MrE's Algebra Blog

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 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!

SNOW DAY #2

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 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!

Algebra 1a: Chapter 7 Review for Benchmark #5 Test Tomorrow

Algebra Benchmark #5 Review was worked on in class on Wednesday and Thursday. These problems are just EXACTLY the same as on the Benchmark, with just SLIGHTLY different numbers. Make sure that you KNOW HOW to do these problems!!

BRING YOUR NOTES!

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 1a: Chapter 7, Lesson 10, page 344.

Reasoning Strategies

You can use the following strategies on the reasoning strategy problems in Chapter 7-10.

• Draw a diagram
• Make an organized list
• Look for a pattern
• Try, test and revise
• Use logical reasoning
• Simplify the problem
• Write an equation
• Make a table
• Work backward

Algebra 1a: 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`.

PRACTICE MAKES PERFECT!!

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 1a: Chapter 7, Lesson 6, page 328.

Finding an Equation of a Line

There are 2 ways that we can find an equation of a line. However, we need to have at least 2 points `(x,y)` or 1 point `(x,y)` and the slope, `m` provided. If we have those things, we can find the equation by the 2 methods below, you choose the one you like:

A. Using the slope-intercept equation, `y=mx+b`

1. (Don’t forget, for this to work, you have to be given the slope, `m`, and at least 1 `(x,y)` point.)
2. If you have the slope, `m`, plug it in for `m` above and pick the `(x,y)` that correspond to the point given.
3. In the equation then, you have the `y`, `m` and `x` known.
4. All you have to do is solve for `b`, the y-intercept.
5. Solve for `b`, then plug in the `b` and `m` into the slope-intercept equation.

WARNING: IF you are not given the slope, then you are given 2 points. Given the 2 points, find the slope `m` with the equation `m=(y_2−y_1)/(x_2−x_1)`, then proceed as in step 2 above.

With this method, you have to solve for b, the y-intercept.

B. Using the point-slope equation (which is a derivation of the slope definition), `(y−y_1)=m(x−x_1)`

1. (Don’t forget, for this to work, you need the slope and 1 point or at least 2 points from which you can find the slope.)
2. (Notice too, that there is NO LONGER a `y_2` and `x_2`, just a `y` and `x`. LEAVE IT THAT WAY!)
3. If you have the slope `m`, use it. If you have 2 points, then find the slope – like the WARNING above.
4. Choose 1 of the `(x, y)` points to use for `(x_1, y_1)` and plug in the values that you know (`x_1`, `y_1`, and `m`).
5. Solve the equation for `y` and remember that you have to distribute on the right side!

With this method you have to you the distribution method on the right. You DO NOT find the `b` or y-intercept.

Either method works, you choose what is most comfortable for YOU!

Here is a link to both methods from purplemath.com.

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

Algebra 1a: Chapter 7, Lesson 5, page 323

Equations and Slope

An equation `y=mx+b` is called the slope-intercept equation of a line. The slope is `m` and the y-intercept is `b`. Without having to plot points, or make a T chart, we can easily determine the slope as the coefficient in front of the `x` variable and the y-intercept as `(0, b)`, the constant in the slope-intercept equation.

If the equation is not of the slope-intercept form, solve for `y` to isolate it, just like we have done in the past. The key is to have the `y` on one side of the equation and the `x` and its coefficient and the constant `b` on the other side. Usually, you have to add/subtract terms first, then multiply/divide by the coefficient in front of the `y`.

You can easily plot an equation. Start with the `(0, b)` y-intercept and then use the slope definition of `m=(rise)/(run)` to move up/down and then left/right on the graph paper as determined by the values of the rise and run. Remember to watch the signs of the rise and run.

For example, find the slope of:

`2x + 3y = 7`

first subtract `2x` from both sides

`2x – 2x + 3y = -2x +7`

to give us

`3y = -2x + 7`

divide both sides by `3` to isolate the `y`

`y = (-2/3)x + 7/3`

and the slope is then `(-2/3)` and the y-intercept is `(0, 7/3)`

See these examples from purplemath.com too! Here are others to help you graph equations given the slope m and the y-intercept b.

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