Michael D. answered • 05/01/19
Patient Math tutor specializing in High School Pre-AP/AP courses.
The first thing we'll need to do is to substitute the values with their respective variables. So, the expression ab - (b/a) would be (2/3)(-6) - (-6/(2/3)) where (2/3) is our "a" and (-6) is our "b". We see that this expression has a number of operators; Multiplication, subtraction, and division, so we will tackle these using our order of operations.
Next, we will start to evaluate our expression using our order of operations. Which brings us to our first obstacle; parentheses. to do this we see that this is a division problem. This expression would read "Negative 6 divided by two thirds; or (-6 / 2/3). However we are told that when we are dividing and the denominator, the number on the bottom of the fraction, in this case (2/3) is a fraction, then we switch the operation to multiplication and flip the fraction. So, the expression (-6) / (2/3) would then be changed to (-6)x(3/2).
When we evaluate this type of problem (one where there is an integer (-6) and a fraction and we are asked to multiply, we simply multiply across (numerators with numerators and denominators with denominators)) Since (-6) is a whole number, we can express this as (-6/1). Thus the new expression would read (-6/1) X (3/2) now we simply multiply across and we get (-18/2) but see that this can be reduced to (-9).
But we are not done yet, we've only solved half of the problem. We were asked to evaluate ab - (b/a). We've only evaluated (b/a). So, lets solve the rest.
recall that a = 2/3 and b = -6; so, again, we will simply plug in (2/3)(-6) - (-9). Order of operations tell us to do the multiplication operation next so we will evaluate a x b or (2/3) x (-6).
Multiply across to get (-12/3) reduced to -4.
Plug in and solve the final part of the expression
ab or (-4) which we just found out, minus b/a or (-9) which we found out previously when we started this problem leads to the expression;
-4 - (-9) = 5
I know this one was kind of lengthy, if you have any more questions, please feel free to message me directly :)
