Question Number 170838 by pablo1234523 last updated on 01/Jun/22
$$\mathrm{if}\:{a}<{b},\:\mathrm{show}\:\mathrm{that}\:{a}<\frac{{mb}+{na}}{{m}+{n}}<{b} \\ $$ $${a},{b},{m},{n}\:\mathrm{are}\:\mathrm{arbitrary}\:\mathrm{constants} \\ $$
Answered by chengulapetrom last updated on 04/Jun/22
$$\mathrm{if}\:{a}<{b},\:\mathrm{show}\:\mathrm{that}\:{a}<\frac{{mb}+{na}}{{m}+{n}}<{b} \\ $$ $${a},{b},{m},{n}\:\mathrm{are}\:\mathrm{arbitrary}\:\mathrm{constants} \\ $$ $${na}<{nb} \\ $$ $${ma}<{mb} \\ $$ $${na}+{ma}<{mb}+{na} \\ $$ $${a}\left({m}+{n}\right)<{mb}+{na} \\ $$ $${a}<\frac{{mb}+{na}}{{m}+{n}}.............\left({i}\right) \\ $$ $${na}+{mb}<{nb}+{mb} \\ $$ $${mb}+{na}<{b}\left({m}+{n}\right) \\ $$ $$\frac{{mb}+{na}}{{m}+{n}}<{b}...............\left({ii}\right) \\ $$ $${combining}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{yields} \\ $$ $${a}<\frac{{mb}+{na}}{{m}+{n}}<{b} \\ $$
Commented bypablo1234523 last updated on 07/Jun/22
$$\mathrm{thanks}\:\mathrm{sir}. \\ $$ $$\mathrm{but}\:\mathrm{if}\:{n}<\mathrm{0}\:\mathrm{then}\:{na}>{nb} \\ $$ $$\mathrm{then}\:\mathrm{it}\:\mathrm{will}\:\mathrm{be}\:\mathrm{incorrect} \\ $$