Day 114 – February 27

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!

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