Let old/initial price of 1 egg \(=\# x\)
Let no of eggs #120 would buy (at that time) be \(y_{1}\)
l egg \(=\# x\)
\(y_{1}=\# 120\)
\(y_{1}=\# \frac{120}{x}\)rom the question. \(y_{1}-5=y_{2}\)rom the question.
\begin{aligned}
y_{1}-5 &=1 \\
\text { i.e. } \frac{120}{x}-5 &=\frac{120}{x+2} \\
\frac{120-5 x}{x} &=\frac{120}{x+2}
\end{aligned}
\begin{aligned}
(120-5 x)(x+2) &=120 x \\
120 x+240-5 x^{2}-10 x &=120 x \\
x-5 x^{2}-10 x+240 &=0
\end{aligned}
divide all through by 5
\(x^{2}+2 x-48=0\)
by factorizing
\begin{aligned}
&x^{2}+8 x-6 x-48=0 \\
&x(x+8)-6(x+8)=0 \\
&x-6=0 x+8=0 \\
&x=6 x=-8
\end{aligned}
since the value of money cannot be negative.
x=6
Therefore.
\begin{aligned}
\text { the new price/present price } &=+2 \\
&=\# 6+\# 2 \\
&=\# 8
\end{aligned}