MrE's Algebra Blog

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.

Algebra 1a: Chapter 9, Lesson 5, page 417.

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

Algebra 1a: Chapter 9, Lesson 4, page 413.

Inequalities and Absolute Value

If the inequality with absolute values looks like: `| A | < b`, then we solve the conjunction `-b < A < b`. Think of a number line, and the solution will be within the bounds of `-b` and `b`. This also works with `≤`.

If the inequality with absolute values look like: `| A | > b`, then we solve the disjunction `A < -b` OR `A > b`. On the number line, these solutions look like arrows on the outside of the values `-b` and `b`. This works for `≥` as well.

Click here for some examples!

Two of tonight’s homework problems solved by MrE are here! Just click it

Algebra 1a: Chapter 9, Lesson 3, page 411.

Equations and Absolute Value

To solve an equation of the form `| A | = b`, solve the disjunction `A = b` OR `A = −b`. You will have 2 equations to solve with the right side of the second equation having the opposite sign of the first equation’s right side.

REMEMBER by definition, the solution of `| A | ≠ a` NEGATIVE NUMBER! So … the solution to these type of problems is the NULL SET! or the symbol `∅` !

Here is a link to examples!

Two of tonight’s homework problems solved by MrE are here! Just click it!

Algebra 1a: CST Testing Review!

Algebra 1a notes will resume after testing is OVER!

Algebra 1a: CST Review

WK7 (April 12): Area of Triangles, Parallelograms and Trapezoids

Triangle: `A = (bh)/2` or `A= 1/2(bh)` where `b` is the base and `h` is the vertical height. Sometimes it is useful to move the triangle around (rotate the page) to easily see the base and height.

Parallelogram: `A=bh`, where `b` is the base and `h` is the vertical height. This formula is similar to the area of a rectangle!

Trapezoid: `A=(h(a+b))/2`, or `A=1/2(h)(a+b)` where there are 2 bases, `a` and `b` and `h` is the vertical height. This formula has the divide by 2, just like the triangle.

Worksheet 12-1, page 100 has 15 problems to practice with! You might find an electronic calculator useful.

WK6 (April 11): Area and Perimeter of Triangles

The area of a triangle, `A = (bh)/2` where `b` is the base and `h` is the vertical height.

The perimeter of a triange is the sum of all 3 sides, just add them up to find the distance AROUND the triangle!

Worksheet D-56 and CC-51 are good practice. CC-51 provides both sides `a` and `b`, you have to solve for `c`, the hypotenuse (`c=sqrt(a^2+b^2)`)!

Algebra 1a: CST Review

WK5 (April 1): The area of a circle is defined as `A = πr^2`. The relationship between the radius `r` and the diameter `d` is:

`2r=d` or `d/2=r`

Remember on the 12-2 worksheet (page 101), that `π = 3.14`

WK4 (March 31): Plotting x-y graphs. X axis is horizontal and Y axis is vertical.

Plot the points on worksheet E-68 for the secret message!

WK3 (March 30): Remember, this only works for right triangles,

`c^2 = a^ + b^2` or `a^2 = c^2 – b^2` or `b^2 = c^2 – a^2`

`c = sqrt(a^2 + b^2)`

Worksheet D-73 is all about finding `c`, the hypotenuse!