Question

4m + 6n = 54 3m + 2n = 28 D = 4 6 3 2 = (4 × 2) - (6 × 3) = 8 - 18 ∴ D =_____ Dm = 54 6 28 2 = (54 × 2) - (6 × 28) = 108 -168 ∴ Dm = _____ Dn = 4 54 3 28 = (4 ×28) - (54 × 3) = 112 - 162 ∴ Dn = _____ By Cramer's rule m = Dm D = _____ = 6 and 12 n 12 2 | | | | | | n = _____ = −50 −10 = 5 ∴ m = _____, n = _____is the solution of given simultaneous equations

Answer 1

SOLUTION

GIVEN

4m + 6n = 54

3m + 2n = 28

TO DETERMINE

To fill in the blanks using Cramer's rule

EVALUATION

Here the given system of equations are

4m + 6n = 54

3m + 2n = 28

Thus we get

$\displaystyle \sf{D= \begin{vmatrix} 4& 6 \\ 3 & 2 \end{vmatrix} }$

$\displaystyle \sf{ \implies \: D = (4 \times 2) - (3 \times 6) }$

$\displaystyle \sf{ \implies \: D = 8 - 18 }$

$\displaystyle \sf{ \implies \: D = - 10}$

∴ D = - 10

$\displaystyle \sf{D_ m= \begin{vmatrix} 54& 6 \\ 28 & 2 \end{vmatrix} }$

$\displaystyle \sf{ \implies \: D_m = (54 \times 2) - (28 \times 6) }$

$\displaystyle \sf{ \implies \: D_m = 108 - 168 }$

$\displaystyle \sf{ \implies \: D_m = - 60}$

$\displaystyle \sf{ \therefore \: \: \: D_m = - 60}$

$\displaystyle \sf{D_ n= \begin{vmatrix} 4& 54 \\ 3 & 28 \end{vmatrix} }$

$\displaystyle \sf{ \implies \: D_n = (4 \times 28) - (54 \times 3) }$

$\displaystyle \sf{ \implies \: D_n = 112 - 162}$

$\displaystyle \sf{ \implies \: D_n = - 50}$

$\displaystyle \sf{ \therefore \: \: D_n = - 50}$

By Cramer's rule we get

$\displaystyle \sf{m = \frac{D_ m}{D} = \frac{ - 60}{ - 10} = 6 }$

$\displaystyle \sf{n = \frac{D_ n}{D} = \frac{ - 50}{ - 10} = 5 }$

∴ m = 6 , n = 5 is the solution of given simultaneous equations

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