From the standard equation (h, k) is the center of the circle, r is the radius
Example:
If given the equation:
x ² + y ² – 2 x + 4y - 2 = 0
First you must complete the square:
( x ² – 2 x + 1) + (y ² + 4y + 4) = 2 - 1 - 4
(x – 1) ² + (y + 2 ) ² = 11
Therefore, the center of the circle is (1,- 2 )
The radius is √11
But, what about if you’re given the center of the circle and the diameter and you are to find the equation?
For example:
The center of the circle is (4, -10) and the radius is 6
All you have to do is substitute those numbers into the standard equation.
(4, -10) (h, k)
(x – h) ² + (y – k) ² = r ²
(x – 4) ² + (y + 10) ² = 36
Circle Equations Ex. Diagram
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Distance Equations Finding distance can be tricky, but this section will show you just how to solve them.
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Distance Equations HOW TO SOLVE Example: Line1 0 = 4x - 2y - 8 y = 2x - 4 Line2 4y = 8x + 8 y = 2x + 2 Find the horizontal distance for Line1. Go to ‘x’ axis if you can determine the distance between the two intercepts. 3 6 Line2 Line1 Find the vertical distance for Line2. Go to ‘y’ intercepts and find the difference between them. (1, 4) (1, -2) The shortest distance distance formula |Ax + By + C| A + B 2 2 * Continued on next page
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Distance Equations Continued .. When you use the formula, use the ‘x’ and ‘y’ of the perpendicular line. Line1 0 = 4x - 2y - 8 A = 4 B = -2 C = -8 L = |4(1) + -2(4) + -8| 4 + -2 1 2 2 = |4 – 8 – 8| 16 + 4 2 2 = |-12| 20 20 20 = 12 20 20 Line2 0 = 2x - y + 2 A = 2 B = -1 C = 2 L = |2(1) + -1(-2) +2| 2 + -1 2 2 2 = |2 + 2 + 2| 5 = |6| 5 5 5 = 6 5 5
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Two Variable Systems In this section, we will show you how to solve two variable systems.
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Two Variable Equations * Remember that if there are 2 variables, it requires 2 equations Three ways to solve ..
Graph
Look for point of intersection
Both equations are valid at those ‘x’ and ‘y’ values
2) Elimination Example: 3x - 2y = 6 -2y + x = 4 Step 1: Place both equations in same form so variables align 3x - 2y = 6 (the ‘x’ and ‘y’ align) x - 2y = 4 Step 2: Multiply both equations by an appropriate value to eliminate one of the variables 3x - 2y = 6 (the ‘y’ can eliminate) x - 2y = 4 Continued on next page
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Two Variable Equations Continued .. Step 3: Either add or subtract to eliminate 3x - 2y = 6 - x - 2y = 4 2x 0 = 2 Algebraically solve for ‘x’ 2x = 2 x = 1 Step 4: Substitute the value back into the original equation to solve for second variable 3x - 2y = 6 3(1) - 2y = 6 3 - 2y = 6 -2y = 3 y = -3/2 * Solution is (1, -3/2)
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Three Variable Systems A repeat of two variable systems, but an extra equation is added. A different solution is presented.
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3 Variable Systems … Oh, three variables, three equations… LOVELY! HOW TO SOLVE; 1) -9x + 5y – 8z = -61 2) -4x + y – 2z = -7 3) 7x – 5y + 11z = 96 Now, to solve this, we must make two more equations ; so using equations 1&2, we can make equation four like so ; -9x + 5y – 8z = -61 5(-4x + y – 2z = -7) 11x + 2z = -26 <- multiplied the equation by 5 so that at least one of the variables have the same value [in this case ‘’y’’] so that we can remove one of the variables to find one of the remaining two. Using equations 1&3, we can make equation 5 ; **Note ; When making equations 4 & 5 be sure that you are eliminating the same variable. -9x + 5y – 8z = -61 +7x – 5y + 11z = 96 -2x + 3z = 35 RULE ; 3 Variables, 3 Equations 4) 11x + 2z = -26 5) -2x + 3z = 35 Now that we have these two, we can now substitute. 3(11x + 2z = -26) 33x + 6z =-78 37x = - 148 2(-2x + 3z = 35) - (-4x + 6z = 70) 37 37 Therefore X = -4 Next, we can find the other variables by subbing in ‘’x’’, for either equations 4 or 5. 11(-4) + 2z = -26 -44 + 44 + 2z = -26 +44 2z = 18 Z = 9 ……… . 2 2 So far, we have variables Z and X. Now we can go back to one of our original equations and sub in these values to find the final variable and solve the system. -9(-4) = 5y – 8(9) = -61 36 - 36 + 5y – 72 + 72 = -61 + 72 - 36 5y = -25 y = -5 Thus, we have solved the 3 variable system with a solution of (-4, -5, 9) Not hard when you think about it… just time consuming.
