How do you solve $3 m + n = 2$ $3m+n=2$ and $m - 2 n = 3$ $m-2n=3$ using substitution?

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Answer 1 · 2018-04-19T13:15:39

$m = 1, n = - 1$ $m = 1, n=-1$

1 ) $3 m + n = 2$ $3m + n = 2$
2 ) $m - 2 n = 3$ $m - 2n = 3$

1(x2) ) $6 m + 2 n = 4$ $6m + 2n = 4$

Add them together:

$7 m = 7$ $7m = 7$
$m = 1$ $m = 1$

Sub $m$ $m$ into an equation

$1 - 2 n = 3$ $1 - 2n = 3$
$- 2 n = 2$ $-2n = 2$
$n = - 1$ $n = -1$

How do you solve 3m+n=23m+n=2 and m−2n=3m-2n=3 using substitution?