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ax+by=a+b÷23x+5y=4 |
| Answer» The given system of equations is{tex}ax + by = \\frac{{a + b}}{2}{/tex}\xa0...(1)3x + 5y = 4 ...(2)From (1), we get2(ax + by) = a + b{tex}\\Rightarrow{/tex}\xa02ax + 2by - (a + b) = 0 ...(3)From (2), we get; 3x + 5y -4 = 0 ...(4)We know that if,\xa0a1x +\xa0b1y + c1\xa0= 0...........(i)a2x + b2y + c2\xa0= 0............(ii)then by cross multiplication method,{tex}\\frac x{b_1c_2-c_1b_2}=-\\frac y{a_1c_2-c_1a_2}=\\frac1{a_1b_2-a_2b_1}{/tex}Here,comparing equation (3) & (4) with the above formula, we get:-a1 = 2a, b1 = 2b, c1 = -(a + b)a2 = 3, b2 = 5, c2 = -4Now applying cross-multiplication method, we have ;{tex} \\frac{x}{{2b \\times ( - 4) - [ - (a + b) \\times 5}]}{/tex}\xa0{tex} = \\frac{{ - y}}{{2a \\times ( - 4) - \\left[ { - (a + b)} \\right] \\times 3}}{/tex}\xa0{tex}= \\frac{1}{{2a \\times 5 - 2b \\times 3}}{/tex}{tex} \\Rightarrow \\frac{x}{{ - 8b + 5(a + b)}} = \\frac{{ - y}}{{ - 8a + 3(a + b)}}{/tex}\xa0{tex} = \\frac{1}{{10a - 6b}}{/tex}{tex}\\Rightarrow \\frac{x}{{ - 8b + 5a + 5b}} = \\frac{{ - y}}{{ - 8a + 3a + 3b}}{/tex}\xa0{tex} = \\frac{1}{{10a - 6b}}{/tex}{tex}{/tex}Now, from 1st & 3rd part;{tex}\\frac{x}{{5a - 3b}} = \\frac{1}{{10a - 6b}}{/tex}{tex} \\Rightarrow x = \\frac{{5a - 3b}}{{10a - 6b}} = \\frac{{5a - 3b}}{{2(5a - 3b)}} = \\frac{1}{2}{/tex}Again, from 2nd & 3rd part;{tex}\\frac{{ - y}}{{ - 5a + 3b}} = \\frac{1}{{10a - 6b}}{/tex}{tex} \\Rightarrow - y = \\frac{{ - 5a + 3b}}{{2(5a - 3b)}}{/tex}{tex} \\Rightarrow y = \\frac{{ - ( - 5a + 3b)}}{{2(5a - 3b)}}{/tex}{tex} = \\frac{{5a - 3b}}{{2(5a - 3b)}}{/tex}{tex} \\Rightarrow y = \\frac{1}{2}{/tex}Hence, {tex}x = \\frac{1}{2},\\;y = \\frac{1}{2}{/tex}\xa0is the solution to the given system of equations. | |