# Obtain an expression for pressure difference [closed]

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 StevensAug 15 at 14:59

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