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Same thing here C-R....substitute the   x,y  in to the equation to see if they satisfy they eq....NOTE:  more than one set may work....

10 x 3/3   +  2/3   =   30/3  + 2/3   =  (30+2)/3  =  32/3

thank you ElectricPavlov 

False.    If you have the equation  and   the graph of the equation....all of the points will satisfy the equation.

All you have to do is substitute   -2  for x and  2 for y in to the equations and see which one 'works'   ...try it !

Thank you Link, (though it was my friend who wanted a quick answer for a problem, so thats my reason for change of text )

thank you Link

no. there isnt a diagram. 

Transpose the formula for ‘A’ and find the value of A when
k1 = k2 = 0.2

k3 = 0.5 

x1 = 40 x 10-3

x2 = 125 x 10-3

x3 = 60 x 10-3

Rt = 500

$\begin{array}{|rcll|} \hline R_T &=& \dfrac{x_1}{k_1A} + \dfrac{x_2}{k_2A} + \dfrac{x_3}{k_3A} \\\\ &=& \dfrac{1}{A}\left(\dfrac{x_1}{k_1 } + \dfrac{x_2}{k_2 } + \dfrac{x_3}{k_3 }\right)\\ \\ \mathbf{A} & \mathbf{=} & \mathbf{\dfrac{1}{R_T}\left(\dfrac{x_1}{k_1} + \dfrac{x_2}{k_2} + \dfrac{x_3}{k_3}\right)} \\\\ &=& \dfrac{1}{500}\cdot\left(\dfrac{40 \cdot 10^{-3}}{0.2} + \dfrac{125 \cdot 10^{-3}}{0.2} + \dfrac{60 \cdot 10^{-3}}{0.5}\right) \\\\ &=& \dfrac{10^{-3}}{500}\cdot\left(\dfrac{40}{0.2} + \dfrac{125}{0.2} + \dfrac{60}{0.5}\right) \\\\ &=& \dfrac{10^{-3}}{500}\cdot\left(\dfrac{40\cdot 10}{ 2} + \dfrac{125\cdot 10}{2} + \dfrac{60\cdot 10}{5}\right) \\\\ &=& \dfrac{10^{-3}\cdot 10}{500}\cdot\left(\dfrac{40}{ 2} + \dfrac{125}{2} + \dfrac{60}{5}\right) \\\\ &=& \dfrac{10^{-3}\cdot 10}{5\cdot 10^2}\cdot\left(\dfrac{40}{ 2} + \dfrac{125}{2} + \dfrac{60}{5}\right) \\\\ &=& \dfrac{10^{-3}}{50} \cdot (20 + 62.5 + 12 ) \\\\ &=& \dfrac{94.5\cdot 10^{-3}}{50} \\\\ \mathbf{A} & \mathbf{=} & \mathbf{1.89\cdot 10^{-3}} \\ \hline \end{array}$

Looks good  I appreciate all the effort you put into helping me. 

Ok here it is without knowing the formula at all.

I think my hand sketching needs practice though  LOL