Solving $3x - 40 \cdot 2 = x + 5$
Matthew Harrington 
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
3 Answers
$\begingroup$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.
$\endgroup$ 3 $\begingroup$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 $42.5$).
$\endgroup$ $\begingroup$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 $3x - 40 \cdot 2 = 3x - 80 = x + 5$
$2x-80=5$
$2x = 85$
$x = \frac{85}{2}$
$\endgroup$
