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| 1. |
ax +by = cbx + ay = 1+ cBy elimination method |
| Answer» The given pair of equations isax + by = c ...(1)bx + ay = 1 + c ...(2){tex}\\Rightarrow{/tex} ax + by - c = 0 ...(3){tex}\\Rightarrow{/tex} bx + ay - (1 + c) = 0 ...(4)To solve the equations by the cross multiplication method, we draw the diagram below:Then,{tex}\\frac{x}{{(b)( - (1 + c)) - (a)( - c)}}{/tex}{tex} = \\frac{y}{{( - c)(b) - ( - (1 + c))(a)}}{/tex}{tex} = \\frac{1}{{(a)(a) - ({b})(b)}}{/tex}{tex}\\Rightarrow \\frac{x}{{ - b - bc + ac}} = \\frac{y}{{ - bc + a + ac}} = \\frac{1}{{{a^2} - {b^2}}}{/tex}{tex} \\Rightarrow x = \\frac{{ - b - bc + ac}}{{{a^2} - {b^2}}}{/tex}{tex}y = \\frac{{ - bc + a + ac}}{{{a^2} - {b^2}}}{/tex}Hence, the solution of the given pair of linear equations is{tex}x = \\frac{{ - b - bc + ac}}{{{a^2} - {b^2}}},\\;y = \\frac{{ - bc + a + ac}}{{{a^2} - {b^2}}}{/tex}Verification, Substituting{tex}x = \\frac{{ - b - bc + ac}}{{{a^2} - {b^2}}},\\;y = \\frac{{ - bc + a + ac}}{{{a^2} - {b^2}}}{/tex}We find that both the equations (1) and (2) are satisfied as shown below:ax + by {tex} = a\\left( {\\frac{{ - b - bc + ac}}{{{a^2} - {b^2}}}} \\right) + b\\left( {\\frac{{ - bc + a + ac}}{{{a^2} - {b^2}}}} \\right){/tex}{tex} = \\frac{{ - ab - abc + {a^2}c - {b^2}c + ab + abc}}{{{a^2} - {b^2}}} = c{/tex}bx + ay {tex}= b\\left( {\\frac{{ - b - bc + ac}}{{{a^2} - {b^2}}}} \\right) + a\\left( {\\frac{{ - bc + a + ac}}{{{a^2} - {b^2}}}} \\right){/tex}{tex}= \\frac{{ - {b^2} - {b^2} + abc - abc + {a^2} + {a^2}c}}{{{a^2} - {b^2}}}{/tex}This verifies the solution. | |