Question 813784
Something that is not often made clear about absolute value is that, in addition to what absolute value means (distance from zero), it <i>also</i> counts as a grouping symbol. So one should go through the operations inside of an absolute value <i>before</i> you find the absolute value. This is the key to your problem.<br>
{{{(abs(3(-5))-abs(1-7))/abs(6(2))}}}
So we start by performing the operations inside the absolute values:
{{{(abs(-15)-abs(-6))/abs(12)}}}
Next we find the absolute values:
{{{(15-6)/12}}}
Next simplify the numerator:
{{{9/12}}}
Reduce
{{{3/4}}}