ax +by = c bx + …

The given pair of equations is
ax + 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.