides to clear fractions or decimals, if necessary.
2. Collect like terms on each side, if necessary.
3. Use the addition property to move the variable to one side and all other terms to the other side of the equation.
4. Collect like terms again, if necessary
5. Add or subtract to isolate the variable and finally
6. Use the multiplication or division or reciprocal properties to solve for the variable.

Algebra: Chapter 9, Lesson 4, page 413.

Inequalities and Absolute Value

If the inequality with absolute values looks like: , then we solve the conjunction . Think of a number line, and the solution will be within the bounds of ${\displaystyle -b}$ and ${\displaystyle b}$. This also works with ${\displaystyle \le }$.

If the inequality with absolute values look like: ${\displaystyle \left|A\right|>b}$, then we solve the disjunction OR ${\displaystyle A>b}$. On the number line, these solutions look like arrows on the outside of the values ${\displaystyle -b}$ and ${\displaystyle b}$. This works for ${\displaystyle \ge }$ as well.

Click here for some examples!

Two of tonight’s homework problems solved by MrE are here! Just click it

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Algebra 1a: Chapter 3, Lesson 11, page 158.

More Expressions and Equations

Remember, the sum of an integer and the next integer can be represented by ${\displaystyle x}$ and ${\displaystyle (x+1)}$ or ${\displaystyle x+(x+1)}$ or ${\displaystyle 2x+1}$.

The sum of consecutive (comes right after each other) odd OR even integers can be expressed as ${\displaystyle x}$ and ${\displaystyle (x+2)}$ or ${\displaystyle x+(x+2)}$ or ${\displaystyle 2x+2}$.

If you get confused, just make a little table, like ${\displaystyle 3}$, ${\displaystyle 4}$, ${\displaystyle 5}$, ${\displaystyle 6}$, ${\displaystyle 7}$ and ${\displaystyle 8}$ and see where the variable ${\displaystyle n}$ would line up if the numbers were hidden.

Here is a link that shows a few examples too.

Two of tonight’s homework problems solved by MrE are here! Just click it!

Algebra: Chapter 9, Lesson 3, page 411.

Equations and Absolute Value

To solve an equation of the form ${\displaystyle \left|A\right|=b}$, solve the disjunction ${\displaystyle A=b}$ OR ${\displaystyle A=-b}$. You will have 2 equations to solve with the right side of the second equation having the opposite sign of the first equation’s right side.

REMEMBER by definition, the solution of ${\displaystyle \left|A\right|\ne a}$ NEGATIVE NUMBER! So … the solution to these type of problems is the NULL SET! or the symbol ${\displaystyle \varnothing }$ !

Here is a link to examples!

Two of tonight’s homework problems solved by MrE are here! Just click it!

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Algebra 1a: Chapter 3, Lesson 11, page 158.

More Expressions and Equations

Remember, the sum of an integer and the next integer can be represented by ${\displaystyle x}$ and ${\displaystyle (x+1)}$ or ${\displaystyle x+(x+1)}$ or ${\displaystyle 2x+1}$.

The sum of consecutive (comes right after each other) odd OR even integers can be expressed as ${\displaystyle x}$ and ${\displaystyle (x+2)}$ or ${\displaystyle x+(x+2)}$ or ${\displaystyle 2x+2}$.

If you get confused, just make a little table, like ${\displaystyle 3}$, ${\displaystyle 4}$, ${\displaystyle 5}$, ${\displaystyle 6}$, ${\displaystyle 7}$ and ${\displaystyle 8}$ and see where the variable ${\displaystyle n}$ would line up if the numbers were hidden.

Here is a link that shows a few examples too.

Two of tonight’s homework problems solved by MrE are here! Just click it!

Algebra: Chapter 9, Lesson 2, page 405.

Compound Sentences

Conjunctions and Disjunctions are like Intersections and Unions (respectively) from lesson 9-1.

A disjunction of 2 statements is formed by connecting them with the word “or”. A disjunction is true when one OR both statements are true.

A conjunction of 2 statements is formed by connecting them with the word “and”. A conjunction is true when both statements are true.

Two of tonight’s homework problems solved by MrE are here! Just click it!

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Algebra 1a: Chapter 3, Lesson 10, page 152.

Using Percent
The ratio of numbers to 100 is called percent. Percent means “per one hundred”. We use whickity-whack divide, the method that Ms. Phillips taught us in 7th grade. Here is my podcast that describes the process!

Two of tonight’s homework problems solved by MrE are here! Just click it!

Algebra: Chapter 9, Lesson 1, page 400.

Sets, Intersections and Unions

A set is a well-defined collection of objects called members or elements.

• Roster notation LISTS the members of the set.
• Set-Builder Notation gives a DESCRIPTION of how the set is built.

The intersection of 2 sets ${\displaystyle A}$ and ${\displaystyle B}$, written ${\displaystyle A\cap B}$ is the set of all members that are COMMON to both sets. We say ” A intersection B”.

The union of 2 sets ${\displaystyle A}$ and ${\displaystyle B}$, written ${\displaystyle A\cup B}$ is the set of all members that are in ${\displaystyle A}$ or ${\displaystyle B}$ or in both. If an intersection is EMPTY, we say the intersection is the empty set which is symbolized as ${\displaystyle \varnothing }$.

All of these concepts are described here too with examples!

Two of tonight’s homework problems solved by MrE are here! Just click it!

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Algebra 1a: Chapter 3, Lesson 10, page 152.

Using Percent
The ratio of numbers to 100 is called percent. Percent means “per one hundred”. We use whickity-whack divide, the method that Ms. Phillips taught us in 7th grade. Here is my podcast that describes the process!

