Day 111 – February 22

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.

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