y=
y=2
Hence the point of intersection of the linear equations is (-1, 2)
2) Taking another example of linier equation with constants forming an A.P. such as
2x+4y=6
4x+9y=14
In this case we can double the first equation to make one of the terms equal in both the equation
4x+8y=12
4x+9y=14
Now in this case both the x terms are positive and cannot be subtracted so we change all the signs of one of the equation.
-4x-8y=-12
4x+9y=14
Now the terms in the equation can be subtracted which makes it
9y-8y=2
y=2
Now value of x can be found by replacing y by 2
2x+4(2) =6
2x+8=6
2x=-2
x=-1
Hence again I found that again the point of intersection of the linear equations is (-1, 2)
The third example would be
4x+11y=18
10x+y=-8
In this case we are suppose to multiply the second equation by 11 to make both the y terms equal
4x+11y=18
110x+11y=-88
Now we can eliminate the y term by changing signs of one of the equation
4x+11y=18
-110x-11y=88
4x-110x+ (11y-11y) =18+88
-106x=106
x=
x=-1
Therefore if x=-1 the value of y can be found by replacing the term x with -1
4x+11y=18
4(-1) +11y=18
-4+11y=18
11y=18+4
11y=22
y=
y=2
Hence the point of intersection of the linear equations is (-1, 2) again.
3x+9y=15
2x+13y=24
In this case when one the term is not the multiple of other both the equations are to be multiplied by the term value of other
For example in this case if we want to find out the x term first then we multiply the first equation by 2 and the second equation by 3 which will make it
(3x+9y=15)2
(2x+13y=24)3
=
6x+18y=30
6x+39y=72
Now we can change the signs of one of the equation like we did previously
-6x-18y=-30
6x+39y=72
(-6x+6x)+ (-18y+39y) = (72-30)
21y=42
y=
y=2
Now we can find the value of x by replacing the term y with 2
3x+9(2)=15
3x+18=15
3x=-3
x=
x=-1
Hence the point of intersection of the linear equations is (-1, 2)
a conjecture which can be made from these examples can be obtained by using the arithmetic series and the linear equation formed by the terms of A.P. The arithmetic sequence can be defined
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-1b
Thus following the formula we first need to find a-1 for which we can use our GDC calculators.
but unfortunately the GDC shows magnitude error which means a-1 does not exist hence the determinant of the matrix is 0
The possibilities of the determinant to be 0 are only if there is no solution or infinite solutions exists for the given system of 3x3 equations.
No solution is possible only if the planes are parallel
and infinite solutions are possible if the planes coincide or the planes intersect at multiple points.
This puzzle can be solved using row reduced echelon form which can be done using the GDC
In row reduced echelon form if the last column of the matrix has all the digits 0 then the planes intersect in a line
If the last digit of the last column of the matrix is another number but not 0 then the planes are parallel, hence no solution exists.
The last option of coincide planes is not possible because we did not take any multiple equations
The row reduced echelon form of the matrix found from GDC is as follows:
As we can see the last column of the matrix is 0 this means that the planes intersect in a line.
Now to see whether the solutions are infinite for all the A.P. due to the planes intersecting in one line we can take another example:
2x+6y+10z=14
5x+3y+z=-1
10x+15y+20z=25
Of which the matrix would be
a=
x=
b=
When tried to find a-1 the GDC displayed magnitude error which again brings us to the same conclusion that the determinant is 0 and we need do depend on the row reduced echelon for the solution
The row reduced echelon form of this would be:
This is the exactly same conclusion we got for the previous question the planes intersect at the same line
Just to be sure we can take one more example
5x-1y-7z=-13
11x+22y+33z=44
6x+12y+18z=24
When written as matrix it comes to:
a=
x=
b=
When checked in the GDC once again the a-1 which gets us to same conclusion of infinite solutions. Thus we need to find the row reduced echelon form which according to the GDC would be:
As the last column has all the digits 0 the planes intersect at the same line another thing noticeable is that the row reduced echelon form of all the equations we took was same.
And to derive equation of the line from the matrix we can form equations from the line equal to one of the variable
Therefore we can derive:
1x+0y-1z=-2
x-z=-2
z=x+2
Another equation can be
0x+1y+2z=3
Y+2z=3
y-3=2z
z=
Therefore the equation of the line would be
=
=
A conjecture which can be made from these examples can be obtained by using the arithmetic series and the linear equation formed by the terms of A.P. The arithmetic sequence can be defined
a, a+d, a+2d, a+3d,............and SO ON
a linear equation can be written as:
ax+ (a+d) y+ (a+2d) z= a+3d
a(x+y+z)+d(y+2z)= a+3d
In this case a(x+y+z)=a and d(y+2z)=3d
Therefore if d(y+2z)=3d then y+2z=3 and if a(x+y+z)=a then x+y+z=1
So if
y+2z=3
y=-2z+3
and
-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
