I'm doing this course online and unfortunately it did not explain the steps.
Find z if: $$x = 2z + 1, \quad x = y, \quad y = 4z + 2.$$

The solution:

$4z + 2 = 2z + 1$

$4z − 2z + 2 = 2z − 2z + 1$ //why did the teacher put 4z on one side and 2z on the other?

$2z + 2 = 1$

$2z + 2 − 2 = 1 − 2$

$2z = − 1$

$2z/2 = −1/2$

$z = −1/2$

Can someone try to explain how this works? Sorry if im stupid but I have not done math in forever. thanks

  • 1
    $\begingroup$ Since $y=x$ you basically have $z$ in terms of $y$ in two different ways. So you can set them equal to each other and then you have an equation just in $z$ that you can then solve for $z$ by getting all $z$ terms to one side and everything else to the other side. $\endgroup$ – Gregory Grant Jan 5 '16 at 7:23
  • 1
    $\begingroup$ Exactly where is your problem? Any equality, such as $4z+2=2z+1$, says that two things or expressions are equal. When you do the same operation on both sides of the equality sign, the results will be equal to each other. By choosing the operations wisely, you end up with an equality saying that $z$ is equal to some number. $\endgroup$ – Per Manne Jan 5 '16 at 7:44

You have:

$$\begin{align} 4z+2&=2z+1\\ 4z−2z+2&=2z−2z+1& \text{why did the teacher put $4z$ on one side and $2z$ on the other?} \end{align}$$

It's like this, assuming you can see colors.

$$\begin{align} \color{blue}{4z+2}&=\color{blue}{2z+1}\\ \color{blue}{4z}\color{red}{−2z}\color{blue}{+2}&=\color{blue}{2z}\color{red}{−2z}\color{blue}{+1}& \text{the red terms were added to each side} \end{align}$$

Answering your question in the side comment, the $4z$ and $2z$ in the second line are just coming from the first line. It's the instances of ${}-2z$ that are new in the second line.


We are given that $x=2z+1$, $y=4z+2$ and $x=y$

so lets rearrange and just write $x=2z+1$ and $x=4z+2$ ( since $x=y$)

and so $x=2z+1=4z+2=y$

$4z+2=2z+1$ $2z+2=1$ $2z=-1$ $z=\frac{-1}{2}$


We basically just use that we are given that x=y, and from that we can simplify the equations


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