If the pressure between the cross-sections of a pipe is equal, will $P_1$ and $P_2$ be cancelled out of Bernoulli's Equation?

Question

a). Water is flowing through a pipe of two circular cross- sections area $A_1$ and $A_2$ lying horizontally as shown in figure below. The pressure difference between the cross-section $A_1$ and $A_2$ is 7500 Pascals. If the velocity of the water through cross-section $A_1$ is $3.25\ \mathrm{m/s}$, what would be the velocity of the water through cross-section $A_2$?

Using Bernoulli's Equation, I tried rearranging the equation so that it can be used to answer a problem like the one above. $$P_1 + \rho gh_1 + 1/2 \rho v^2_1 = P_1 + \rho gh_2 + 1/2 \rho v^2_2$$ Height is constant. Therefore equation will be: $$P_1 + 1/2 \rho v^2_1 = P_2 + 1/2 \rho v^2_2$$ My question: Since the pressure is constant, will $P_1$ and $P_2$ be removed from the equation?

Answer 1

Summary of discussion on this question between AugieJavax98, philip_0008 and sammy_gerbil :

$P_1$ is not equal to $P_2$. The question says that there is a "pressure difference" between $A_1$ and $A_2$. The height $h$ is assumed to be the same at both ends (the pipe is lying horizontally). The difference in area causes a difference in flow speed (continuity equation) and a difference in pressure (Bernoulli equation).

The illustration suggests that $A_2 > A_1$. Therefore $P_2 > P_1$. However, this leads to an imaginary value for $v_2$, the speed through $A_2$.

However, the illustration could be misleading. The text does not state that $A_2 > A_1$. Assuming instead that $A_1 > A_2$ and therefore $P_1 > P_2$ leads to a real (and realistic) value for $v_2$ :

$$P_1 + \frac{1}{2} \rho v^2_1 = P_2 + \frac{1}{2} \rho v^2_2$$ $$P_1 - P_2 = \frac{1}{2} \rho (v^2_2 - v_1^2)$$ $$7500Pa = \frac{1}{2} (1000kg/m^3)(v^2_2 - (3.25m/s)^2)$$ which gives $$v=5.06m/s$$.