Solve the given system of three linear equati | vedclass

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(A) Given equations:$(i) 7x + 6y + 4z = 122$$(ii) 4x + 5y + 3z = 88$$(iii) 9x + 2y + z = 78$Step $1$: Eliminate $z$ using equations $(ii)$ and $(iii)$

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if $x < y = z$

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(A) Given equations:$(i) 7x + 6y + 4z = 122$$(ii) 4x + 5y + 3z = 88$$(iii) 9x + 2y + z = 78$Step $1$: Eliminate $z$ using equations $(ii)$ and $(iii)$.Multiply $(iii)$ by $3$: $27x + 6y + 3z = 234$Subtract $(ii)$ from this: $(27x + 6y + 3z) - (4x + 5y + 3z) = 234 - 88$$23x + y = 146 \dots (iv)$Step $2$: Eliminate $z$ using equations $(i)$ and $(iii)$.Multiply $(iii)$ by $4$: $36x + 8y + 4z = 312$Subtract $(i)$ from this: $(36x + 8y + 4z) - (7x + 6y + 4z) = 312 - 122$$29x + 2y = 190 \dots (v)$Step $3$: Solve for $x$ and $y$ using $(iv)$ and $(v)$.Multiply $(iv)$ by $2$: $46x + 2y = 292$Subtract $(v)$ from this: $(46x + 2y) - (29x + 2y) = 292 - 190$$17x = 102 \Rightarrow x = 6$Substitute $x = 6$ into $(iv)$: $23(6) + y = 146 \Rightarrow 138 + y = 146 \Rightarrow y = 8$Step $4$: Find $z$ using $(iii)$.$9(6) + 2(8) + z = 78 \Rightarrow 54 + 16 + z = 78 \Rightarrow 70 + z = 78 \Rightarrow z = 8$Thus,$x = 6, y = 8, z = 8$. Therefore,$x < y = z$.

Solve the given system of three linear equations and select the correct relationship between $x, y,$ and $z$ from the given options.
$I. 7x + 6y + 4z = 122$
$II. 4x + 5y + 3z = 88$
$III. 9x + 2y + z = 78$

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Solve the given system of three linear equations and select the correct relationship between $x, y,$ and $z$ from the given options.$I. 7x + 6y + 4z = 122$$II. 4x + 5y + 3z = 88$$III. 9x + 2y + z = 78$