Solving $3x - 40 \cdot 2 = x + 5$ What did I do wrong? 



*

*Start problem

*got rid of the 2 on the left side by reversing the multiplication

*simplified - got rid of the two on the left side and divided on the right to get 2.5

*undoing the subtraction of 40 by adding 40 to both sides

*simplified - got rid of the 40 on the left side and added 40 to 2.5 on the right. 

*Undoing the multiplication of 3 on the left side by dividing by 3 on both sides. 

*Simplifying - got rid of the 3 on the left, and divided on the right

*undid addition of x on right side by subtracting x on both sides

*simplified ------ confusion  

 A: Your mistake is in step 2. If you're going to divide either side of the equation by $2$, you have to divide the entire equation on both sides. You did:
$$3x-\dfrac{40 \times 2}{2} = x+\dfrac{5}{2}$$
However, this is incorrect. The correct way to carry this out would be:
$$\dfrac{3x - 40 \times 2}{2} = \dfrac{x+5}{2}$$
But that's not really the easiest way to solve this. Try subtracting $x$ from both sides first. That leaves us with:
$$2x-40 \times 2=5$$
Try to take it from here.
A: You divided some but not all of each side of your equation in step 2. 
The result of dividing by 2 in step 2 should be $(3x-40\cdot 2)/2 = (x+5)/2$ which is $(3/2)x-40 = x/2+5/2$ which is $1.5x-40=0.5x+2.5$.
The next mistake occurs in line 7 where you have $...=\frac{x+42.5}{3}$ and then on the next line you have $x+42.5/3$ where you should have $x/3+42.5/3$ (the 3 divides both $x$ and $42.5$).
A: You didn't divide the x on the right by 2 and didn't treat the multiplication on the left appropriately
It should go more like this
$3x - 40 \cdot 2 = 3x - 80 = x + 5$
$2x-80=5$
$2x = 85$
$x = \frac{85}{2}$
