Question

One year ago a mother was six times as old as her daughter presently the age of the mother is numerically equal to the squre of the age of her daughter find the daughter is present age (in years )

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$let \: here \: \\ the \: present \: age \: of \: mother \: be \: x \: years \\ and \: that \: of \: her \: daughter \: \\ be \: y \: years \\ \\ to \: find : \\ present \: ages \: of \: mother \: and \: \\ her \: daughter \\ \\ so \: according \: to \: first \: scenario \\ x - 1 = 6(y - 1) \\ x - 1 = 6y - 6 \\ x - 6y - 1 + 6 = 0 \\ x - 6y + 5 = 0$

$according \: to \: second \: condition \\ x = y {}^{2} \: \: \: \: \: \: \: \: (1)$

$thus \: then \: accordingly \\ y {}^{2} - 6y + 5 = 0 \\ y {}^{2} - 5y - y + 5 = 0 \\ y(y - 5) - 1(y - 5) = 0 \\ (y - 1)(y - 5) = 0 \\ \\ y - 1 = 0 \: \: or \: \: y - 5 = 0 \\ \\ y = 1 \: \: or \: \: y = 5$

$but \: if \: we \: take \\ y = 1 \\ then \: \: since \: here \\ x = y {}^{2} = (1) {}^{2} = 1 \\ as \: age \: of \: mother \: cannot \: be \: equal \\ to \: here \: daughter \: age \: \\ y = 1 \: \: i s\: absurd \\ \\ y = 5 \\ \\ thus \: then \\ x = (5) {}^{2} = 25$