Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 104777 by Anindita last updated on 23/Jul/20

Answered by Aziztisffola last updated on 23/Jul/20

3.5+4.5=8 + + 9.5−3.5=6 13 8

$$\mathrm{3}.\mathrm{5}+\mathrm{4}.\mathrm{5}=\mathrm{8} \\ $$$$+\:\:\:\:\:\:\:\:\:+ \\ $$$$\mathrm{9}.\mathrm{5}−\mathrm{3}.\mathrm{5}=\mathrm{6} \\ $$$$\mathrm{13}\:\:\:\:\:\:\:\:\mathrm{8} \\ $$

Answered by 1549442205PVT last updated on 24/Jul/20

We have the following system of four equations: { ((x+y=8(1))),((z−t=6(2))),((x+z=13(3))),((y+t=8(4))) :} From (1)(4)we get x=t=8−y(5) subtract two eqs.(2)and(3)we get x+t=7⇒2x=2t=7(by (5))⇒x=t=3.5 (4)⇒y=8−3.5=4.5,(2)⇒z=6+t=9.5 Thus,we get (x,y,z,t)=(3.5;4.5;9.5;3.5) determinant (((3.5),+,(4.5),=,8),(+,,+,,),((9.5),−,(3.5),=,6),(=,,=,,),((13),,8,,))

$$\mathrm{We}\:\mathrm{have}\:\mathrm{the}\:\mathrm{following}\:\mathrm{system}\:\mathrm{of}\:\mathrm{four} \\ $$$$\mathrm{equations}:\begin{cases}{\mathrm{x}+\mathrm{y}=\mathrm{8}\left(\mathrm{1}\right)}\\{\mathrm{z}−\mathrm{t}=\mathrm{6}\left(\mathrm{2}\right)}\\{\mathrm{x}+\mathrm{z}=\mathrm{13}\left(\mathrm{3}\right)}\\{\mathrm{y}+\mathrm{t}=\mathrm{8}\left(\mathrm{4}\right)}\end{cases} \\ $$$$\mathrm{From}\:\left(\mathrm{1}\right)\left(\mathrm{4}\right)\mathrm{we}\:\mathrm{get}\:\mathrm{x}=\mathrm{t}=\mathrm{8}−\mathrm{y}\left(\mathrm{5}\right) \\ $$$$\mathrm{subtract}\:\mathrm{two}\:\mathrm{eqs}.\left(\mathrm{2}\right)\mathrm{and}\left(\mathrm{3}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{x}+\mathrm{t}=\mathrm{7}\Rightarrow\mathrm{2x}=\mathrm{2t}=\mathrm{7}\left(\mathrm{by}\:\left(\mathrm{5}\right)\right)\Rightarrow\mathrm{x}=\mathrm{t}=\mathrm{3}.\mathrm{5} \\ $$$$\left(\mathrm{4}\right)\Rightarrow\mathrm{y}=\mathrm{8}−\mathrm{3}.\mathrm{5}=\mathrm{4}.\mathrm{5},\left(\mathrm{2}\right)\Rightarrow\mathrm{z}=\mathrm{6}+\mathrm{t}=\mathrm{9}.\mathrm{5} \\ $$$$\boldsymbol{\mathrm{Thus}},\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{get}}\:\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{z}},\boldsymbol{\mathrm{t}}\right)=\left(\mathrm{3}.\mathrm{5};\mathrm{4}.\mathrm{5};\mathrm{9}.\mathrm{5};\mathrm{3}.\mathrm{5}\right) \\ $$$$\begin{vmatrix}{\mathrm{3}.\mathrm{5}}&{+}&{\mathrm{4}.\mathrm{5}}&{=}&{\mathrm{8}}\\{+}&{}&{+}&{}&{}\\{\mathrm{9}.\mathrm{5}}&{−}&{\mathrm{3}.\mathrm{5}}&{=}&{\mathrm{6}}\\{=}&{}&{=}&{}&{}\\{\mathrm{13}}&{}&{\mathrm{8}}&{}&{}\end{vmatrix} \\ $$

Answered by Rasheed.Sindhi last updated on 24/Jul/20

determinant (( [(( x)),((2x+6=13)),(( x=3 1/2_(−) )) ],→_+ , [(( 8−x)),((8−3 1/2 )),(( 4 1/2)) ],8),(( ↑+),,( ↓+),),( [(( x+6)),(( 3 1/2+6 )),((=9 1/2)) ],←_− , [((8−(8−x))),(( x)),(( 3 1/2)) ],6),((−−−−−−−),−,(−−−−−−),−),(( 13),,( 8),))

$$\begin{vmatrix}{\begin{bmatrix}{\:\:\:\:\:\:\:\:\:\:\:{x}}\\{\mathrm{2}{x}+\mathrm{6}=\mathrm{13}}\\{\underset{−} {\:{x}=\mathrm{3}\:\mathrm{1}/\mathrm{2}}\:\:\:\:\:\:\:\:}\end{bmatrix}}&{\underset{+} {\rightarrow}}&{\begin{bmatrix}{\:\:\:\:\:\mathrm{8}−{x}}\\{\mathrm{8}−\mathrm{3}\:\mathrm{1}/\mathrm{2}\:\:}\\{\:\:\:\:\mathrm{4}\:\mathrm{1}/\mathrm{2}}\end{bmatrix}}&{\mathrm{8}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\uparrow+}&{}&{\:\:\:\:\:\:\:\:\:\:\downarrow+}&{}\\{\begin{bmatrix}{\:\:\:\:\:\:\:\:\:{x}+\mathrm{6}}\\{\:\:\:\:\:\mathrm{3}\:\mathrm{1}/\mathrm{2}+\mathrm{6}\:\:\:\:}\\{=\mathrm{9}\:\mathrm{1}/\mathrm{2}}\end{bmatrix}}&{\underset{−} {\leftarrow}}&{\begin{bmatrix}{\mathrm{8}−\left(\mathrm{8}−{x}\right)}\\{\:\:\:\:\:\:\:\:\:{x}}\\{\:\:\:\:\mathrm{3}\:\mathrm{1}/\mathrm{2}}\end{bmatrix}}&{\mathrm{6}}\\{−−−−−−−}&{−}&{−−−−−−}&{−}\\{\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{13}}&{}&{\:\:\:\:\:\:\:\:\:\:\:\mathrm{8}}&{}\end{vmatrix} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com