Simple algebra formula for which I can't find the right answer

Question

I have the formula $y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$, and I should work to $y = \frac{1}{2}\cdot z \cdot (z + 1)$.

Somebody showed me how it's done:

$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$y + (z + 1) = \frac{1}{2} \cdot ((z + 1) \cdot (z + 2))$
$y + (z + 1) = \frac{1}{2} \cdot (z^2 + 3z + 2)$
$y + (z + 1) = \frac{1}{2}(z^2) + \frac{1}{2}(3z) + \frac{1}{2}(2)$
$y + (z + 1) = \frac{1}{2}(z^2) + 1\frac{1}{2}z + 1$
$y = \frac{1}{2}(z^2) + 1\frac{1}{2}z + 1$ - z - 1
$y = \frac{1}{2}(z^2) + \frac{1}{2}z$
$y = \frac{1}{2}z(z + 1)$

Great! But, my try went completely wrong, and I don't understand what I'm doing wrong:

$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$y + (z + 1) = \frac{1}{2} \cdot ((z + 1) \cdot (z + 2))$
$y + (z + 1) = \frac{1}{2} \cdot (z^2 + 3z + 2)$
$y = \frac{1}{2} \cdot z^2 + 3z + 2 - z - 1$
$y = \frac{1}{2} \cdot z^2 + 2z + 1$
$y = \frac{1}{2} \cdot (z^2 + 2z + 1)$
$y = \frac{1}{2}(z^2) + \frac{1}{2}(2z) + \frac{1}{2}1$
$y = \frac{1}{2}(z^2) + z + \frac{1}{2}$

But from this last step, I can't get anywhere near $y = \frac{1}{2}z(z + 1)$, and I do not understand what I did wrong.

Answer 1

$$y + (z + 1) = \frac{1}{2} (z + 1) (z + 2)$$ $$2y+2(z+1)=(z+1)(z+2)$$ $$2y=(z+1)(z+2)-2(z+1)$$ $$2y=(z+1)(z+2-2)$$ $$2y=z(z+1)$$ $$y=\frac{1}{2}z(z+1)$$

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Where is your error? From $$y + (z + 1) = \frac{1}{2} (z^2 + 3z + 2)$$ you can say $$y+(z+1)=\frac{1}{2}z^2+\frac{3}{2}z+1$$ where instead you only multiplied $z^2$ by $1/2$.

Answer 2

Your step $y=\frac12 z^2+3z+2-z-1$ is wrong because $k(a+b)=ka+kb\neq ka+b$. A faster way to tackle the question is this.

$y+z+1=\frac12 (z+1)(z+2)$

$y=\frac12 (z+1)(z+2)-(z+1)$

$y=(z+1)(\frac 12 z+1-1)$

$y=\frac12 z(z+1)$

Answer 3

In your step 4, you forgot to divide (3z+2) by 2

Answer 4

There is an error in the fourth line of your second set of equations:- $$y = \frac{1}{2} \cdot z^2 + 3z + 2 - z - 1$$ should be $$y = \frac{1}{2} \cdot (z^2 + 3z + 2 \color{red}{- 2z - 2})\\\Rightarrow y = \frac{1}{2} \cdot (z^2 + z)=\frac{1}{2}\cdot z(z+1)$$