MrE's Algebra Blog

Algebra: Chapter 10, Lesson 5, page 445 – DAY 1 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.

Unlike Lesson 4, we have differing denominator terms and we must find a common denominator of both rational expressions.

Given that p, q, r and s are rational numbers or expressions, the easiest way for addition and subtraction is with this equation: `(p/q) + (r/s)` converts to become `(ps + rq)/(qs)` for addition or `(ps – rq)/(qs)` for subtraction. We are sort of criss-crossing like we did for ratios!

Sometimes, we need to find the LCM (least common multiple of the denominators. These steps will help:

• Factor each expression
• For the product using each factor the greatest number of times it occurs.
• Just go slow and be methodical.

SHOW ALL THE WORK and you’ll be fine, try shortcuts and you’ll regret it.

Purplemath has a good explanation and plenty of examples here too. Just make sure to scroll down to the middle of the page and the next page as well.

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

Tonight’s homework problem number 73 is  solved by MrE are here! Just click it!

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Algebra 1a: Chapter 8, Lesson 1, page 358.

Solving Systems of Equations by Graphing

A set of equations for which a common solution is sought is called a SYSTEM OF EQUATIONS. A solution of a system of 2 equations in 2 variables (x, y) is an ordered pair that makes both equations true.

Take 2 linear equations and graph them (with at least 2 points for each linear equation) and where they INTERSECT is a “SOLUTION” to BOTH equations.

Pretty simple to do, but it can be time consuming in that you have to have graph paper and a ruler and some time ….

Here is a link with LOTS of examples from purplemath.com. It goes on for 2 pages so make sure that you see them both!

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

SNOW DAY #2

Algebra: Chapter 10, Lesson 4, page 442.

Adding and Subtracting Rational Expressions with Like Denominators

To add or subtract rational expressions with like denominators, add or subtract the numerators, and write the sum or difference over the common denominator. DON’T FORGET TO SIMPLIFY/FACTOR THE FINAL ANSWERS!

For example:

`(4m)/3+(5m)/3=(4m+5m)/3=(9m)/3=3m`

and another example,

`(2x^2+3x-7)/(2x+1)+(x^2+x-8)/(2x+1)`

Since the denominators are the same, we can just smash the numerator together being mindful of the signs to give us `=(2x^2+3x-7+x^2+x-8)/(2x+1)` which becomes when we combine like terms `=(3x^2+4x-15)/(2x+1)`

Factoring the numerator we arrive at a final answer of: `((x+3)(3x-5))/(2x+1)`

Here is a link from purplemath but it is more geared toward tomorrow’s lesson with different denominators.

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

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Algebra: Chapter 8, Lesson 1, page 358.

Solving Systems of Equations by Graphing

A set of equations for which a common solution is sought is called a SYSTEM OF EQUATIONS. A solution of a system of 2 equations in 2 variables (x, y) is an ordered pair that makes both equations true.

Take 2 linear equations and graph them (with at least 2 points for each linear equation) and where they INTERSECT is a “SOLUTION” to BOTH equations.

Pretty simple to do, but it can be time consuming in that you have to have graph paper and a ruler and some time ….

Here is a link with LOTS of examples from purplemath.com. It goes on for 2 pages so make sure that you see them both!

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

Algebra: Chapter 10, Lesson 3, page 439.

Dividing Rational Expressions

We divide rational expressions using the same techniques we used in Chapter 10-2. The only difference is that we have to change the division sign in front of the 2nd term to a multiplication sign and FLIP (use the reciprocal) the second term upside down.

For example, `(8n^5)/3 ÷ (2n^2)/9` becomes

`=(8n^5)/3⋅9/(2n^2)` which simplifies to `=(72n^5)/(6n^2)=12n^3

Remember to factor the DIFFERENCE of 2 SQUARES (binomial squares) and TRINOMIAL SQUARES as well as use the BOX METHOD of FACTORING.

Here is a link to some more examples from purplemath.com

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

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Algebra 1a: Chapter 7 Algebra Benchmark (#5)

Algebra: Chapter 10, Lesson 2, page 436.

