Saved Bookmarks
| 1. |
Solve each of the following systems of equations by method of cross -multiplication NB |
| Answer» The given system of equations arex + ay = bSo, x + ay - b = 0 ........ (i)And ax - by = cSo, ax - by - c = 0 .......... (ii)The given equations are in form ofa1x + b1y + c1 = 0and a2x + b2y + c2 = 0Compare (i) and (ii),we geta1 = 1, b1 = a, c1 = -ba2 = a, b2 = -b, and c2 = -cBy cross-multiplication, we get{tex}\\Rightarrow \\frac{x}{{a \\times ( - c) - ( - b) \\times ( - b)}}{/tex}\xa0{tex}= \\frac{{ - y}}{{1 \\times ( - c) - ( - b) \\times a}}{/tex}\xa0{tex}= \\frac{1}{{1 \\times ( - b) - a \\times a}}{/tex}{tex}\\Rightarrow \\frac{x}{{ - ac - {b^2}}} = \\frac{{ - y}}{{ - c + ab}} = \\frac{1}{{ - b - {a^2}}}{/tex}Now,{tex}\\frac{x}{{ - ac - {b^2}}} = \\frac{1}{{ - b - {a^2}}}{/tex}{tex}\\Rightarrow x = \\frac{{ - ac - {b^2}}}{{ - b - {a^2}}}{/tex}{tex}\\Rightarrow x = \\frac{{ - \\left( {{b^2} + ac} \\right)}}{{ - \\left( {{a^2} + b} \\right)}}{/tex}{tex} = \\frac{{{b^2} + ac}}{{{a^2} + b}}{/tex}and,{tex}\\frac{{ - y}}{{ - c + ab}} = \\frac{1}{{ - b - {a^2}}}{/tex}{tex} \\Rightarrow - y = \\frac{{ab - c}}{{ - \\left( {{a^2} + b} \\right)}}{/tex}{tex} \\Rightarrow y = \\frac{{ab - c}}{{{a^2} + b}}{/tex}Hence, {tex}x = \\frac{{ac + {b^2}}}{{{a^2} + b}},y = \\frac{{ab - c}}{{{a^2} + b}}{/tex}\xa0is the solution of the given system of the equations. | |