Hey Everyone , Its Deep

In our Today's Class , first we did Systems of Equations Word Problems . Mr.p told us and tell us how to do the sum .Here is the example that Mr.P told us on Smart board.

   Example:-3. The sum of the digits of a certain two digit number is 12. Reversing the digits decreases the number by 54. What is the number?

  Solution:-

                 we let the first number be x and second number be y .

so , according to the statement 

the first equation will be 

                       x+y=12

 and by understanding the statement 

                  10y+x+54=10x+y

                   10y-y+x-10x =-54

so by solving the statement 

             we got the second equation  and it will be

                              9y-9x =-54

by solving this both equations by any one of the method 

                       we got ,

        18y =54 

              y=3

so by putting the value 

we got 

                x+y=12 

                   x+3=12 

                   x=9 

so the number will be 93

                        And then after this we did systems of nonlinear equations and it includes mostly the stuff we did in the beginning of this chapter and some stuff related to conics .

        The General Equation for a Conic Section:

Ax² + Bxy + Cy² + Dx + Ey + F = 0

              The related stuff here i m showing u 

                         

• Solve the following system:
y = 2x² + 3x + 4 y = x² + 2x + 3

As before, I'll set these equations equal, and solve for the values of x:		2x2 + 3x + 4 = x2 + 2x + 3 x2 + x + 1 = 0
Using the Quadratic Formula:
But I can't graph that negative inside the square root! What's going on here?

The lines do not intersect. Since there is no intersection, then there is no solution. That is, this is an inconsistent system. My final answer is: no solution: inconsistent system.

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In general, the method of solution for general systems of equations is to solve one of the equations (you choose which) for one of the variables (again, you choose which). Then you plug the resulting expression into the other equation for the chosen variable, and solve for the values of the other variable. Then you plug those solutions back into the first equation, and solve for the values of the first variable. Here are some additional examples:  Copyright © 2006-2009 Elizabeth Stapel All Rights Reserved

• Solve the following system:
y = –x – 3 x² + y² = 17
Graphically, this system is a straight line crossing a circle centered at the origin:
There appear to be two solutions. I'll proceed algebraically to confirm this impression, and to get the exact values. Since the first equation is already solved for y, I will plug "–x – 3" in for "y" in the second equation, and solve for the values of x:
x² + y² = 17 x² + (–x – 3)² = 17 x² + (–x – 3)(–x – 3) = 17 x² + (x² + 6x + 9) = 17 2x² + 6x + 9 = 17 2x² + 6x – 8 = 0 x² + 3x – 4 = 0 (x + 4)(x – 1) = 0 x = –4, x = 1
When x = –4,  y = –x – 3 = –(–4) – 3 = 4 – 3 = 1 When x = 1,  y = –x – 3 = –(1) – 3 = –4 Then the solution consists of the points (–4, 1) and (1, –4).

Note the procedure: I solved one of the equations (the first equation looked easier) for one of the variables (solving for "y=" looked easier), and then plugged the resulting expression back into the other equation. This gave me one equation in one variable (the variable happened to be x), and a one-variable equation is something I know how to solve. Once I had the solution values for x, I back-solved for thecorresponding y-values. I emphasize "corresponding" because you have to keep track of which y-value goes with which x-value. In the example above, the points (–4, –4) and (1, 1) are not solutions. Even though I came up with x = –4 and 1 and y = –4 and 1, the x = –4 did not go with the y = –4, and thex = 1 did not go with the y = 1. Warning: You must match the x-values and y-values correctly!

• Solve the following system of equations:
y = (¹/₂)x – 5 y = x² + 2x – 15
Since both equations are already solved for y, I'll set them equal and solve for the values of x:
(¹/₂)x – 5 = x² + 2x – 15     x – 10 = 2x² + 4x – 30             0 = 2x² + 3x – 20             0 = (2x – 5)(x + 4) x = ⁵/₂, x = –4
When x = ⁵/₂:
y = (¹/₂)x – 5 = (¹/₂)(⁵/₂) – 5 = ⁵/₄ – ²⁰/₄ = – ¹⁵/₄ = –3.75
When x = –4:
y = (¹/₂)x – 5 = (¹/₂)(–4) – 5 = –2 – 5 = –7
Then the solutions are the points ( ⁵/₂, ^–15/₄ ) and (–4, –7).

Graphically, the above system looks like this: