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maths portfolio 1

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Introduction

Maths Portfolio- linear equation

The aim of the portfolio is to investigate the system of linear equation where the system constant have well known mathematical pattern the first linear equation we will consider is :

x+2y=3

2x-y=-4

The first equation in the linear equation (x+2y=3) has a consecutive term for the values which is 1(x), 2(y) and 3(the answer). They also form an arithmetic progression with the series 1, 2 and 3 with a common difference of 1 (3-2=1 and 2-1=1). The second equation does not have consecutive terms but has and arithmetic progression with the series 2, -1, and -4 with the common difference of -3      (-1-2=-3 and -4-1=-3). When the equations  are  solved:

x+2y=3

2x-y=-4

= x+2y=3

4x-2y=-8

In this step the second equation is doubled so that the y term in both the equation is equal and can be eliminated

x+4x=-5

after the elimination of the y term add the x terms and the answers to the equations this will from and equation with only one term in this case it will be

5x=-5

This can be solved and value of x can be determined

x x=-1

This will give the value of x, -1 in this case. Now using the value of x term find the value of y term by inserting the x value in anyone of the equation for example:

-1+2y=3

This will form an equation

2y=3+1

Middle

a, a+d, a+2d, a+3d............and SO ON

a linear equation can be written as:

ax+ (a+d) y= a+2d

In the same way we can write a linier equation using 2 equations of such kind

ax+ (a+d) y= a+2d

bx+ (b+e) y=b+2e

Now as we did before we can multiply the variables with the equations in this case multiple the first equation with b and the second equation with a

(ax+ (a+d) y= a+2d)x b

(bx+ (b+e) y=b+2e)x a

after multiplying we will get:

axb+ (ab+db) y= ab+2db

axb+ (ab+ea)y=ab+2ea

after doing this elimination can take place doing which we will get

dby-eay=2db-2ea

(db-ea)y= (db-ea)2

y= y=2

Now like done previously the value of x can be found by plugging in the value of y.

bx+ (b+e) y=b+2e

bx+ (b+e) 2=b+2e

bx+ 2b+2e=b+2e

bx+ 2b=b+2e-2e

bx+ 2b=b

bx=b-2b

bx=-b

x= x=-1

as clearly proven in the conjecture x=-1 and y=2 and from this we can make an estimation that in any linear equation with constants forming an  arithmetic series , the point of intersection is always (-1,2) Now this was a 2x2 linear equation but if we consider a 3x3 equations let us investigate  the solutions :

x+2y+3z=4

2x+5y+8z=11

8x+3y-z=-6

Representing equations in matrix form: AX = B

a = x= b= The solution is x= a-1

Conclusion

-2z=y-3

z= and then

x+y+z=1

x+-2z+3+z=1

x+3-z=1

z=x+3-1

z=x+2

Therefore the equation of line on the plane would be = = In the same way we can write a linier equation using other 2 equations of such kind

bx+ (b+e) y+ (b+2e)z= a+3e

cx+(c+k)y+ (c+2k)z= c+3k

In the end we could observe that all the 3 planes contain same line = = , hence they meet at this line.

Now let’s look at some other 2x2 linier equations like:

x+2y= 4

5x-y= In these types of equations if we notice closely the terms follow G.P.

This means that the second number is multiple of the first and so on

These equations can be rewritten in line format y=ax+b

If we change the equations they would look like:

x+2y= 4

=

2y= 4-x

y= y= 2- in this case a= and b=2

And the second equation would be

5x-y= y= 5x In this case a=5 and b= A noticeable relation between both the equations is that the product of ab is same in both the equation

In the first equation

a x b x 2

=-1

And in the second equation

a x b x 5

=-1

Now let’s try taking some more of such examples and plot on the graph  From the graph we can observe that there is a horizontal parabola formed due to no points passing through a region. We can prove this theoretically by solving the equations of the linear pair in G.P.

Y=mx- Y= mx- y=nx- y=mx- y = - and y = - y= -x and y= -x

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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