Question

Which values of "P" and "Q" result in an equation with no solution.

-68x+P=Qx+34

choose all answers that apply


A: P=68 and Q= - 68

B: P=34 and Q= - 68

C: P= -68 and Q= - 68

D: P= -34 and Q= -68

Answer 1

`-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=(34-P)/(68+Q)\qquad\text{multiply both sides by (-1)}\\\\x=(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\\eq0\ \wedge\ 68+Q=0\\\\P\\eq34\ \wedge\ Q=-68`

`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\ (34-34)/(68-68)=(0)/(0)\\\\Then\ the\ equation\ has\ infinitely\ many\ solutions.`