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.