I have this but the answer it is not correct and i do not have the right answer.
If $X \sim N(\mu_X,\sigma_X^2)$, $Y \sim N(\mu_Y,\sigma_Y^2)$ and $X$ and $Y$ are independent, then: $AX+BY \sim N(A \mu_X + B \mu_Y, A^2 \sigma_X^2+B^2 \sigma_Y^2)$
Let $2X+Y$ be $W$ and $X-0.5Y$ be $V$ then $W\sim N(3,25)$ and $V\sim N(0.5, 1.75)$.
If $P(W\le a) = P(V\ge a)$ normalizing we have $P\left( Z\le \frac{a-3}{5}\right) = P\left( Z\ge \frac{a-0.5}{\sqrt{1.75}}\right)$
If $P(X > b)=P(X<c)$ then $b = -c$
Considering this we have that $\frac{a-0.5}{\sqrt{1.75}} = - \frac{a-3}{5}$
Solving for $a$ we have that $a = 1.023051428$
Calculating the probabilities $P\left( Z\le \frac{a-3}{5}\right) = P\left( Z\ge \frac{a-0.5}{\sqrt{1.75}}\right)$ seems to be right but it is not.
I am missing something?
As mentioned in the question's comments, the variance is calculated incorrectly. I'd like to add another observation that could prove useful for future readers. OP made $V=X-0.5Y$ because the second condition has $\geq 4a$. So the change in variables in the right side of the equation is:
$$P(4V\geq4a)=P(V\geq a)$$
Also, with the corrections $V \sim N( \frac{1}{2}, \frac{25}{4})$ (it's better if the algebra is done with fractions).Then, make a normalization of both $V$ and $W$, which is simply turning each variable into a variable $Z \sim N( 0, 1)$. So, following the property $P(X\leq m)=P(X\geq n) \Rightarrow m=-n$ for the normal distribution (because of symmetry) we have:
$$\frac{a-\mu_{V}}{\sigma_{V}}=\frac{a-\mu_{W}}{\sigma_{W}}$$
Replace the values for each r.v. and you'll get that:
$$a=\frac{3}{4}$$