Can you Pass Oxford University Admission Exam? ✍️🖋️📘💙

[@9허공]

a(b + c) = 300 & b(a + c) = 400 & c(a + b) = 500 ---(1) adding 3 equations => 2(ab + bc + ca) = 1200 => ab + bc + ca = 600 ---(2) from (1) and (2) => bc = 300 & ca = 200 & ab = 100 ---(3) multiplying 3 equations, (abc)^2 = 6*10^6 => abc = ±√6*10^3 ---(4) from (3) and (4) , a = ±10*√6/3 & b = ±10*√6/2 & c = ±10*√6 => a + b + c = ±55/3*√6

[@IsaacMakhabane]

Your work awesome. You are eligible of producing independent students. You use everything learners can understand. Keep it up.

[@yuliacrisp1932]

There is much easier way of solving it, it doesn't involve quadratic equation.

[@shannonmcdonald7584]

You are the master. Wow.

[@superacademy247]

Thanks for sharing your feedback 💯🙏🤩💕. I'm glad it was helpful 🙏🤩😍🥰✅

[@RyanLewis-Johnson-wq6xs]

a(b+c)=300 b(a+c)=400 c(a+b)=500 a+b+c=(±55Sqrt[6])/3=±18.3 recurring Sqrt[6]=(18 1/3) Sqrt[6]

[@ManojkantSamal]

^=read as to the power *=read as square root As per question a(b+c)=300...eqn1 b(a+c)=400....eqn2 C(a+b)=500...eqn3 Eqn1 -eqn3 ab+ac-ac-bc=300-500 ab-bc=-200 b(a-c)=-200....eqn4 Eqn2/eqn4 b(a+c)/{b(a-c)}=400/(-200) (a+c)/(a-c)=(-2) (a+c)=-2(a-c) a+c=-2a+2c a+2a=2c-c C=3a a=c/3......eqn5 Eqn2 - eqn3 ab+bc-ac-bc=400-500 ab-ac=-100 a(b-c)=-100......eqn6 Eqn1/eqn6 a(b+c)/{a(b-c)}=300/-100 (b+c)/(b-c)=-3 b+c=-3b+3c b+3b=3c-c 4b=2c b=2c/4=c/2.....eqn7 Put eqn1 & eqn2 in eqn3 C(a+b)=500 C{(c/3)+(c/2)}=500 C(5c/6)=500 5c^2=500×6=3000 C^2=3000/5=600 *(c^2)=*600 C=10.*6 a=c/3 =(10.*6)/3 b=c/2 =(10.*6)/2 =5.*6 a+b+c={(10.*6)/3}+(5.*6)+(10*6) ={10.*6)+(15.*6)+30.*6)}/3 =(55.*6)/3

[@Марія-75]

Я восхищаюсь, спасибо за учение и старание

[@superacademy247]

I'm glad you found it helpful 🙏💕💯. Thanks for your support and feedback 💯🙏🔥

[@davidseed2939]

at first i can see ab=100, ac=200, bc=300 eliminating ab from the first two equations gives c(b-a)=100 subs into eqn 3 100(a+b)=500b -500a 6a=4b etc so c=3a=2b, so c( a+b)=500 becomes c(c/3+c/2)=500 5/6 c^2=500, c^2=600, c =10 sqrt(6) a+b+c =c+c/3+c/2 = (11/6)c =110sqrt(6)/6

[@ericchan7128]

I square up the 3 equation first a(b+c) =300 ...(1) b(a+c) =400 ...(2) c(a+b) =500 ...(3) since, 300^2 +400^2=500^2 , get the equation [a(b+c)]^2 +[b(a+c)]^2 = [c(a+b)]^2 a^2(b+c)^2 +b^2(a+c)^2 = c^2(a+b)^2 a^2(b^2+c^2+2bc)+ b^2(a^2+c^2+2ac )^2 = c^2(a^2+b^2 +2ab) (a^2b^2)+(a^2c^2)+2(a^2bc)+ (a^2b^2)+(b^2c^2)+(2ab^2c ) = (a^2c^2)+(b^2c^2) +(2abc^2) (a^2b^2)+2(a^2bc)+ (a^2b^2)+(2ab^2c ) = (2abc^2) 2(a^2b^2)+2(a^2bc)+(2ab^2c ) = (2abc^2) (a^2b^2)+(a^2bc)+(ab^2c ) = (abc^2) (ab)+(ac)+(bc ) = (c^2) ...(4) summing up ...(1)...(2)...(3) a(b+c) b(a+c) c(a+b) =300 +400 +500 2[(ab)+(ac)+(bc ) ] =1200 [(ab)+(ac)+(bc ) ] =600 ...(4)=...(5) => (c^2) =600 c = sqrt(600) = 10sqrt(6) ...(1)+...(2) a(b+c) + b(a+c) =300 +400 2ab+ac + bc =700 2ab+(a + b) [10sqrt(6)]=700 ...(5) from (3) c(a+b) =500 (a+b) =50 /sqrt(6) ...(6) sub into ...(5) 2ab+(a + b) [10sqrt(6)]=700 2ab+[50 /sqrt(6)] [10sqrt(6)]=700 ab=100...(7) sub into ...(1) a(b+c) =300 ab+ac =300 100 +a(10sqrt(6)) =300 a(sqrt(6)) =20 a =20/(sqrt(6)) sub into ...(7) ab=100 b[20/(sqrt(6))] =100 b =100/[20/(sqrt(6))] =5(sqrt(6)) a +b+ c =20/(sqrt(6)) +5(sqrt(6)) +10sqrt(6) =20/(sqrt(6)) +15(sqrt(6)) =110/(sqrt(6))

[@Jsvs-q2b]

В конце запутались. Проще можно: 10sq(2/3)+10sq(3/2)+10sq(6)

[@9허공]

10 => ±10

[@davidshen5916]

X=AB， Y=BC， Z=CA， X+Z=300， X+Y=400， Y+Z=500， X+Y+Z=600， X=100， Y=300，Z=200，XYZ=6*10^6=（ABC）^2，ABC=-+1000Sqrt（6）

[@AnilSonkar-ly5ex]

Vimp