If $f:\mathbb{R}\rightarrow \mathbb{{R}}$ and $f(x)$ satisfying $f(2x+3)+f(2x+5) = 2$ . The period of $f(x)$ If $f:\mathbb{R}\rightarrow \mathbb{{R}}$ and $f(x)$ be a function satisfying $f(2x+3)+f(2x+5) = 2$. Then period 
of function $f(x)$ is.
$\bf{My\; Solution::}$ Let $(2x+3) = t\;,$ Then equation is $f(t)+f(t+2) = 2$
Now Replace $t\rightarrow (t+2)\;,$ we get $f(t+2)+f(t+4) = 2$
So from these two equation, we get $f(t+4) = f(t)$
So period of function $f(t)$ is $=4$
But answer Given as $=2$  . But in Solution it is given as $ = 2$
I did not understand where I have done Wrong.
Help me
Thanks

$\displaystyle f(2x+3) + f(2x+5) = 2$
Replace $x$ with $x+1$:
$\displaystyle f(2x+5) + f(2x+7) = 2$
Subtracting and rearranging, $\displaystyle f(2x+3) = f(2x+7)$
Hence $\displaystyle f(2x+3) = f((2x+3)+4)$
At this point, you can conclude that the period is $4$. If this is difficult to see, you can replace $(2x+3)$ with $t$, as the OP has done.
It might be tempting to do this:
$\displaystyle f(2(x+\frac{3}{2})) = f(2(x+\frac{7}{2}))$
and conclude that the period is $\displaystyle \frac{7}{2} - \frac{3}{2} = 2$.
But this would be a mistake. This ($2$) is the period of $f(2x)$. The period of $f(x) = 4$.
If you're still unconvinced, here's another example. If we define $f(x)=\sin 2x$, then the period of $f(x)$ is $\pi$. But if we define $f(2x)=\sin 2x$, then the period of $f(x)= \sin x$ is $2\pi$, even if the period of $\sin 2x=f(2x)$ remains $\pi$. This is what's happening in this case. The period of $f(2x)$ is $2$, but the period of $f(x)$ is actually $4$.
