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Question Number 90225 by jagoll last updated on 22/Apr/20

x+y+z = 4 z+t+x = −3 y+z+t = 4 t+x+y = 1 find the value of x^2 +y^2 +z^2 +t^2

x+y+z=4z+t+x=−3y+z+t=4t+x+y=1findthevalueofx2+y2+z2+t2

Commented by niroj last updated on 22/Apr/20

x+y+z= x+y=4−z.....(i) t+x+y=1...(iv) from (i)&(iv) t+4−z=1 t−z=−3....(v) (ii)+(iii) z+t+x=−3 ((y+z+t=4)/( 2t+2z+x+y=1)) 2t+2z+4−z=1 (∵x+y=4−z) 2t+z=−3...(vi) (v)+(vi) t−z=−3 ((2t+z=−3)/(3t=−6)) t=−2 t^2 =4 put the value of t in (v) t−z=−3 t+3=z z=3−2=1 z^2 =1 put the value of t,z in (ii) z+t+x=−3 1−2+x=−3 x=−3+1 x=−2 x^2 =4 put the value of x & z in(i) x+y+z=4 −2+y+1=4 y= 4+1=5 y^2 =25 ∴ x^2 +y^2 +z^2 +t^2 =4+25+1+4=34.

x+y+z=x+y=4−z.....(i)t+x+y=1...(iv)from(i)&(iv)t+4−z=1t−z=−3....(v)(ii)+(iii)z+t+x=−3y+z+t=42t+2z+x+y=12t+2z+4−z=1(∵x+y=4−z)2t+z=−3...(vi)(v)+(vi)t−z=−32t+z=−33t=−6t=−2t2=4putthevalueoftin(v)t−z=−3t+3=zz=3−2=1z2=1putthevalueoft,zin(ii)z+t+x=−31−2+x=−3x=−3+1x=−2x2=4putthevalueofx&zin(i)x+y+z=4−2+y+1=4y=4+1=5y2=25∴x2+y2+z2+t2=4+25+1+4=34.

Commented by Prithwish Sen 1 last updated on 22/Apr/20

adding all 4 equations x+y+z+t = 2......(v) now (v)−(i) t = 2 (v)−(ii) y=5 (v)−(iii) x= −2 (v) − (iv) z = 1 ∴ x^2 +y^2 +z^2 +t^2 = 4+25+1+4 = 34

addingall4equationsx+y+z+t=2......(v)now(v)−(i)t=2(v)−(ii)y=5(v)−(iii)x=−2(v)−(iv)z=1∴x2+y2+z2+t2=4+25+1+4=34

Commented by jagoll last updated on 22/Apr/20

thank you all

thankyouall

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