for what value of m the system of equations is 3x+4y=12 and [m+n]x+2[m+n]y=5m-1
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Here a1=3 ,b1=4, c1=12
a2=(m+n), b2=2(m-n), c2=5m -1
Using a1/a2 = b1/b2 = c1/c2
3/(m+n) ---- 1 = 4/2(m-n) ---- 2 = 12/(5m-1) ------3
From 1&2
3/(m+n) = 4/2(m-n)
Cross multiplying
3(m-n) = 2(m+n)
3m - 3n = 2m + 2n
3m - 2m = 3n + 2n
m = 5n -------- (i)
similarly,equating 1&3
3/(m+n) = 12/(5m-1)
1/(m+n) = 4/(5m-1)
by Cross multiplying
5m – 1 = 4(m+n)
5m – 1 = 4m + 4n
5m – 4m -4n = 1
m – 4n = 1 ----------- (ii)
Substituting (i) in (ii)
5n -4n = 1
n = 1
m = 5n
m = 5(1)
m = 5
therefore, m = 5 and n = 1
himlu:
thankyou very much
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