Day 62 – November 23

Algebra: 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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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 `x` and `(x+1)` or `x+(x+1)` or `2x+1`.

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

If you get confused, just make a little table, like `3`, `4`, `5`, `6`, `7` and `8` and see where the variable `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!


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