Answer:
a=6, b=-5
a=12, b=-5
Step-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.
