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Question Number 187817 by normans last updated on 22/Feb/23

Answered by mr W last updated on 22/Feb/23

Commented by mr W last updated on 22/Feb/23

c+d=9/3=3 d+e=7/2 c+e=(9+7)/5=16/5 ⇒2d=3+7/2−16/5 ⇒d=33/20 ⇒c=3−33/20=27/20 ⇒e=7/2−33/20=37/20 b+c=Y/4 b+d=(Y+9)/7 b+e=(Y+9+7)/9 e+d−2c=(Y+9)/7+(Y+16)/9−2Y/4 37/20+33/20−2×27/20=9/7+16/9−31Y/126 ⇒Y=46/5 ⇒b=46/5/4−27/20=19/20 a+b=X/5 a+c=(X+Y)/9 b−c=X/5−X/9−Y/9 19/20−27/20=4X/45−46/5/9 ⇒X=7 ✓

$${c}+{d}=\mathrm{9}/\mathrm{3}=\mathrm{3} \\ $$$${d}+{e}=\mathrm{7}/\mathrm{2} \\ $$$${c}+{e}=\left(\mathrm{9}+\mathrm{7}\right)/\mathrm{5}=\mathrm{16}/\mathrm{5} \\ $$$$\Rightarrow\mathrm{2}{d}=\mathrm{3}+\mathrm{7}/\mathrm{2}−\mathrm{16}/\mathrm{5}\: \\ $$$$\Rightarrow{d}=\mathrm{33}/\mathrm{20} \\ $$$$\Rightarrow{c}=\mathrm{3}−\mathrm{33}/\mathrm{20}=\mathrm{27}/\mathrm{20} \\ $$$$\Rightarrow{e}=\mathrm{7}/\mathrm{2}−\mathrm{33}/\mathrm{20}=\mathrm{37}/\mathrm{20} \\ $$$${b}+{c}={Y}/\mathrm{4} \\ $$$${b}+{d}=\left({Y}+\mathrm{9}\right)/\mathrm{7} \\ $$$${b}+{e}=\left({Y}+\mathrm{9}+\mathrm{7}\right)/\mathrm{9} \\ $$$${e}+{d}−\mathrm{2}{c}=\left({Y}+\mathrm{9}\right)/\mathrm{7}+\left({Y}+\mathrm{16}\right)/\mathrm{9}−\mathrm{2}{Y}/\mathrm{4} \\ $$$$\mathrm{37}/\mathrm{20}+\mathrm{33}/\mathrm{20}−\mathrm{2}×\mathrm{27}/\mathrm{20}=\mathrm{9}/\mathrm{7}+\mathrm{16}/\mathrm{9}−\mathrm{31}{Y}/\mathrm{126} \\ $$$$\Rightarrow{Y}=\mathrm{46}/\mathrm{5} \\ $$$$\Rightarrow{b}=\mathrm{46}/\mathrm{5}/\mathrm{4}−\mathrm{27}/\mathrm{20}=\mathrm{19}/\mathrm{20} \\ $$$${a}+{b}={X}/\mathrm{5} \\ $$$${a}+{c}=\left({X}+{Y}\right)/\mathrm{9} \\ $$$${b}−{c}={X}/\mathrm{5}−{X}/\mathrm{9}−{Y}/\mathrm{9} \\ $$$$\mathrm{19}/\mathrm{20}−\mathrm{27}/\mathrm{20}=\mathrm{4}{X}/\mathrm{45}−\mathrm{46}/\mathrm{5}/\mathrm{9} \\ $$$$\Rightarrow{X}=\mathrm{7}\:\checkmark \\ $$

Commented by normans last updated on 22/Feb/23

nice solution

$$\:{nice}\:{solution} \\ $$

Answered by mr W last updated on 23/Feb/23

Commented by mr W last updated on 23/Feb/23

B_2 =(3/5)X_2 C_2 =(2/5)X_2 ((A_1 +X_1 )/X_1 )=(((4+5)/5))^2 =((81)/(25)) ((B_1 +A_1 +X_1 )/X_1 )=(((3+4+5)/5))^2 =((144)/(25)) ((C_1 +B_1 +A_1 +X_1 )/X_1 )=(((2+3+4+5)/5))^2 =((196)/(25)) ⇒(C_1 /X_1 )=((196−144)/(25))=((52)/(25)) ⇒(B_1 /X_1 )=((144−81)/(25))=((63)/(25)) B=B_1 +B_2 =((63)/(25))X_1 +(3/5)X_2 =9 ...(i) C=C_1 +C_2 =((52)/(25))X_1 +(2/5)X_2 =7 ...(ii) 63X_1 +15X_2 =225 52X_1 +10X_2 =175 ⇒X_1 =((225×10−175×15)/(63×10−52×15))=2.5 ⇒X_2 =((225×52−175×63)/(15×52−10×63))=4.5 ⇒X=X_1 +X_2 =2.5+4.5=7 ✓

