Table of Contents 2 Variable Systems 3 Variable Systems Linear Inequalities Rational

Inequalities Absolute

Inequalities Circle Equations Distance Equations

Linear I nequalitiesLinear I nequalitiesLinear I nequalities

Luis EnriquezLuis Enriquez

Solving Linear I nequalities

Think about how you've done inequalities on thenumber line. For instance, they'd ask you to graphsomething like x > 2. How did you do it? You woulddraw your number line, fi nd the "equals" part (in thiscase, x = 2), mark this point with the appropriatenotation (an open dot or a parenthesis, indicating that thepoint x = 2 wasn't included in the solution), and then you'dshade everything to the right, because "greater than" meant"everything off to the right". The steps f or linear inequalitiesare very much the same.

Graph the solution to y < 2x + 3.

J ust as f or number-line inequalities, the fi rst step is to find the "equals“part. I n this case, the "equals" part is the line y = 2x + 3. There are a coupleways you can graph this: you can use a T-chart, or you can graph f rom the yintercept and the slope. Either way, you get a line that looks like this:

• Now we're at the point where your book gets complicated, with talk of "test points" and such. When you did those one-variable inequalities (like x < 3), did you bother with "test points", or did you just shade one side or the other? I gnore the "test point" stuff , and look at the original inequality: y<2x + 3.

• You've already graphed the "or equal to" part (it's just the line); now you're ready to do the "y less than" part. I n other words, this is where you need to shade one side of the line or the other. Now think about it: I f you need y LESS THAN the line, do you want ABOVE the line, or BELOW? Naturally, you want below the line. So shade it in:

• Graph the solution to 2x – 3y < 6.

• First, solve f or y:

2x – 3y < 6

–3y < –2x + 6

y > ( 2/ 3 )x – 2

• [Note the fl ipped inequality sign in the last line. Don't f orget to fl ip the inequality if you multiply or divide through by a negative!] Nowyou need to fi nd the "equals" part, which is the line y = ( 2/ 3 )x – 2. I t looks like this:

• But this is what is called a "strict" inequality. That is, it isn't an "or equals to" inequality; it's only "y greater than". When you had strict inequalities on the number line (such as x < 3), you'd denote this by using a parenthesis (instead of a square bracket) or an open [unfi lled] dot (instead of a closed [fi lled] dot). I n the case of these linear inequalities, the notation for a strict inequality is a dashed line. So the border of the solution region actually looks like this:

• By using a dashed line, you still know where the border is, but you also know that it isn't included in the solution. Since this is a "y greater than" inequality, you want to shade above the line, so the solution looks like this:

Rational InequalitiesEdward Duong

Rational Inequalities x-1 . < 0

(x-2)(x+)

1.Factor numerator and denominator

2.Determine the crucial numbers

(crucial number are the restricted values and the values that will make the numerator 0)

The crucial numbers here are 2, -3 , and 1

3. Place crucial numbers on a number line

4. Test Each Interval

(- , -3)(1,2)

Examples of Rational Inequalities

E.I

x > 0 here it shows that x over x-5 is greater

X-5 than 0.• Now find out the crucial numbers which are the numbers that make

the numerator and denominator equal to 0. The crucial numbers here are 0 and 5

• Now we can put the crucial numbers on the number line.

Example of Rational InequalitiesE.I

x > 0 here it shows that x over x-5 is greater

X-5 than 0.• Now find out the crucial numbers which are the numbers

that make the numerator and denominator equal to 0. The crucial numbers here are 0 and 5

• Now we can put the crucial numbers on the number line.

Continued on the next page……

Example of Rational Inequalities

• The 0 and 5 on the number line represents the crucial numbers

• Now test each of the intervals and you get your answer

• (- ,0),(5, )

Absolute Inequalities Jon

Absolute-Value Inequalities

Absolute value: "| x | is the distance of x from zero." | –2 | = | 2 | = 2

Example:

-Absolute-Value Inequalities : | x | < 3 :

-The number 1 will work, as will –1; the number 2 will work, as will –2. But 4 will not work, and neither will –4, because they are too far away. Even 3 and –3 won't work (though they're right on the edge). But 2.99 will work, as will –2.99. In other words, all the points between –3 and 3, but not actually including –3 or 3, will work in this inequality. Then the solution looks like this:

-Translating this into symbols, you find that the solution is –3 < x < 3.

This pattern always holds: Given the inequality | x | < a, the solution is always of the form –a < x < a. Even when the problems get more complicated, the pattern still holds.

Absolute-Value Inequalities

Solve | 2x + 3 | < 6

a. | 2x + 3 | < 6 –6 < 2x + 3 < 6 [this is the pattern for "less than"]–6 –3 < 2x + 3 –3 < 6 –3 –9 < 2x < 3 –9/2 < x < 3/2

b. then the solution to | 2x+ 3 | < 6 is the interval –9/2 < x< 3/2.

Solve | 2x – 3 | > 5.

a. | 2x – 3 | > 5 2x – 3 < –5 or 2x – 3 > 5 [this is the pattern for "greater than"]2x < –2 or 2x > 8 x < –1 or x > 4

b. the solution to | 2x – 3 | > 5 consists of the two intervals x < –1 and x > 4.

Absolute-Value Inequalities

x < 19 or x > 24

a. Take a look at the endpoints. Nineteen and 24 are five units apart. Half of five is 2.5. So you want to adjust the inequality so it relates to –2.5 and 2.5, instead of 19 and 24. Since 19 –(–2.5) =21.5 and 24 – 2.5 = 21.5, you'll need to subtract 21.5 all around:

b. x < 19 or x > 24 x – 21.5 < 19 – 21.5 or x – 21.5 > 24 – 21.5 x – 21.5 < –2.5 or x – 21.5 > 2.5

c. Since this is the "greater than" format, the absolute-value inequality will be of the form "absolute value of something is greater than or equal to 2.5". You can convert this nicely to | x – 21.5 | > 2.5

CirclesKevin Dao

1 2 3 4-4 -3 -2 -1

4

3

2

1

-1

-2-3

-4

y

x

Formula for a circle is: (x - h)² + (y - k)² = r²or

x² + y² = r²

Center of circle = (h , k)Radius = r

Equation for the graph on the left is: (x-1)² + (y-1)² = 1² Center = (1,1)Radius = 1

Ex. x ² + y ² - 2x + 4y + 3 = 0(x ² - 2x) + (y ² + 4y) = -3(x ² - 2x + 1) + (y ² + 4y + 4) = -3 + + 1 + 4(x -1) ² + (y + 2) ² = 2Center = (1 , -2)Radius = √2

DistanceKevin Dao

Steps to finding distance1. You must first find the horizontal

distance -by using the “x” intercepts and finding the differences between each line, which in this case would be 2- (-1) = 3 units

2. You must find the vertical distance-by using the “y” intercepts and finding the differences between each line, which in this case would be 2- (-4) = 6 units

Continued on next slide….

x

y

l₁= 2y= 4x + 4

l₂= y=2x - 4

l ₁

l₂

Steps to finding distanceContinued…

3. You must now find the shortest distance using the ┴ distance formula│A x ₁ + B y ₁ + C│

√ A² + B²- You must pick 1 point on each line and use the ┴ distance formula on each of the points.

P (x₁ , y₁)(1 , 4) │2(1) + (-1)(4)-4│

y= 2x – 2 √2²+(-1²)= 2x – y – 4 =│6│ . √5 6√5

A B C √5 √5 5

P(1 , -2) │2(1) + (-1)(-2)+2│y= 2x + 2 √2²+(-1²)= 2x – y + 2 = 6 . √5 6√5

√5 √5 5

=

=