Multiplying Rational Expressions

To multiply rational numbers, we multiply the numerators and multiply the denominators. We multiply rational expressions in the same way.

For example, we have the following examples:

`–2/(2y+6)⋅3/(y–5)`

First multiply the numerators and the denominators so that  is looks like

`(-2⋅3)/((2y+6)(y-5))` which becomes `(-2⋅3)/((2)(y+3)(y-5))` and finally is `=-3/((y+3)(y-5))`!

Another example is:

`4/(5x^2)⋅(x-2)/(2x^3)`

Multiplying numerators and denominators again, we have it becoming

`(4(x-2))/(10x^5)` which turns into `(2(x-2))/(5x^5)`

Here is a link from purplemath.com as well with more information and examples.

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

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Algebra 1a: Chapter 7 Review for Algebra Benchmark (#5)

Algebra Benchmark #5 Practice and Review were started in class. The end of the Chapter Review is a good overview and practice of the entire chapter.

Remember, to solve problems using the slope-intercept formula, `y = mx + b`, you need to have (or solve first for) the slope, then using one of the ordered pairs given `(x, y)` solve for the y-intercept, b.

Given the slope `m`, and the y-intercept `b`, we can develop the equation by plugging in the values for the slope and the y-intercept.

An equation perpendicular to the given one will have to have its slope be the negative reciprocal for the product to be -1. In other words, `m_1 * m_2 = -1`

A new equation that has to be parallel to the given one, must have its slope be exactly the same, `m_1 = m_2`!

Algebra: Chapter 10, Lesson 1, page 432.

Simplifying Rational Expressions

A rational expression is a quotient of 2 polynomials. A rational expression always indicates division. A rational expression is in simplest form when the numerator and denominator have NO COMMON factors other that `1` or `−1`.

Factor the numerator and the denominator and see what terms can be cancelled. For example:

`(5x-10)/(5x)=(5(x−2))/(5x)`. We can cancel the `5/5` in the numerator and denominator, leaving us with `=(x-2)/x`

and

`(y^2+3y+2)/(y^2-1)`

The numerator using the box, factors to `(y+1)((y+2)` and the denominator being the difference of 2 squares at `(y+1)(y-1)`, we have:

`=((y+1)(y+2))/((y+1)(y-1))`. In this example, then we can cancel the `(y+1)/(y+1)` (in the numerator and denominator), leaving us with just `=(y+2)/(y-1)`

AND, 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 1a: Chapter 7 Review for Algebra Benchmark (#9)

Algebra Skills Practice 19/20 and the Benchmark #2 Practice and Review were started in class. The end of the Chapter Review is a good overview and practice of the entire chapter.

Remember, to solve problems using the slope-intercept formula, `y = mx + b`, you need to have (or solve first for) the slope, then using one of the ordered pairs given `(x, y)` solve for the y-intercept, b.

Given the slope `m`, and the y-intercept `b`, we can develop the equation by plugging in the values for the slope and the y-intercept.

An equation perpendicular to the given one will have to have its slope be the negative reciprocal for the product to be -1. In other words, `m_1 * m_2 = -1`

A new equation that has to be parallel to the given one, must have its slope be exactly the same, `m_1 = m_2`!

Algebra: Chapter 6 Review

Make sure that your notes are up-to-date.  Can you do the BOX METHOD? Here is a link to my podcast on iTunes about the BOX and CROSS METHODS.

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Algebra 1a: Chapter 7 Review for Algebra Benchmark (#9)

Algebra Skills Practice 19/20 and the Benchmark #2 Practice and Review were started in class. The end of the Chapter Review is a good overview and practice of the entire chapter.

Remember, to solve problems using the slope-intercept formula, `y = mx + b`, you need to have (or solve first for) the slope, then using one of the ordered pairs given `(x, y)` solve for the y-intercept, b.

Given the slope `m`, and the y-intercept `b`, we can develop the equation by plugging in the values for the slope and the y-intercept.

An equation perpendicular to the given one will have to have its slope be the negative reciprocal for the product to be -1. In other words, `m_1 * m_2 = -1`

A new equation that has to be parallel to the given one, must have its slope be exactly the same, `m_1 = m_2`!