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Linear Inequalities In mathematics , an inequality is a statement about the relative size or order of two objects.
In inequalities, the four main types of inequality you’ll see are the above. Greater than, less than, greater than or equal to, and lesser than or equal to equations. The difference between them determines what kind of sign diagram you would use for solving the equation, as well as what kind of line you draw. If the equation is either < or > then the line is dashed. If it is ≤ or ≥ then the line you draw on the graph is solid. linear inequalities
The equations of inequalities are equations of boundary lines used to find the solution of the inequality. Their solutions are not certain points but rather entire areas, its limits defined by the ‘boundary line’. When faced with a linear inequality you will always treat it as a simple y = mx + b equation. Example.
y < x + 2
To find the solution of this inequality you must graph the boundary line, and then test one point to see if it fits the equation. If the point makes the equation true, then the side of the boundary line that the point was found on, is the solution. If the point makes the equation false, then it is the opposite side that is the solution.
Now on the Left hand side we have the equation graphed in a y =mx+b format, and from the example equation, we get the line illustrated. As mentioned before, the next step is to test one point anywhere on the graph. Usually this point is the origin (0,0) for the sake of ease. y > x + 2 0 > 2 As you can see, the point ( 0, 0 ) makes the equation false (because 0 is not greater then 2) so the solution to the inequality is the area, on the side of the boundary line, opposite to where ( 0, 0 ) is located. This is seen on the graph to the right. *Note* when putting an equation into y = mx+b format, you may have to divide by -1. When you do this however, you must reverse the symbol ( < becomes >, and > becomes < ) linear inequalities
In addition to single equation inequalities you also may find two inequalities in the same problem. Now if faced with
such a question don’t panic, you will follow the exact same steps to solve the inequality, except figuring out the solution Is slightly harder.
Example.
Sketch the region defined by y < x – 2, and 5x – 2y < 10
The first step here, is to put the second equation, into y = mx + b appropriate format.
5x – 2y < 10
-2y < -5x + 10
Y > 5/2x – 5
With that out of the way, we can go ahead and graph the two
Equations. These can be seen on the near right.
The line marked as red designates the line as y < x -2, while the line marked at yellow designates the line as y > 5/2x -5. Now, following the same procedure as before, we’ll test the point ( 0, 0 ), this time with both equations. Y < x – 2 Y > 5/2x - 5 (0) < (0) – 2 (0) > 5/2 (0) - 5 0 < 2 0 > - 5 Now that we’ve plugged in the coordinates of the origin, we can look for the solution. With the first inequality the point ( 0, 0) makes the statement true (because 0 is indeed, less than 2). So the first part of the solution is on the same side of the line y = x – 2 as the origin. Next we look at the second equation. Since the origin made the equation true once again we know that the second part of the solution is on the same side of the line y = 5/2x -5 as the origin. Now, since we’re looking for the area covered by both, it’s only where the two overlap , that we will shade, as the solution. This is shown in the graph on the left. linear inequalities
When dividing or multiplying by -1 to get an inequality ready for graphing, remember to switch the symbol (< converts to >, and > converts to <)
Pay special attention to the wording of problems. Questions that say, “Find the solution to the inequalities a and b” signifies the intersection of the two inequalities areas. If the question is worded “Find the solution to the inequalities a or b” the word or signifies the unity of the solution areas of the inequalities.
Don’t overcomplicate them! Treat linear inequalities as y = mx + b equations first. Then work out the solutions areas.
Remember that when an inequality uses < or > then the line you draw for the graph is dashed. If it is ≥ or ≤ then the line you draw would be solid.
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Rational Inequalities We are given this ; X – 2 (x -4)(x+6) < 0 How to solve this you ask? Four Simple Steps. FIRST We factor the Numerator and Denominator SECOND , Determine The CRITICAL Numbers THIRD , Place Critical Numbers on a Number Line LASTLY, Test each Interval **Note ; CRITICAL means the restricted values & the values that will make the numerator zero . Critical Values in this inequality are 2, 4 and -6 With those done, we can test values. *Test any value from each section Testing -11 we get values of - - = - Area / Testing 0 - - = + Area - - + - + - Testing 3 + + = - Area / Testing 11 + = + Area - + - + + **Note that the < in the equation means we’re looking for a negative area & vice versa. If something like X > 1 appears, just rearrange it to X - 1 > 0 (X-4) (X+3) (X-4) (X+3) Follow through Normally X(X+3) – (X-4) X ² +3X – X – 4 X ² + 2X – 4 (X-4) (X+3) (X-4) (X+3) (X-4) (X+3) Then, Just plot them on a number line and Test the area’s. So our solution is at ( - ∞ , -3)(4, ∞ ) So our solution is at ( - ∞ , -6)(2, 4 )
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