Slide 14 could be done similarly to slide 5.
Here’s my solution for slide 15, in the table CL:
eq1=parseEquation(this.cellContent(1,1)).differenceFunction(“x”,“y”)
eq2=parseEquation(this.cellContent(1,2)).differenceFunction(“x”,“y”)
correct: eq1.evaluateAt(-3,-3)=0 and eq2.evaluateAt(-8,-4)=0
parseEquation splits an equation into the left and right side. differenceFunction() takes the left side - right side (essentially setting an equation equal to zero). (“x”,“y”) defines what variables are being used. (lhs and rhs are other functions that output at the left hand side and right hand side respectively). The correctness check evaluates whether two solutions are correct. You can use when-otherwise conditionals to give more specific feedback.