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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[m+n]x+2[m+n]y=5m-1, either in [m+n]x or in 2[m+n]y, one value mush and should have m-n in place of m+n in order to solve... other wise the variables get cancel initially, n question will not get sloved

Answers

2016-04-30T23:23:52+05:30

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 

3 5 3
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