For a unique value of 'μ' and 'λ ' the system of equations given $$byx+y+z=6x+2y+3z=142x+5y+$$λ$$z=$$μhas infinitely many solutions, then μ−λ$$4=_____$$

$$y+2z=8y+($$λ−$$6)z=$$μ−$$281=2$$λ−$$6=8$$μ−$$28$$λ$$=8$$;μ$$=36$$μ−λ$$4=284=7$$

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For a unique value of ' $μ$ ' and ' $λ$ ' the system of equations given by
$\begin{array}{l} x + y + z = 6 \\ x + 2 y + 3 z = 14 \\ 2 x + 5 y + λz = μ \end{array}$
has infinitely many solutions, then $\frac{μ - λ}{4} =_____$

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detailed solution

$\begin{array}{l} y + 2 z = 8 \\ y + (λ - 6) z = μ - 28 \\ 1 = \frac{2}{λ - 6} = \frac{8}{μ - 28} \\ λ = 8; μ = 36 \\ \frac{μ - λ}{4} = \frac{28}{4} = 7 \end{array}$

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For a unique value of 'μ' and 'λ ' the system of equations given byx+y+z=6x+2y+3z=142x+5y+λz=μhas infinitely many solutions, then μ−λ4=_____

Unlock the full solution & master the concept.

Ready to Test Your Skills?

free study material

For a unique value of ' $μ$ ' and ' $λ$ ' the system of equations given by
$\begin{array}{l} x + y + z = 6 \\ x + 2 y + 3 z = 14 \\ 2 x + 5 y + λz = μ \end{array}$
has infinitely many solutions, then $\frac{μ - λ}{4} =_____$