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Question Number 50328 by peter frank last updated on 15/Dec/18
Answered by peter frank last updated on 16/Dec/18
1)x+2y−3z=a...(i)2x+6y−11z=b...(ii)x−2y+7z=c....(iii)eliminatexx+2y−3z=a...(i)2x+6y−11z=b...(ii)−2y+5z=2a−b....(iv)eliminatexx+2y−3z=a...(i)x−2y+7z=c....(iii)4y−10z=a−c....(v)eliminateyandz−2y+5z=2a−b....(iv)4y−10z=a−c....(v)5a−2b−c=0
2)x+y−z=1....(i)2x+3y−ay=3....(ii)x+ay+3z=2....(iii)eliminatexx+y−z=1....(i)2x+3y−ay=3....(ii)y+(a+2)z=1....(iv)eliminatexx+y−z=1....(i)x+ay+3z=2....(iii)y(a−1)+4z=1....(v)eliminateyy+(a+2)z=1....(iv)y(a−1)+4z=1....(v)(a+2)(a−1)−4]z=a−2(a2+a−6)z=a−2......
Commented by peter frank last updated on 16/Dec/18
howdoigetvalueofa?pleasehelp
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