How do you solve $9 m + n = 20$ $9m+n=20$ and $m - 4 n = 31$ $m-4n=31$ using substitution?

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Answer 1 · 2018-05-09T04:59:35

$m = 15, n = - 4$ $color(blue)(m = 15, n = -4$

$9 m + n = 20, Eqn (1)$ $9m + n = 20, " Eqn (1)"$

$m - 4 n = 31, Eqn (2)$ $m - 4n = 31, " Eqn (2)"$

$m = 31 + 4 n$ $m = 31 + 4n$

Substituting value of m in terms of n in Eqn 91),

$9 \cdot (31 + 4 n) + n = 20$ $9 * (31 + 4n) + n = 20$

$279 + 36 n + n = 20$ $279 + 36n + n = 20$

$37 n = - 259$ $37n = -259$

$n = - \frac{259}{37} = - 4$ $color(blue)(n = -259/37 = -4$

Substituting value of n in Eqn (2),

$m - 4 \cdot - 4 = 31$ $m - 4 * -4 = 31$

$m = 31 - 16 = 15$ $color(blue)(m = 31 - 16 = 15$

How do you solve 9m+n=209m+n=209m+n=20 and m-4n=31m−4n=31m-4n=31 using substitution?