MrE's Algebra Blog

Algebra: Chapter 10, Lesson 10, page 469.

Complex Rational Expressions

To simplify a complex rational expression, multiply the numerator and denominator by an expression equivalent to `1`. The expression selected should use the least common multiple of any denominator found in the numerator or denominator of the complex rational expression.

Sometimes it is easier to just work with the numerator and the denominator separately AND THEN, combine them with their division. Problems like 23-29 are HARD, look at my solutions for my method. You may have a different approach and that is OK!

Here is a link from purplemath with more examples.

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

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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, `.05n + .10d = 2.05`. You can then multiply both sides by `100` to clear the decimals.

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

Algebra: Chapter 10, Lesson 9, page 465 (day #2).

Dividing Polynomials

To divide a polynomial by a monomial, divide each term by that monomial.

For example:

`(6x^2 + 3x – 2)/3` becomes `(6x^2)/3 + (3x)/3 -2/3` or simplifying terms, we have

`2x^2 + x – 2/3`

If the divisor is not a monomial, then we need to use long (I call it synthetic from my ‘oldness’) division. It is just like regular division but takes a little getting use to.

A tip is that the quotient should be in descending order and you need to include the missing terms by sticking in ZEROES.

If the quotient is `x^3-x+1`, I would re-write it as `x^3+0x^2+0x^1+1`. Practice makes perfect. Examples 3, 4 and 5 on pages 465 and 466 are pretty easy to follow.

Here is a link from purplemath.com for more clarification.

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

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Algebra 1a: Chapter 8, Lesson 5, page 380 (day #2).

Motion Problems

D-R-T Problems (otherwise known as DIRT problems). `D` is the distance, `R` is the rate (or speed like miles per hour [mph] or kilometers per hour [kph] or feet per second [ft/sec]) and `T` is the time. Be careful and make sure that if the rate is in mph, then the time has to be in hours. If the rate is in feet/sec, then the time has to match and be in seconds as well.

For those type of problems that have 2 people starting off, remember that the first person’s time is `t`, but the second person (who leaves later) actually has `t – x` time where x is the time delay of the second person leaving. For example if one person leaves at `t` and the second leaves 2 hours later, then the second person’s time is `t – 2`.

REMEMBER, diagrams are great to get you to understand what you are looking for. The tables in the book are also good techniques. The more pictures or diagrams you have, the better chance you have of understanding what steps you have to go through! GO SLOW!!

Here are some links to excellent D-R-T examples at the PurpleMath website. Its easier to link to these than show you the same thing!

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

Algebra: Chapter 10, Lesson 9, page 465.

Dividing Polynomials

To divide a polynomial by a monomial, divide each term by that monomial.

For example:

`(6x^2 + 3x – 2)/3` becomes `(6x^2)/3 + (3x)/3 -2/3` or simplifying terms, we have

`2x^2 + x – 2/3`

If the divisor is not a monomial, then we need to use long (I call it synthetic from my ‘oldness’) division. It is just like regular division but takes a little getting use to.

A tip is that the quotient should be in descending order and you need to include the missing terms by sticking in ZEROES.

If the quotient is `x^3-x+1`, I would re-write it as `x^3+0x^2+0x^1+1`. Practice makes perfect. Examples 3, 4 and 5 on pages 465 and 466 are pretty easy to follow.

Here is a link from purplemath.com for more clarification.

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

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Algebra 1a: Chapter 8, Lesson 5, page 380.

Motion Problems

D-R-T Problems (otherwise known as DIRT problems). `D` is the distance, `R` is the rate (or speed like miles per hour [mph] or kilometers per hour [kph] or feet per second [ft/sec]) and `T` is the time. Be careful and make sure that if the rate is in mph, then the time has to be in hours. If the rate is in feet/sec, then the time has to match and be in seconds as well.

For those type of problems that have 2 people starting off, remember that the first person’s time is `t`, but the second person (who leaves later) actually has `t – x` time where x is the time delay of the second person leaving. For example if one person leaves at `t` and the second leaves 2 hours later, then the second person’s time is `t – 2`.

REMEMBER, diagrams are great to get you to understand what you are looking for. The tables in the book are also good techniques. The more pictures or diagrams you have, the better chance you have of understanding what steps you have to go through! GO SLOW!!

