Hence the one and only quadratic which 18 9 7 ts the

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Unformatted text preview: tem. Do its friendly integer coefficients belie something more sinister? We note that if we multiply both sides of the first equation by 3 and the both sides of the second equation by −2, we are ready to eliminate the x 452 Systems of Equations and Matrices 6x − 12y = 18 + (−6x + 12y = −18) 0= 0 We eliminated not only the x, but the y as well and we are left with the identity 0 = 0. This means that these two different linear equations are, in fact, equivalent. In other words, if an ordered pair (x, y ) satisfies the equation 2x − 4y = 6, it automatically satisfies the equation 3x − 6y = 9. One way to describe the solution set to this system is to use the roster method2 and write {(x, y ) : 2x − 4y = 6}. While this is correct (and corresponds exactly to what’s happening graphically, as we shall see shortly), we take this opportunity to introduce the notion of a parametric solution. Our first step is to solve 2x − 4y = 6 for one of the 1 1 variables, say y = 2 x − 3 . For each value of x, the formula y = 2 x − 3 determines the 2 2 corresponding y -value of a solution. Since we have no restriction on x, it is called a free 1 variable. We let x = t, a so-calle...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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