The Heat is On

Since the boiling point of water is $100°C$ and the freezing point is $0°C$ , I will use that as my ToR (temperature of reference).

Satvik scale: Seeing how Satvik's scale shows $90°S$ for $100°C$ , and $-30°S$ for $0°C$ , we can have our slope as:

$m=\frac{90-(-30)}{100-0} = 1.2$

$b=-30$ since $-30°S = 0°C$

$y=1.2x-30$

Agnishom scale: The same concept applies as above, so I'm going to cut to the chase:

$y=0.8x+20$

And then we find when our two equations are equal... $y=\boxed{120}$ .

$\Rightarrow$ $S =\frac{120}{80} \times (A -20) -30$

$\Rightarrow$ $A =\frac{80}{120} \times (S +30) + 20$

$\Rightarrow$ $Let:$ $S =A = x$

$\Leftrightarrow$ $\frac{120}{80}$ $\times$ $(x -20)-30 = \frac{80}{120}$ $\times$ $(x +30)+20$

$\Leftrightarrow$ $\frac{120}{80}x - \frac{120}{80}$ $\times$ $20 - 30$ $= \frac{80}{120}x + \frac{80}{120}$ $\times$ $30 + 20$

$\Leftrightarrow$ $\frac{120}{80}x - \frac{80}{120}x = \frac{80}{120}$ $\times$ $30 + \frac{120}{80}$ $\times$ $20 + 50$

$\Leftrightarrow$ $x (\frac{12}{8}-\frac{8}{12}) = 20 + 30 +50$

$\Leftrightarrow$ $x (\frac{3}{2}-\frac{2}{3}) = 100$

$\Leftrightarrow$ $x (\frac{9}{6}-\frac{4}{9}) = 100$

$\Leftrightarrow$ $\frac{5}{6}x = 100$

$x = \boxed{120}$

Satvik scale
Ys = (90+30)x - 30

Agniston scale

Ya= (100-20)x +20

eliminating x in above we get (Ys +30)/(Ya-20)= 120/80

Since we look same value in both scales Ys=Ya

it results Ys=Ya= 120

The rate of temperature change for Agnishom to Satvik is 80:120 or 2:3 since that's the span of their readings.

The conversion formula from S to A is (S* 2/3) + 40 = A

The conversion formula from A to S is (A-40) * 3/2 = S

To find temperature at which both scales show same reading we use S=A in any of the the above equations and solve.

(S *2/3) +40 = S

S= A = 120 degrees.