Two of tonight’s homework problems solved by MrE are here! Just click it!

Algebra: Chapter 8 Benchmark!

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Algebra 1a: Chapter 3, Lesson 9, page 148.

Proportions

A ratio of 2 quantities is a comparison, often expressed as a fraction. An equation that states that 2 ratios are equal is called a proportion. I prefer to just criss-cross, or cross multiply proportional problems, but the book’s way is OK too.

For example,

${\displaystyle \frac{x}{63}=\frac{2}{9}}$, I solve by criss-crossing. That becomes

${\displaystyle x\cdot 9=2\cdot 63}$ or ${\displaystyle 9x=63\cdot 2}$ or ${\displaystyle 9x=126}$

and dividing both sides by 9 to clear the x, gives us ${\displaystyle x=14}$.

Here is a link from purplemath.com that has some more examples.

Two of tonight’s homework problems solved by MrE are here! Just click it!

Algebra: Chapter 8 Review

Test on Friday, make sure that your notes are all up-to-date. The Chapter 8 Review Packet for homework is a great prep for the test.

Make sure that you REMEMBER HOW TO GRAPH 2 EQUATIONS OR SUBSTITUTE ONE FOR ANOTHER OR ADD/SUBTRACT 2 EQUATIONS TO SOLVE FOR EITHER X OR Y.

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Algebra 1a: Chapter 3, Lesson 9, page 148.

Proportions

A ratio of 2 quantities is a comparison, often expressed as a fraction. An equation that states that 2 ratios are equal is called a proportion. I prefer to just criss-cross, or cross multiply proportional problems, but the book’s way is OK too.

For example,

${\displaystyle \frac{x}{63}=\frac{2}{9}}$, I solve by criss-crossing. That becomes

${\displaystyle x\cdot 9=2\cdot 63}$ or ${\displaystyle 9x=63\cdot 2}$ or ${\displaystyle 9x=126}$

and dividing both sides by 9 to clear the x, gives us ${\displaystyle x=14}$.

Here is a link from purplemath.com that has some more examples.

Two of tonight’s homework problems solved by MrE are here! Just click it!

Algebra: Chapter 8 Review

Test on Friday, make sure that your notes are all up-to-date. The Chapter 8 Review Packet for homework is a great prep for the test.

Make sure that you REMEMBER HOW TO GRAPH 2 EQUATIONS OR SUBSTITUTE ONE FOR ANOTHER OR ADD/SUBTRACT 2 EQUATIONS TO SOLVE FOR EITHER X OR Y.

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Algebra 1a: Chapter 3, Lesson 8, page 145.

Solving Equations Involving Absolute Value

The absolute value of a number, ${\displaystyle \left|x\right|}$ is its distance from zero on the number line. Remember that the absolute value is always positive. Treat absolute value equations just like those without absolute value and then solve them as normal. At the VERY END, put the absolute value symbols back in and see if the answer has ANOTHER solution.

Remember too, that we cannot have an absolute value be NEGATIVE. In these cases, there is NO SOLUTION. For example:

${\displaystyle \left|x\right|+2=12}$

${\displaystyle \left|x\right|+2+(-2)=12+(-2)}$, we subtract 2 from both sides to isolate the variable, x

${\displaystyle \left|x\right|=10}$, we simplify the right side and finally,

${\displaystyle x=10}$ or ${\displaystyle x=-10}$ are the solutions

This is a good purplemath.com link with examples.

Two of tonight’s homework problems solved by MrE are here! Just click it!

Algebra: Chapter 8, Lesson 6, page 387.

Digit and Coin word problems.

Just remember to write the coin problems with the d (dime), q (quarter), n (nickel) preceeded by the value of the coin remembering that the d, q or n stand for the number of that type of coin. For example, ${\displaystyle .05n+.10d=2.05}$. You can then multiply both sides by ${\displaystyle 100}$ to clear the decimals.

Remember too, that any 2-digit number can be expressed as ${\displaystyle 10x+y}$ where ${\displaystyle x}$ is the digit in the tens place and ${\displaystyle y}$ is the digit in the one (units) place. For example, the number ${\displaystyle 23}$ can be written as ${\displaystyle 10\cdot 2+3}$. If we reverse the digits in the original number, the new number can be expressed as ${\displaystyle 10y+x}$. The reverse of ${\displaystyle 23}$, ${\displaystyle 32}$ can be written as ${\displaystyle 10\cdot 3+2}$.

Here is a link for some examples of coin problems and here is a link for digit type problems (about 1/2 the way down the page)!

Two of tonight’s homework problems solved by MrE are here! Just click it!

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Algebra 1a: Chapter 3, Lesson 8, page 145.

Solving Equations Involving Absolute Value

The absolute value of a number, ${\displaystyle \left|x\right|}$ is its distance from zero on the number line. Remember that the absolute value is always positive. Treat absolute value equations just like those without absolute value and then solve them as normal. At the VERY END, put the absolute value symbols back in and see if the answer has ANOTHER solution.

Remember too, that we cannot have an absolute value be NEGATIVE. In these cases, there is NO SOLUTION. For example:

${\displaystyle \left|x\right|+2=12}$

${\displaystyle \left|x\right|+2+(-2)=12+(-2)}$, we subtract 2 from both sides to isolate the variable, x

${\displaystyle \left|x\right|=10}$, we simplify the right side and finally,

${\displaystyle x=10}$ or ${\displaystyle x=-10}$ are the solutions

This is a good purplemath.com link with examples.

Two of tonight’s homework problems solved by MrE are here! Just click it!