Consider the equation 1/4 (2a-20x)=3+bx.Which combination of values for a and b yields no solution for x? Select the two answers that apply. NEEED HELP AGAIN PLEASE 12 POINTSGroup of answer choicesa=6 b=-5a=6 b=-20a=12 b=-5a=12 b=-20a=2 b=-5

Answer:a=6, b=-5a=12, b=-5Step-by-step explanation:We have the equation[tex]\frac{1}{4}(2a-20x)=3+bx[/tex]we will try to resolve for [tex]x[/tex] in terms of [tex]a[/tex] and [tex]b[/tex]Then[tex]\frac{1}{4}(2a-20x)=3+bx\\\frac{a}{2}-5x=3+bx\\\frac{a}{2}-3=5x+bx\\\frac{a-6}{2}=(5+b)x\\\frac{a-6}{2}\frac{1}{5+b}=x\\\frac{a-6}{10+2b}=x[/tex]Note that if the denominator is 0, then the division is undefined and therefore there would be no solution for x.Then, the denominator is 0 if [tex]10+2b=0\\b=-5[/tex].Then, for the pairs a=6, b=-5 and a=12, b=-5, there is no solution for x.