Algebra: Chapter 6 Review

Make sure that your notes are up-to-date.  Can you do the BOX METHOD? Here is a link to my podcast on iTunes about the BOX and CROSS METHODS.

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Algebra: Chapter 7, Lesson 8, page 338, day #3.

Parallel and Perpendicular Lines

Parallel lines by definition have the same slope. So, for the equation of 2 lines, all we have to do is figure out what the slope is of them both. If they have the same slope, then they are parallel. Check too, however, to make sure that both lines have DIFFERENT y-intercepts. If they have the same slope and y-intercept, then they are the same line, one on top of the other.

Perpendicular lines are lines that intersect at 90° or are at right angles to each other. By definition, the slopes of 2 lines that are perpendicular, when multiplied together, have a resultant product of −1.

Remember, the slope-intercept formula to find the slope, `m`: `y = mx + b`

You MAY have to solve the equation lines for `y`, isolating it to see what the slope, `m`, is as well as the y-intercept, `b`.

Here is a link from purplemath too with more explanation and examples.

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

Algebra: Chapter 6, Lesson 9, page 291.

Using Equations that Factor

There are 3 steps to using the problem solving guidelines below to help solve problems involving writing and solving an equation. They are:

1. Understand the problem

• What am I trying to find out
• What data am I given
• Have I ever solved a similar problem?

2. Develop and carry out a PLAN

• What strategy might I use to solve the problem?
• How can I correctly carry out the strategy I selected?

3. Find the ANSWER and CHECK

• Does the proposed solution check?
• What is the answer to the problem?
• Does the answer seem reasonable?
• Have I stated the answer clearly?

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

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Algebra: Chapter 7, Lesson 8, page 338, day #2.

Parallel and Perpendicular Lines

Parallel lines by definition have the same slope. So, for the equation of 2 lines, all we have to do is figure out what the slope is of them both. If they have the same slope, then they are parallel. Check too, however, to make sure that both lines have DIFFERENT y-intercepts. If they have the same slope and y-intercept, then they are the same line, one on top of the other.

Perpendicular lines are lines that intersect at 90° or are at right angles to each other. By definition, the slopes of 2 lines that are perpendicular, when multiplied together, have a resultant product of −1.

Remember, the slope-intercept formula to find the slope, `m`: `y = mx + b`

You MAY have to solve the equation lines for `y`, isolating it to see what the slope, `m`, is as well as the y-intercept, `b`.

Here is a link from purplemath too with more explanation and examples.

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

Algebra: Chapter 6, Lesson 8, page 286, day #2.

Solving Equations by Factoring

The principle of zero products states that: for any rational numbers a and b, if the product `ab = 0`, then `a=0` or `b=0` or both `a` and `b = 0`.

Factor and solve equations by using the following:

1. Get `0` on one side of the equation using the addition property
2. FACTOR the expression on the other side of the equation
3. Set each factor equal to zero
4. Solve each equation.

If we have an equation with `0` on one side and a factorization on the other side, we can solve the equation by finding the values that make the factors `0`. This is even easier!

Since we are solving quadratic equations (they are NOT linear because they have an exponent that is squared, `x^2`), we can have at MOST 2 solutions. We can have NO solutions, 1 solution or 2 solutions.

Remember to use all the techniques you know to factor!

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

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Algebra: Chapter 7, Lesson 8, page 338.

Parallel and Perpendicular Lines

Parallel lines by definition have the same slope. So, for the equation of 2 lines, all we have to do is figure out what the slope is of them both. If they have the same slope, then they are parallel. Check too, however, to make sure that both lines have DIFFERENT y-intercepts. If they have the same slope and y-intercept, then they are the same line, one on top of the other.

Perpendicular lines are lines that intersect at 90° or are at right angles to each other. By definition, the slopes of 2 lines that are perpendicular, when multiplied together, have a resultant product of −1.

Remember, the slope-intercept formula to find the slope, `m`: `y = mx + b`

You MAY have to solve the equation lines for `y`, isolating it to see what the slope, `m`, is as well as the y-intercept, `b`.

Here is a link from purplemath too with more explanation and examples.

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