Question

18yh answer please guys​

Answer 1

$let \: the \:no.be \: x \\ other \: no \: be \: y \\ given \: in \: equation \: that \: \: \: x = 3y \: \: \\ {eq1} \\ \: \ but \: in \: second \: equation \: \: \: \: x - y = 26 \\ so \: now \\ from \: these \: eqs \: we \: get \: \\ 3y - y = 26 \: \\ y = 13 \\ x = 3 \times 13 = 39$