Question

Which expressions are equivalent to 2 (three-fourths x + 7) minus 3 (one-half x minus 5)? Check all that apply.

Answer 1

Answer:

$2(\frac{3}{4}x+7)+(-3)(\frac{1}{2}x+(-5))$

$2(\frac{3}{4}x)+2(7)+(-3)(\frac{1}{2}x)+(-3)(-5)$

Step-by-step explanation:

The original expression given in the text is

$2(\frac{3}{4}x+7)-3(\frac{1}{2}x-5)$  (1)

And we want to check to what other expressions is equivalent. First of all, we solve it by writing explicitely each term:

$\frac{3}{2}x+14-\frac{3}{2}x+15$ (2)

Let's verify each of the other expressions separately. For the first one:

$2(\frac{3}{4}x+7)+(-3)(\frac{1}{2}x+(-5))$

We see that this is equivalent to expression (1), since the first half is identical, while in the second one, the combination "+-" can be simply written as "-", so we get

$2(\frac{3}{4}x+7)-3(\frac{1}{2}x-5)$

Which is equivalent to (1).

For the 2nd one:

$2(\frac{3}{4}x)+2(7)+3(\frac{1}{2}x)+3(-5)$

This is not equivalent. In fact, here we have applied the distributive property to each term: however, the 3rd and 4th term are not correct, because the (3) must be negative (-3), as in the original expression.

If we write it explicitely in fact, we get

$\frac{3}{2}x+14+\frac{3}{2}x-15$

Which is different from (2).

For the 3rd one:

$2(\frac{3}{4}x)+2(7)+(-3)(\frac{1}{2}x)+(-3)(-5)$

This one is equivalent. In fact, here we have applied the distributive property correctly. By solvign each term we get:

$\frac{3}{2}x+14-\frac{3}{2}x+15$

Answer 2

Answer:

B, C, and A

hope this helps :)