Question

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 POINTS


Group of answer choices


a=6 b=-5


a=6 b=-20


a=12 b=-5


a=12 b=-20


a=2 b=-5

Answer 1

a=6, b=-5

a=12, b=-5

Explanation:

We have the equation

`(1)/(4)(2a-20x)=3+bx`

we will try to resolve for
`x` in terms of
`a` and
`b`

Then

`(1)/(4)(2a-20x)=3+bx\\(a)/(2)-5x=3+bx\\(a)/(2)-3=5x+bx\\(a-6)/(2)=(5+b)x\\(a-6)/(2)(1)/(5+b)=x\\(a-6)/(10+2b)=x`

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

`10+2b=0\\b=-5`.

Then, for the pairs

a=6, b=-5 and a=12, b=-5, there is no solution for x.