Plugging In: Using Good Numbers

If 9 more than $$3x$$ is three-quarters of $$y$$, what is the value of $$y$$ in terms of $$x$$?

Incorrect. [[snippet]] __Plug In__ $$x=3$$ to get $$\frac{3}{4(3+9)}$$. This is a fraction and nowhere near your goal of 24. __POE__ and move on.

Incorrect. [[snippet]] Did you __Plug In__ $$x=0$$ and get $$y=12$$? Although comfortable to work with, 0 may cause the answer choices to look alike. Be sure to check all five answer choices before you take your pick. Change the numbers to change the way the answer choices look.

How did you solve this problem?

Incorrect. [[snippet]] __Plug In__ $$x=3$$ to get $$3\cdot 3+9=18$$. This does not match your goal of 24, so __POE__ and move on.

Incorrect. [[snippet]] __Plug In__ $$x=3$$ to get $$3\cdot \frac{3}{4} - 9$$. This is a fraction and nowhere near your goal of 24. __POE__ and move on.

While it may have worked for you this time, it may not work the next time. The level of difficulty of that problem is no more than *medium*. If you solved it algebraically, you should be able to solve it using __Plugging In__ in the same amount of time. If this is not the case, it is because you are not proficient enough with __Plugging In__. Be sure to __Plug In__ more often so that you'll be able to ace the harder questions as well.

That's great. Identifying a __Plugging In__ problem is the first step. Solving is the second. Be sure to use __Plugging In__ as often as you can so that you will be able to use it in harder questions as well.

Correct. [[snippet]] __Plug In__ $$x=3$$ into each answer choice: >A) $$~\frac{3}{4(x+9)} = \frac{3}{4(3+9)} = \color{red}{\frac{3}{48}}$$ >B) $$~3(x+4) = 3(3+4) = \color{red}{21}$$ >C) $$~4x+12 = 4(3)+12 = \color{green}{24}$$ >D) $$~3x+9 = 3(3)+9 = \color{red}{18}$$ >E) $$~\frac{3}{4}x - 9 = \frac{3}{4}(3)-9 = \color{red}{\frac{9}{4}-9}$$ Since answer choice C equals the goal, that is the correct answer.

$$\frac{3}{4(x+9)}$$

$$3(x+4)$$

$$4x+12$$

$$3x+9$$

$$\frac{3}{4}x - 9$$

I created an equation and then solved it algebraically.

I used __Plugging In__. What else?

Continue