Question

solve 4m - 2n + 4 = 0 ; 3n + 4m - 16 = 0 by Cramer's method.​

Answer 1

Answer:

Solution:-

given by:- 

\begin{lgathered}4m - 2n = - 4...........(1) \\ 4m + 3n = 16............(2)\end{lgathered}4m−2n=−4...........(1)4m+3n=16............(2) 

Solving eq(1) and eq(2) by cramer's method 

\begin{lgathered}Δ = \binom{4 \: \: \: \: - 2}{4 \: \: \: \: \: \: 3} = 12 + 8 = 20 \\ \\ Δx = \binom{ - 4 \: \: \: \: - 2 }{16 \: \: \: \: \: \: \: 3} = - 32 + 12 = - 20 \\ Δy = \binom{4 \: \: \: - 4}{4 \: \: \: \: \: \: 16} = 64 + 16 = 80 \\ x = \frac{Δx}{Δ} = \frac{ - 20}{20} = - 1 \\ y = \frac{Δy}{Δ} = \frac{80}{20} = 4\end{lgathered}Δ=(434−2)=12+8=20Δx=(163−4−2)=−32+12=−20Δy=(4164−4)=64+16=80x=ΔΔx=20−20=−1y=ΔΔy=2080=4 

here ( x , y ) = (-1 ,4) 

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