System of Equations

HallsofIvy said:
I'm not sure what you mean by "this". What I did was use the basic properties of algebra- which are really basic properties of numbers. And they are true for any equation or system of equations.
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I mean the instructions you gave me above. Can I do the same for other system of equations?
 
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JeffM said:
\(\displaystyle ax + by = c \implies -adx - bdy = -cd.\)

\(\displaystyle dx + ey = f \implies +adx + aey = +af.\)

\(\displaystyle SO\ adx + aey - adx - bdy = af - cd \implies aey - bdy = y(ae - bd) = af - cd \implies\)

\(\displaystyle y = \dfrac{af - cd}{ae - bd}.\)

\(\displaystyle SO\ ax + by = c \implies ax + b\left(\dfrac{af - cd}{ae - bd}\right) = c \implies ax = c - \dfrac{abf - bcd}{ae - bd} = \dfrac{ace - bcd - abf + bcd}{ae - bd} = \dfrac{a(ce - bf)}{ae - bd}\implies\)

\(\displaystyle x = \dfrac{ce - bf}{ae - bd}.\)

Let's check

\(\displaystyle ax + by = a\left(\dfrac{ce - bf}{ae - bd}\right) + b\left(\dfrac{af -cd}{ae - bd}\right) = \dfrac{ace - abf + abf - bcd}{ae - bd} = \dfrac{ace - bcd}{ae - bd} = \dfrac{c(ae - bd)}{ae - bd} = c.\)

Good so far.

\(\displaystyle dx + ey = d\left(\dfrac{ce - bf}{ae - bd}\right) + e\left(\dfrac{af -cd}{ae - bd}\right) = \dfrac{cde - bdf + aef - cde}{ae - bd} = \dfrac{aef - bdf}{ae - bd} = \dfrac{f(ae - bd)}{ae - bd} = f.\)

Yes, it works in general.
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Okay. All of this confuses me...
 
Tiahara: Please do not use the caret symbol to "point" to another post. If you want to reference something in particular on these boards, either use the quoting feature or spell it out!

The caret symbol is only for texting exponents, as in x^2.
 
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