Algebra: Chapter 10, Lesson 10, page 469.
Complex Rational Expressions
To simplify a complex rational expression, multiply the numerator and denominator by an expression equivalent to `1`. The expression selected should use the least common multiple of any denominator found in the numerator or denominator of the complex rational expression.
Sometimes it is easier to just work with the numerator and the denominator separately AND THEN, combine them with their division. Problems like 23-29 are HARD, look at my solutions for my method. You may have a different approach and that is OK!
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 9, Lesson 5, page 417 – DAY #2
Inequalities in 2 variables
Given an inequality, treat it as an equality and using the `x` and `y` intercepts, find the solution to the equality. Plot it on your graph paper.
- If the inequality is just a `<`, or `>` problem, then the boundary line (the line you draw connecting the dots) will itself be dotted or dashed. This mean that the points on the line are NOT part of the solution.
- If the inequality has a `≤` or `≥`, then the line will be solid, signifying that the line is part of the solution.
There are 2 ½ planes on the graph, one side of the boundary line that belongs to the solution set (this side will be shaded as part of the solution) and the other side of the line that does not satisfy the inequality.
Now to figure out what ½ plane to shade, pick a point [I like to pick `(0, 0)` or `(1, 1)`] and try those `(x, y)` values in the inequality.
- If the point chosen makes the inequality TRUE, then shade that part of the plane.
- If the point chosen does not satisfy the inequality, then shade the OPPOSITE side ½ plane.
The textbook is actually pretty good in this area, see pages 417-419 for good examples. Purplemath.com has these examples as well.
Two of tonight’s homework problems solved by MrE are here! Just click it