Here are some links to excellent D-R-T examples at the PurpleMath website. Its easier to link to these than show you the same thing!

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

Algebra: Chapter 10, Lesson 8, page 460.

Mixture Problems

Whether they are acid mixes or coffee bean blends or mixed nuts, we treat these problems the same. `x` will stand for the amount of one of the mixes and the (total amount – x) is the amount of the other mix. Convert percentages to decimals (remember `D2P`?) and use monetary cost per pound as decimals (like 2 dollars and 4 cents is equal to 2.04) and you’ll be fine.

`CA + CA = CA` (oops!)

where:

`C` = concentration (in decimals, convert from percentages)

`A` = amount

The problems CAN be solved with 2 variables, `x` and `y`, BUT it requires 2 equations and substitution – I think it is just twice the work of developing one equation with just 1 variable. For these problems, the book and I agree on the easiest method!

Here is a great Purplemath link.

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

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Algebra 1a: Chapter 8, Lesson 4, page 373.

Using systems of equations

Steps to solve systems of equations can be:

• Graph the equations
• Substitution of one variable in terms of the other
• Adding or subtracting the 2 equations

Before you start attacking the word problem, make sure you have a plan of attack such as:

• Understand the problem
• Develop and carry out a PLAN
• Find the ANSWER and CHECK

Click here for a pretty good link from purplemath.com with examples.

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

Algebra: Chapter 10, Lesson 7, page 455.

Using Rational Equations

We are working on DRT problems and “WORK” `1/x` problems. These say, if it takes one person 2 hours to do a job and another person 3 hours, how long does it take them both, working together to do the job.

Here is a link for DRT problems and here is a link for  these work type of problems.

Remember, for the work problems, think of setting up the problem on a per hour basis. For example, if it takes one person 3 hours to paint a wall and another 5 hours, then in 1 hour, each person can do `1/3` and `1/5` of the project. If you put them together, then it would take `t` time, BUT, if you only worry about `1` hours’ worth with both working together, then in `1` hour they can both do `1/t` of the project.

Adding this up, we would get an equation that looks like:

`1/3 + 1/5 = 1/t`

You know how to solve that by first finding the LCM (in this example it is `3⋅5⋅t` or `15t`)

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

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Algebra 1a: Chapter 8, Lesson 4, page 373.

Using systems of equations

Steps to solve systems of equations can be:

• Graph the equations
• Substitution of one variable in terms of the other
• Adding or subtracting the 2 equations

Before you start attacking the word problem, make sure you have a plan of attack such as:

• Understand the problem
• Develop and carry out a PLAN
• Find the ANSWER and CHECK

Click here for a pretty good link from purplemath.com with examples.

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

SNOW DAY #3!

Algebra: Chapter 10, Lesson 6, page 451.

Solving Rational Expressions

We are now solving rational equations, they have an equal sign. With rational expressions on both side, we can sometimes structure these as ratios or proportions. We can solve proportions by criss-crossing, if `a/b = c /d`, then `a*d = b*c`. We just plug in the numerators and denominators where appropriate and work it out.

Remember too, that a quadratic equation has 0, 1 or at most 2 roots or answers. Sometimes, one of the solutions can be considered extraneous or invalid. Usually, it results in the denominator of an expression being = 0. In Algebra I, we don’t know how to handle that but eventually in higher math, you will.

Here is a link to the purplemath site with some more great information about rational equations.

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

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Algebra 1a: Chapter 8, Lesson 3, page 367.

Addition and Subtraction for 2 linear equations.

You can add 2 (or subtract) linear equations together so that one of the variables cancels out. An example would be:

`3x – y = 9` and `2x + y = 6`

If we line them up, one under the other, we have:

`3x – y = 9`

`2x + y = 6`

Adding them together, we see that the sum looks like `3x + 2x – y + y = 9 + 6`

or

`5x = 15`

and solving for `x` makes it `x = 3`. If `x = 3`, then we can plug it into EITHER original equation, I’ll use the second one and we can solve for `y`.

So…  `2x + y = 6`

becomes `2*3 + y = 6` or `6 + y = 6` or `y = 0`. The ordered pair solution is then `(3, 0)`!

