In the following equation, a and b are both integers. a(3x - 8) = b - 18x. How do you find the value of a?

Answer 1

deceptively "easy" problem, correct answer a=-6

a(3x-8)= b-18x = - 18x +b

divide both sides by 3x-8 to get

a = (-18x +b)/(3x-8)

now comes the trick as you say a and b are integers

3x goes into -18x with quotient -6 with remainder = 0 = b-+48. b = -48

a = -6 is the surprise answer, but required given a and b are integers

to be an integer 3x-8 has to evenly divide into the numerator with no remainder

helps if you know how to do long division with algebra, very similar to arithmetic

quotient -6 + 0 remainder, 3x+8 = divisor, -18b = dividend

3x + 8 | -18x + b

subtract -18 -48

leaving remainder = (b+48)/(3x+8) = 0

b = -48

remainder has to = 0 for a to be an integer

a = (-18x +b)/(3x+8) = -6(3x8)/(3x+8) = -6

Answer 2

1) a(3x -8) = b - 18x. 2) divide both sides by (3x '8) 3) a = b - 18x/3x - 8. I hope this helps!

Answer 3

Distribute the a

Transform so "x" terms are on one side of equation and all other terms on the other side.

Factor out "x"

Divide both side to isolate x