Solve the following pair of equations: $ax\, +\, by\, =\, c$ $bx\, +\, ay\, =\, 1\, +\, c$
- $x\, =\, \displaystyle \frac{ac\, -\, b\, -\, bc}{b^{2}\, -\, a^{2}}$ and $y\, =\, \displaystyle \frac{bc\, -\, a\, -\, ac}{b^{2}\, -\, a^{2}}$
- $x\, =\, \displaystyle \frac{ab\, +\, b\, -\, ac}{a^{2}\, -\, b^{2}}$ and $y\, =\, \displaystyle \frac{bc\, -\, a\, -\, ac}{b^{2}\, -\, a^{2}}$
- $x\, =\, \displaystyle \frac{ac\, -\, b\, -\, bc}{a^{2}\, -\, b^{2}}$ and $y\, =\, \displaystyle \frac{bc\, -\, a\, -\, ac}{b^{2}\, -\, a^{2}}$
- $x\, =\, \displaystyle \frac{c\, -\, bc\, -\, a}{a^{2}\, -\, b^{2}}$ and $y\, =\, \displaystyle \frac{bc\, -\, a\, -\, ac}{b^{2}\, -\, a^{2}}$
Subtracting the two equations: (a-b)x + (b-a)y = c - 1 - c = -1. Adding the two equations: (a+b)x + (a+b)y = 1 + 2c. Solving this linear system yields x = (ac - b - bc)/(a^2 - b^2) and y = (bc - a - ac)/(b^2 - a^2).
Subtract the second equation from the first to eliminate y, yielding ax minus bx plus cy minus ay = c minus 1 minus c. Factoring this difference gives (a-b)x plus (c-a)y equals negative 1. By expressing x in terms of y and substituting back into the first equation, or using the standard cross-multiplication method for linear systems, the algebraic rearrangement results in x equal to the fraction ac minus b minus bc divided by a^2 minus b^2. The corresponding value for y is the fraction bc minus a minus ac divided by b^2 minus a^2.