We may sometimes have to scale (multiply) ONE OR BOTH of the equations to make one of the variables disappear. Here is a link that can help!

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

Algebra: District Benchmark #2

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Algebra 1a: Chapter 8, Lesson 3, page 367.

Addition and Subtraction for 2 linear equations.

You can add 2 (or subtract) linear equations together so that one of the variables cancels out. An example would be:

`3x – y = 9` and `2x + y = 6`

If we line them up, one under the other, we have:

`3x – y = 9`

`2x + y = 6`

Adding them together, we see that the sum looks like `3x + 2x – y + y = 9 + 6`

or

`5x = 15`

and solving for `x` makes it `x = 3`. If `x = 3`, then we can plug it into EITHER original equation, I’ll use the second one and we can solve for `y`.

So…  `2x + y = 6`

becomes `2*3 + y = 6` or `6 + y = 6` or `y = 0`. The ordered pair solution is then `(3, 0)`!

We may sometimes have to scale (multiply) ONE OR BOTH of the equations to make one of the variables disappear. Here is a link that can help!

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

Algebra: Chapter 10, Lesson 5, page 445 – DAY 3 of 3!

This lesson is TOUGH and takes time to do right. We have to bring all of our tools to the problem. Know how to FOIL, recognize binomial squares (the difference of 2 squares), and use the BOX method of FACTORING ACCURATELY.

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Algebra 1a: Chapter 8, Lesson 2, page 362.

Substitution Method

You can solve 2 equations by solving 1 equation for 1 variable and then substituting that equivalent expression in the other linear equation. By doing this, you eliminate one variable. Solve for the remaining variable and then substitute its value in the original equation to find the first variable.

Here is an example of 2 equations to solve, its easier that way:

`x – 2y = 6` and `3x + 2y = 4`

1. Solve the first equation for x, so … `x = 6 + 2y` (added `2y` to both sides)

2. substitute for `x` in the second equation which now looks like: `3(6 + 2y) + 2y = 4`.

3. Distribute the new equation in `y`, it now looks like: `18 + 6y + 2y = 4`

4. combining like terms we have: `18 + 8y = 4` or simplifying

`8y = 4 – 18` … or … `8y = -14` and `y` is finally = `-14/8` or we reduce it to `-7/4`!

5. With `y = -7/4`, we can use the first equation to write:

`x – 2(-7/4) = 6` … or …. `x + 7/2 = 6` … or …. `x = 5/2`!

6. The solution is then, `(x,y)` or `(5/2, -7/4)`, wow!!!!

Here are some more examples done by purplemath.com with substitution!

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

Algebra: Chapter 10, Lesson 5, page 445 – DAY 2 of 3!

This lesson is TOUGH and takes time to do right. We have to bring all of our tools to the problem. Know how to FOIL, recognize binomial squares (the difference of 2 squares), and use the BOX method of FACTORING ACCURATELY.

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Algebra 1a: Chapter 8, Lesson 2, page 362.

Substitution Method

You can solve 2 equations by solving 1 equation for 1 variable and then substituting that equivalent expression in the other linear equation. By doing this, you eliminate one variable. Solve for the remaining variable and then substitute its value in the original equation to find the first variable.

Here is an example of 2 equations to solve, its easier that way:

`x – 2y = 6` and `3x + 2y = 4`

1. Solve the first equation for x, so … `x = 6 + 2y` (added `2y` to both sides)

2. substitute for `x` in the second equation which now looks like: `3(6 + 2y) + 2y = 4`.

3. Distribute the new equation in `y`, it now looks like: `18 + 6y + 2y = 4`

4. combining like terms we have: `18 + 8y = 4` or simplifying

`8y = 4 – 18` … or … `8y = -14` and `y` is finally = `-14/8` or we reduce it to `-7/4`!

5. With `y = -7/4`, we can use the first equation to write:

`x – 2(-7/4) = 6` … or …. `x + 7/2 = 6` … or …. `x = 5/2`!

6. The solution is then, `(x,y)` or `(5/2, -7/4)`, wow!!!!

Here are some more examples done by purplemath.com with substitution!

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