Algebra: Chapter 10, Lesson 10, page 469 – DAY #2
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 6, page 421.
Graphing Systems of Linear Inequalities
We continue Chapter 9-5 techniques and solve 2 inequalities.
For the first equation, we can use any technique we learned from Chapter 7. Usually, the slope-intercept or `x` and `y` intercepts can be used to quickly define the line. Check a simple point like (0, 0) to see if that part of the ½ plane is true. If so, then shade that area.
Do the same for the other inequality and shade the appropriate ½ plane. The IMPORTANT PART is WHERE THE 2 INEQUALITIES OVERLAP THEIR SHADING, IS THE SOLUTION TO BOTH INEQUALITIES.
Remember, boundary lines of the form `<` or `>` are DASHED. The line is NOT part of the solution. Lines of the form `≤` or`≥` are solid because their line ARE part of the solution.
Again, the textbook is pretty good here but here are some more examples from purplemath.com too!
Two of tonight’s homework problems solved by MrE are here! Just click it.