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+// Exa 1.13
+clc;
+clear;
+close;
+// Given data
+Rf= 250;// in kohm
+// Output voltage expression, Vo= -5*Va+3*Vb
+// and we know that for a difference amplifier circuit, 
+// Vo= -Rf/R1*Va + [R2/(R1+R2)]*[1+Rf/R1]*Vb
+// Comparing both the expression, we get
+// -Rf/R1*Va= -5*Va, or
+R1= Rf/5;// in kohm
+disp(R1,"The value of R1 in kohm")
+// and 
+R2= 3*R1^2/(R1+Rf-3*R1)
+disp(R2,"The value of R2 in kohm")
+
+// Note : Answer in the book is wrong