\(\Rightarrow f _1=-40\,cm\)
\(\frac{1}{ f _2}=(1.75-1)\left(\frac{1}{30}\right) \Rightarrow f _2=40\,cm\)
Image from \(L_1\) will be virtual and on the left of \(L_1\) at focal length \(40 \,cm\). So the object for \(L_2\) will be \(80\,cm\) from \(L _2\) which is \(2 f\). Final image is formed at \(80\,cm\) from \(L _2\) on the right. So \(x=120\)