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Question-189208

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Question Number 189208 by mathlove last updated on 13/Mar/23
Answered by MJS_new last updated on 13/Mar/23
easy to see 9+2=11 3+4=7 x=9∧y=4
$$\mathrm{easy}\:\mathrm{to}\:\mathrm{see} \\ $$$$\mathrm{9}+\mathrm{2}=\mathrm{11} \\ $$$$\mathrm{3}+\mathrm{4}=\mathrm{7} \\ $$$${x}=\mathrm{9}\wedge{y}=\mathrm{4} \\ $$
Commented by mathlove last updated on 13/Mar/23
solve it???
$${solve}\:{it}??? \\ $$
Commented by MJS_new last updated on 13/Mar/23
another exampke: solve 3+x=5: 3=1+1+1 5=1+1+1+1+1 how many 1s must we add to get from 3 to 5? 1+1+1+1+1=1+1+1+1+1 1+1=2 the answer is 3+2=5 ⇒ x=2 now try to solve your question
$$\mathrm{another}\:\mathrm{exampke}: \\ $$$$\mathrm{solve}\:\mathrm{3}+{x}=\mathrm{5}: \\ $$$$\mathrm{3}=\mathrm{1}+\mathrm{1}+\mathrm{1} \\ $$$$\mathrm{5}=\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1} \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{1s}\:\mathrm{must}\:\mathrm{we}\:\mathrm{add}\:\mathrm{to}\:\mathrm{get}\:\mathrm{from}\:\mathrm{3}\:\mathrm{to}\:\mathrm{5}? \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}=\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1} \\ $$$$\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$$$\mathrm{the}\:\mathrm{answer}\:\mathrm{is} \\ $$$$\mathrm{3}+\mathrm{2}=\mathrm{5} \\ $$$$\Rightarrow\:{x}=\mathrm{2} \\ $$$$\mathrm{now}\:\mathrm{try}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{your}\:\mathrm{question} \\ $$
Answered by manxsol last updated on 13/Mar/23
x<11 y<7 x=a^(2 ) y=b^2 a^2 +b=11 a+b^2 =7 (7−b^2 )^2 +b=11 49+b^4 −14b^2 =11 b^4 −14b^2 +b+38=0 1 0 −14 1 38 2 2 4 −20 −38 1 2 −10 −19 0 b_1 =2 a_1 ^2 =11−2 =9 b_2 =3.13 a_2 ^2 =11−3.13=7.87 b_3 =−1.84 a_3 ^2 =11+1.84=12.84 b_4 =−3.28 a_4 ^2 =3.77+3.28=7.05 x_1 =a_1 ^2 =9 y_1 =b_1 ^2 =4 −−−−− x_2 =7.87 y_2 =9.13 y<7 × −−−−−−))) x_3 =12.8 x<11 × −−−−−−− x_4 =7.05 y_4 =10.75 y<7 ×
$${x}<\mathrm{11}\:\:\:{y}<\mathrm{7} \\ $$$${x}={a}^{\mathrm{2}\:\:\:\:} \:\:\:{y}={b}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}=\mathrm{11} \\ $$$${a}+{b}^{\mathrm{2}} =\mathrm{7} \\ $$$$\left(\mathrm{7}−{b}^{\mathrm{2}} \right)^{\mathrm{2}} +{b}=\mathrm{11} \\ $$$$\mathrm{49}+{b}^{\mathrm{4}} −\mathrm{14}{b}^{\mathrm{2}} =\mathrm{11} \\ $$$${b}^{\mathrm{4}} −\mathrm{14}{b}^{\mathrm{2}} +{b}+\mathrm{38}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\mathrm{1}\:\:\:\mathrm{0}\:\:\:\:\:−\mathrm{14}\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{38} \\ $$$$\mathrm{2}\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:−\mathrm{20}\:\:\:−\mathrm{38} \\ $$$$\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\mathrm{2}\:\:\:\:\:\:−\mathrm{10}\:\:\:−\mathrm{19}\:\:\:\:\:\mathrm{0} \\ $$$${b}_{\mathrm{1}} =\mathrm{2}\:\:\:\:\:\:{a}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{11}−\mathrm{2}\:\:=\mathrm{9} \\ $$$${b}_{\mathrm{2}} =\mathrm{3}.\mathrm{13}\:\:\:\:\:{a}_{\mathrm{2}} ^{\mathrm{2}} \:=\mathrm{11}−\mathrm{3}.\mathrm{13}=\mathrm{7}.\mathrm{87} \\ $$$${b}_{\mathrm{3}} =−\mathrm{1}.\mathrm{84}\:\:\:{a}_{\mathrm{3}} ^{\mathrm{2}} =\mathrm{11}+\mathrm{1}.\mathrm{84}=\mathrm{12}.\mathrm{84} \\ $$$${b}_{\mathrm{4}} =−\mathrm{3}.\mathrm{28}\:\:\:\:\:{a}_{\mathrm{4}} ^{\mathrm{2}} =\mathrm{3}.\mathrm{77}+\mathrm{3}.\mathrm{28}=\mathrm{7}.\mathrm{05} \\ $$$${x}_{\mathrm{1}} ={a}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{9} \\ $$$${y}_{\mathrm{1}} ={b}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{4} \\ $$$$−−−−− \\ $$$${x}_{\mathrm{2}} =\mathrm{7}.\mathrm{87} \\ $$$${y}_{\mathrm{2}} =\mathrm{9}.\mathrm{13}\:\:\:\:\:\:\:{y}<\mathrm{7}\:\:\:× \\ $$$$\left.−\left.−\left.−−−−\right)\right)\right) \\ $$$${x}_{\mathrm{3}} =\mathrm{12}.\mathrm{8}\:\:\:\:\:\:{x}<\mathrm{11}\:\:× \\ $$$$−−−−−−− \\ $$$${x}_{\mathrm{4}} =\mathrm{7}.\mathrm{05} \\ $$$${y}_{\mathrm{4}} =\mathrm{10}.\mathrm{75}\:\:\:\:\:{y}<\mathrm{7}\:× \\ $$$$ \\ $$

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