We know that the Fahrenheit scale can be converted to the centigrade scale as
"t(^\\circ\\mathrm{C}) = (t(^\\circ\\mathrm{F}) - 32)\\cdot 5\/9."
"t(^\\circ\\mathrm{C})" should be equal to "3t(^\\circ\\mathrm{F})" , therefore we obtain an equation
"3t(^\\circ\\mathrm{F}) = (t(^\\circ\\mathrm{F}) - 32)\\cdot 5\/9."
"27t(^\\circ\\mathrm{F}) = 5t(^\\circ\\mathrm{F})-160, \\;\\; t(^\\circ\\mathrm{F}) = -7\\dfrac{3}{11}^\\circ F \\approx-7.3^\\circ F."
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