Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Solution to Problem 1

De Morgan enclosed his solution to Problem 1 in a letter to Lovelace of November 14, 1840 [LB 170, 14 Nov. 1840, f. 20v]. Beginning with the initial equation
\[\frac{1+b}{1-b}=\frac{1+x}{x}\]
he multiplied both sides by \(1-b\) and \(x\):
\[(1+b)x=(1-b)(1+x).\]Then, distributing
\[x+bx=1+x-b-bx\]cancelling an \(x\) from both sides
\[bx=1-b-bx\]collecting like terms
\[2bx+b=1\]factoring out the \(b\)
\[(2x+1)b=1\]and finally dividing by \(2x+1\) gave
\[b=\frac{1}{2x+1}.\]

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