I am trying to figure out how to do “isCorrect” in such a way that cell(1,2) is not the same equation as in cell(1,1). Here is my coding so far:
#Correct response x+y=59685 or 5x+2.5y=270425
#creates lhs-rhs, so f(x,y)= x+y-59685 and g(x,y)=5x+2.5y-270425
f = parseEquation(this.cellContent(1,1)).differenceFunction(“x”,“y”)
g = parseEquation(this.cellContent(1,2)).differenceFunction(“x”,“y”)
#solutions should = 0
checkA = f.evaluateAt(59685,0)=0 and f.evaluateAt(0,59685)=0
checkB = f.evaluateAt(54085,0)=0 and f.evaluateAt(0,108170)=0
checkC = g.evaluateAt(59685,0)=0 and g.evaluateAt(0,59685)=0
checkD = g.evaluateAt(54085,0)=0 and g.evaluateAt(0,108170)=0
correct: (checkA and checkD) or (checkB and checkC)
isCorrect1 =
isCorrect2 =
Do I need to add some other coding before I get to isCorrect1 and isCorrect2? I’m wanting isCorrect1 to represent the entry in cell(1,1) and isCorrect2 to represent the entry in cell(1,2).