$${B}_{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{5}}{X}_{\mathrm{2}} \\ $$$${C}_{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{5}}{X}_{\mathrm{2}} \\ $$$$\frac{{A}_{\mathrm{1}} +{X}_{\mathrm{1}} }{{X}_{\mathrm{1}} }=\left(\frac{\mathrm{4}+\mathrm{5}}{\mathrm{5}}\right)^{\mathrm{2}} =\frac{\mathrm{81}}{\mathrm{25}} \\ $$$$\frac{{B}_{\mathrm{1}} +{A}_{\mathrm{1}} +{X}_{\mathrm{1}} }{{X}_{\mathrm{1}} }=\left(\frac{\mathrm{3}+\mathrm{4}+\mathrm{5}}{\mathrm{5}}\right)^{\mathrm{2}} =\frac{\mathrm{144}}{\mathrm{25}} \\ $$$$\frac{{C}_{\mathrm{1}} +{B}_{\mathrm{1}} +{A}_{\mathrm{1}} +{X}_{\mathrm{1}} }{{X}_{\mathrm{1}} }=\left(\frac{\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}}{\mathrm{5}}\right)^{\mathrm{2}} =\frac{\mathrm{196}}{\mathrm{25}} \\ $$$$\Rightarrow\frac{{C}_{\mathrm{1}} }{{X}_{\mathrm{1}} }=\frac{\mathrm{196}−\mathrm{144}}{\mathrm{25}}=\frac{\mathrm{52}}{\mathrm{25}} \\ $$$$\Rightarrow\frac{{B}_{\mathrm{1}} }{{X}_{\mathrm{1}} }=\frac{\mathrm{144}−\mathrm{81}}{\mathrm{25}}=\frac{\mathrm{63}}{\mathrm{25}} \\ $$$${B}={B}_{\mathrm{1}} +{B}_{\mathrm{2}} =\frac{\mathrm{63}}{\mathrm{25}}{X}_{\mathrm{1}} +\frac{\mathrm{3}}{\mathrm{5}}{X}_{\mathrm{2}} =\mathrm{9}\:\:\:...\left({i}\right) \\ $$$${C}={C}_{\mathrm{1}} +{C}_{\mathrm{2}} =\frac{\mathrm{52}}{\mathrm{25}}{X}_{\mathrm{1}} +\frac{\mathrm{2}}{\mathrm{5}}{X}_{\mathrm{2}} =\mathrm{7}\:\:\:...\left({ii}\right) \\ $$$$\mathrm{63}{X}_{\mathrm{1}} +\mathrm{15}{X}_{\mathrm{2}} =\mathrm{225} \\ $$$$\mathrm{52}{X}_{\mathrm{1}} +\mathrm{10}{X}_{\mathrm{2}} =\mathrm{175} \\ $$$$\Rightarrow{X}_{\mathrm{1}} =\frac{\mathrm{225}×\mathrm{10}−\mathrm{175}×\mathrm{15}}{\mathrm{63}×\mathrm{10}−\mathrm{52}×\mathrm{15}}=\mathrm{2}.\mathrm{5} \\ $$$$\Rightarrow{X}_{\mathrm{2}} =\frac{\mathrm{225}×\mathrm{52}−\mathrm{175}×\mathrm{63}}{\mathrm{15}×\mathrm{52}−\mathrm{10}×\mathrm{63}}=\mathrm{4}.\mathrm{5} \\ $$$$\Rightarrow{X}={X}_{\mathrm{1}} +{X}_{\mathrm{2}} =\mathrm{2}.\mathrm{5}+\mathrm{4}.\mathrm{5}=\mathrm{7}\:\checkmark \\ $$

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