Liquid Manometer

The question is "For the two-liquid manometer, shown below, obtain an expression for $p_1 – p_2$. Let $\mathcal A$ be the area of each tank and $\mathcal S$ the area of the tube. Let $S_3$ be the relative density of the manometric liquid, $S_2$ the relative density of liquid in the tanks and $S_1$ the relative density in which the pressure difference is to be measured."

I am struggling with line below because I don't understand why it should be $\gamma S_1 (x_1+\Delta y)$ and $\gamma S_1 (x_1-\Delta y)$. Any clarification is appreciated.

$$\begin{aligned} p_1= \gamma S_1 (x_1+\Delta y)+ \gamma S_2(x_3-\Delta y+x/2)-\gamma S_3x&\\ -\gamma S_2(x_3+\Delta y-x/2)-\gamma S_1 (x_1-\Delta y)=p_2 & \end{aligned}$$

Final answer $p_1-p_2=\gamma x(S_3-S_2)$.


closed as off-topic by ZeroTheHero, John Rennie, Jon Custer, stafusa, Aaron Stevens Aug 15 at 14:59

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