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This question can be answered by using basic linear algebra. The two equations give are:



This can be used to set up a matrix equation of the form:


where x is a column vector of the form:

Row 1: x

Row 2: y

and b is a column vector of the form:

Row 1: 12

Row 2: 6

A is a 2 x 2 matrix containing the coefficients of x and y in the two equations:

Row 1: 2, 6

Row 2: 12, 3

This single matrix equation is thereby equivalent to the initial system of equations. If we could manipulate this matrix equation to yield an "A" matrix of the form:

Row 1: 1, 0

Row 2: 0, 1

we are effectively saying 1 x = something in the "b" matrix, and 1 y = something else in the "b" matrix. We could also conceivably end up with something such as:

Row 1: 1, 0

Row 2: 0, 0

which, depending on what remains in the "b" matrix after our manipulations, could mean no solution exists or could mean that there are an infinite number of solutions. Let's start by creating what is called an "augmented matrix" that contains the coefficients of x and y stored in A and the numerical constants stored in b.

[2, -6 | 12]

[12, -3 | 6]

In this notation, a pipe ( | ) is used to separate the "A" matrix elements from the "b" matrix elements. The first row of this augmented matrix can be read across as "2x + -6y = 12", essentially condensing the first equation. The remaining rows can be read in the same fashion. If more equations were being considered, more rows would exist, and if more variable were being considered, more columns would exist, but there is always only one column to the right of the pipe (that is to say the "b" vector is always a one-column vector).

Row operations can be performed to rewrite this matrix in a form called "reduced row echelon form" (RREF). A matrix in RREF has the leftmost element of each row as 1, and all row elements above and below this 1 as zero, with uppermost rows receiving priority. Visually:

[1, 0, 0, 0 | 5]

[0, 1, 0, 0 | 6]

[0, 0, 3, 4 | 4]

is in RREF, while:

[1, 0, 0, 0 | 3]

[0, 1, 4, 3 | 2]

[2, 1, 0, 0 | 2]

is not in RREF. This is not a full explanation of RREF, but numerous online resources exist to more fully expand on this.

Three different row operations can be performed to keep a matrix "row equivalent" to itself. Row equivalency is key to solving this problem. The three row operations are:

1. Movement of entire rows (Row 1 can be switched with Row 2, for instance)

2. Scalar multiplication of rows (All elements of Row 1 can be multiplied individually by some scalar number)

3. Piecewise addition of rows (Column 1 element of Row 1 can be added to Column 1 element of Row 2 to form a new Row 2, leaving Row 1 unaffected)

Returning to the augmented matrix we are using in this problem:

[2, -6 | 12]

[12, -3 | 6]

To get this matrix into RREF, the following row operation sequence can be performed:

1/2 * Row 1:

[1, -3 | 6]

[12, -3 | 6]

-12 * Row 1 + Row 2:

[1, -3 | 6]

[0, 33 | -66]

1/33 * Row 2:

[1, -3 | 6]

[0, 1 | -2]

3 * Row 2 + Row 1:

[1, 0 | 0]

[0, 1 | -2]

This can be read as "1x + 0y = 0" and "0x + 1y = -2", meaning x=0 and y=-2 is the only solution to this system of equations. Meaning this system has only one solution. Graphically, if these two equations were plotted on in an x-y coordinate system, they would be two intersecting lines that intersect at one point. It is easy to see, however, that different lines could have caused different situations. If the two equations were of the same line, simply one lying on top of the other, there would be an infinite number of solutions. This would be represented in RREF by:

[1, 0 | c]

[0, 0 | 0]

where "c" is some number.

Also possible are parallel lines, meaning no solution exists since they never intersect. This would have been represented by:

[1, 0 | c]

[0, 0 | 1]

again, "c" is some number. The final row of such a matrix in RREF is stating an impossibility (0 = 1) and therefore no solution exists.

This same analysis can be extended to infinite numbers of equations with infinite numbers of variables.

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Q: How many solutions does this system have 2x-6y equals 12 x-3y equals 6?
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