[tex]-68x+P=Qx+34\qquad\text{subtract P from both sides}\\\\-68x=Qx+34-P\qquad\text{subtract QX from both sides}\\\\-68x-Qx=34-P\\\\-x(68+Q)=34-P\qquad\text{divide both sides by (68 + Q)}\\\\-x=\dfrac{34-P}{68+Q}\qquad\text{multiply both sides by (-1)}\\\\x=\dfrac{P-34}{68+Q}\\\\\text{The equation has no solution if the numerator is different than 0}\\\text{and the denominator is equal 0.}\\\\P-34\neq0\ \wedge\ 68+Q=0\\\\P\neq34\ \wedge\ Q=-68[/tex]
[tex]Answer:\\A:\ P=68\ and\ Q=-68\\C:\ P=-68\ and\ Q=-68\\D:\ P=-34\ and\ Q=-68\\-------------------\\\\For\ P=34\ and\ Q=-68\ we\ have\ \dfrac{34-34}{68-68}=\dfrac{0}{0}\\\\Then\ the\ equation\ has\ infinitely\ many\ solutions.[/tex]
