Question

find m and n for which the equation 3x-(3m+4)y=8 and (2n+1)x-56=16 has infinitely many solutions​

Answer 1

Answer:

3x-(3m+4)y=8.

Step-by-step explanation:

-(3m+4)y=8-3x

-(7m)y=5x

y=5x+7m

= 12mx.

2) (2n+1)x-56=16.

( 2n+1)x=16+56.

=73.

x=73-2n+1.

n=2.

= 73-5

x=68

Answer 2

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Find m and n for which the equation

3 x - (3m+4) y = 8 and (2n+1) x - 56 y = 16 has infinitely many solutions.

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For two linear equations in two variable having infinite solutions

$\tt \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

so,,

$\tt \frac{3}{(2n + 1)} = \frac{ - (3m + 4)}{ - 56} = \frac{ 8}{16}$

Firstly taking

$\tt \frac{3}{2n + 1} = \frac{8}{16} \\ \\ \tt \frac{3}{2n + 1} = \frac{1}{2} \\ \\ \bf \:cross \: multiplying \\ \\ \tt 6 = 2n + 1 \\ \\ \tt 2n = 5 \\ \\ \boxed{ \tt \: n = \frac{5}{2} }$

Now taking

$\tt \frac{ - (3m + 4)}{ - 56} = \frac{8}{16} \\ \\ \tt \frac{ 3m + 4}{56 } = \frac{1}{2} \\ \\ \bf \: cross \: multiplying \\ \\ \tt 6m + 8 = 56 \\ \\ \tt \: 6m = 56 - 8 \\ \\ \tt \: 6m = 8 \\ \\ \boxed{ \tt \:m = \frac{8}{6} = \frac{4}{3} }$