Question 59753
{{{(x-1)(6ax+9x+4a+6)/(3x+2)(2ax-2a+3x-3)}}}
{{{(x-1)*((6a+9)x+(4a+6))/((3x+2)*((2a+3)x-(2a+3)))}}}
create a new variable called b
b = 2a + 3
2b = 4a + 6
3b = 6a + 9
substituting,
{{{((x-1)*(3bx+2b))/((3x+2)*(bx-b))}}}
b is a factor in the numerator and denominator
{{{((x-1)*b(3x+2))/((3x+2)*b(x-1))}}}
rewrite
{{{(b*(x-1)*(3x+2)) / (b*(x-1)*(3x+2))}}}
All factors in the numerator cancel with all factors
in the denominator, so the expression = 1 answer