a-2b-5c-3d-1146-2a-3b-2c-d-3e-1279-5a-b-2c-d-3e-1555-4a-2b-3c-d-e-1459-a-2b-2c-2d-4e-1194-Find-a-b-c-d-e-

Question Number 228260 by hardmath last updated on 04/Apr/26

$$\begin{cases}{\mathrm{a}\:+\:\mathrm{2b}\:+\:\mathrm{5c}\:+\:\mathrm{3d}\:=\:\mathrm{1146}}\\{\mathrm{2a}\:+\:\mathrm{3b}\:+\:\mathrm{2c}\:+\:\mathrm{d}\:+\:\mathrm{3e}\:=\:\mathrm{1279}}\\{\mathrm{5a}\:+\:\mathrm{b}\:+\:\mathrm{2c}\:+\:\mathrm{d}\:+\:\mathrm{3e}\:=\:\mathrm{1555}}\\{\mathrm{4a}\:+\:\mathrm{2b}\:+\:\mathrm{3c}\:+\:\mathrm{d}\:+\:\mathrm{e}\:=\:\mathrm{1459}}\\{\mathrm{a}\:+\:\mathrm{2b}\:+\:\mathrm{2c}\:+\:\mathrm{2d}\:+\:\mathrm{4e}\:=\:\mathrm{1194}}\end{cases} \\ $$$$\mathrm{Find}:\:\left(\mathrm{a};\mathrm{b};\mathrm{c};\mathrm{d};\mathrm{e}\right)\:=\:? \\ $$

Commented by nikif99 last updated on 06/Apr/26

$${Another}\:{approach}. \\ $$$${Subtracting}\:\left(\mathrm{3}\right)−\left(\mathrm{2}\right)\:{we}\:{get}\:\mathrm{3}{a}−\mathrm{2}{b}=\mathrm{276} \\ $$$${Multiplying}\:\left(\mathrm{1}\right)\:{by}\:\mathrm{8},\:\left(\mathrm{2}\right)\:{by}\:\mathrm{15},\:\left(\mathrm{3}\right)\:{by}\:\mathrm{13}, \\ $$$$\left(\mathrm{4}\right)\:{by}\:−\mathrm{22}\:{and}\:\left(\mathrm{5}\right)\:{by}\:−\mathrm{15}\:{and}\:{adding} \\ $$$${by}\:{parts}\:{we}\:{eliminate}\:{all}\:{of}\:{a},\:{b},\:{c}\:{and}\:{d}. \\ $$$${What}\:{it}\:{rests}\:{is}\:\mathrm{2}{e}=−\mathrm{1440}\Rightarrow{e}=−\mathrm{720} \\ $$$${then}\:{it}\:{is}\:{a}\:{little}\:{easier}. \\ $$

Answered by TonyCWX last updated on 05/Apr/26

$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{5}}&{\mathrm{3}}&{\mathrm{0}}&{\mathrm{1146}}\\{\mathrm{2}}&{\mathrm{3}}&{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}&{\mathrm{1279}}\\{\mathrm{5}}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}&{\mathrm{1555}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{3}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1459}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{2}}&{\mathrm{2}}&{\mathrm{4}}&{\mathrm{1194}}\end{bmatrix} \\ $$$$\begin{array}{|c|}{}\\\hline\end{array}\cancel{ }\underline{\mathrm{Operation}\:\mathrm{I}} \\ $$

Commented by Math1 last updated on 05/Apr/26

$$\mathrm{Is}\:\mathrm{there}\:\mathrm{another}\:\mathrm{simple}\:\mathrm{way}\:\mathrm{to}\:\mathrm{do}\:\mathrm{this}.? \\ $$

Commented by TonyCWX last updated on 06/Apr/26

$$\mathrm{Substitution}\:\mathrm{or}\:\mathrm{Elimination}\:\mathrm{works}\:\mathrm{as}\:\mathrm{well}. \\ $$$$\mathrm{You}\:\mathrm{may}\:\mathrm{try}\:\mathrm{yourself}. \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{too}\:\mathrm{lazy}\:\mathrm{to}\:\mathrm{work}\:\mathrm{them}\:\mathrm{out}.\:\mathrm{XD} \\ $$

Answered by nikif99 last updated on 05/Apr/26

$${Following}\:{Cramer}'{s}\:{rule},\:{the} \\ $$$${determinant}\:{of}\:{coefficients}\:{is} \\ $$$${D}=\begin{vmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{5}}&{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{2}}&{\mathrm{3}}&{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{5}}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{3}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{2}}&{\mathrm{2}}&{\mathrm{4}}\end{vmatrix}=−\mathrm{4} \\ $$$${Replacing}\:{any}\:{variable}\:{with}\:{constants} \\ $$$${we}\:{get} \\ $$$${D}_{{a}} =\begin{vmatrix}{\mathrm{1146}}&{\mathrm{2}}&{\mathrm{5}}&{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{1279}}&{\mathrm{3}}&{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{1555}}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{1459}}&{\mathrm{2}}&{\mathrm{3}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1194}}&{\mathrm{2}}&{\mathrm{2}}&{\mathrm{2}}&{\mathrm{4}}\end{vmatrix}=−\mathrm{3076} \\ $$$${a}=\frac{{D}_{{a}} }{{D}}=\mathrm{769} \\ $$$${Similarly},\:{D}_{{b}} =−\mathrm{4062},\:{D}_{{c}} =\mathrm{7130},\:{D}_{{d}} −\mathrm{9678} \\ $$$$\left({a};{b};{c};{d};{e}\right)= \\ $$$$\left(\mathrm{769};\mathrm{1015}.\mathrm{5};−\mathrm{1782}.\mathrm{5};\mathrm{2419}.\mathrm{5};−\mathrm{720}\right) \\ $$

Commented by TonyCWX last updated on 05/Apr/26

$$\mathrm{Nice}\:\mathrm{work},\:\mathrm{sir}! \\ $$

Terms of Service
Privacy Policy
Contact: [email protected]