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ax-y-z-1-x-ay-z-a-x-y-az-a-2-Find-value-of-x-y-z-in-a-

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Question Number 154849 by naka3546 last updated on 22/Sep/21
ax + y + z = 1 x + ay + z = a x + y + az = a^2 Find value of x, y, z in a .
ax+y+z=1x+ay+z=ax+y+az=a2Findvalueofx,y,zina.
Answered by Mr.D.N. last updated on 24/Sep/21
ax+y+z=1.....(i) x+ay+z=a.....(ii) x+y+az=a^2 ......(iii) (i)−(ii) ax+y+z = 1 ((−x−ay−z= −a)/(ax+y−x−ay=1−a)) or x(a−1)−y(a−1)=1−a or (a−1)(x−y)=1−a or x−y= ((1−a)/(−(1−a))) x−y= −1......(iv) Again, (ii)×a−(iii) ax+a^2 y+az= a^2 ((−x−y−az =−a^2 )/(ax+a^2 y−x−y= 0)) or x(a−1)+y(a^2 −1)=0 x=− ((y(a^2 −1))/((a−1))) x=−y(a+1).......(v) Substitute the eq^n (v) in eq^n (iv) x−y = −1 − y(a+1)−y= −1 −ya−y−y=−1 − ya−2y = −1 − y(a+2)= −1 y= (1/(a+2)) Put the value of y in (v) x=− y(a+1) x= −(1/(a+2))(a+1) x= −((a+1)/(a+2))//. Now, Put the value of x & y in eq^n (i) for value z. ax+y+z= 1 −a((a+1)/(a+2))+(1/(a+2)) +z=1 ((−(a^2 +a))/(a+2))+(1/(a+2))+z=1 z= 1+((a^2 +a)/(a+2)) −(1/(a+2)) z = 1+((a^2 +a−1)/(a+2)) z= ((a+2+a^2 +a−1)/(a+2))= ((a^2 +2a+1)/(a+2)) z = (((a+1)^2 )/(a+2)) ∴ [x = −((a+1)/(a+2)) , y=(1/(a+2)) & z= (((a+1)^2 )/(a+2)) ]//.
ax+y+z=1…..(i)x+ay+z=a…..(ii)x+y+az=a2……(iii)(i)−(ii)ax+y+z=1−x−ay−z=−aax+y−x−ay=1−aorx(a−1)−y(a−1)=1−aor(a−1)(x−y)=1−aorx−y=1−a−(1−a)x−y=−1……(iv)Again,(ii)×a−(iii)ax+a2y+az=a2−x−y−az=−a2ax+a2y−x−y=0orx(a−1)+y(a2−1)=0x=−y(a2−1)(a−1)x=−y(a+1)…….(v)Substitutetheeqn(v)ineqn(iv)x−y=−1−y(a+1)−y=−1−ya−y−y=−1−ya−2y=−1−y(a+2)=−1y=1a+2Putthevalueofyin(v)x=−y(a+1)x=−1a+2(a+1)x=−a+1a+2//.Now,Putthevalueofx&yineqn(i)forvaluez.ax+y+z=1−aa+1a+2+1a+2+z=1−(a2+a)a+2+1a+2+z=1z=1+a2+aa+2−1a+2z=1+a2+a−1a+2z=a+2+a2+a−1a+2=a2+2a+1a+2z=(a+1)2a+2∴[x=−a+1a+2,y=1a+2&z=(a+1)2a+2]//.
Commented by Rasheed.Sindhi last updated on 23/Sep/21
GO^(⌢) O^(⌢) D & Explained Answer!
GO⌢OD⌢&ExplainedAnswer!
Commented by Mr.D.N. last updated on 24/Sep/21
No you are right ✓ before in my explaination has been going some error now its completly done solution sir.
Noyouareright✓beforeinmyexplainationhasbeengoingsomeerrornowitscompletlydonesolutionsir.
Commented by naka3546 last updated on 22/Sep/21
thank you , sir .
thankyou,sir.
Commented by naka3546 last updated on 23/Sep/21
I got that y = (1/(a+2)) ; x = −((a+1)/(a+2)) ; z = (((a+1)^2 )/(a+2)) Am I wrong, sir ?
Igotthaty=1a+2;x=−a+1a+2;z=(a+1)2a+2AmIwrong,sir?

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