How do you solve using elimination of $5 m + 3 n = 1.5$ $5m + 3n = 1.5$ and $- 8 m - 2 n = 20$ $-8m - 2n = 20$ ?

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Answer 1 · 2015-10-31T17:03:49

First, scale the equations such that one of the variables for both equations will have the same coefficient (the sign need not be the same).

$[1] 5 m + 3 n = 1.5$ $[1] 5m + 3n = 1.5$
$[2] - 8 m - 2 n = 20$ $[2] -8m - 2n = 20$

For the above equations, let us multiply $[1]$ $[1]$ by $2$ $2$ and $[2]$ $[2]$ by $3$ $3$

$[1] \Rightarrow 2 (5 m + 3 n = 1.5)$ $[1] => 2(5m + 3n = 1.5)$
$[1] \Rightarrow 10 m + 6 n = 3$ $[1] => 10m + 6n = 3$

$[2] \Rightarrow 3 (- 8 m - 2 n = 20)$ $[2] => 3(-8m -2n = 20)$
$[2] \Rightarrow - 24 m - 6 n = 60$ $[2] => -24m -6n = 60$

Now, let's add equations $[1]$ $[1]$ and $[2]$ $[2]$

$[1] \Rightarrow 10 m + 6 n = 3$ $[1] => 10m + 6n = 3$
$[2] \Rightarrow - 24 m - 6 n = 60$ $[2] => -24m - 6n = 60$

$[1] + [2] \Rightarrow - 14 m = 60$ $[1] + [2] => -14m = 60$
$\Rightarrow m = - \frac{60}{14}$ $=> m = -60/14$

$\Rightarrow m = - \frac{30}{7}$ $=> m = -30/7$

To get $n$ $n$ , simply substitute the value of $m$ $m$ to either $[1]$ $[1]$ or $[2]$ $[2]$ and solve for n

$[1] \Rightarrow 5 m + 3 n = 1.5$ $[1] => 5m + 3n = 1.5$

$[1] \Rightarrow 10 (- \frac{30}{7}) + 6 n = 3$ $[1] => 10(-30/7) + 6n = 3$

$[1] \Rightarrow - \frac{300}{7} + 6 n = 3$ $[1] => -300/7 + 6n = 3$

$[1] \Rightarrow 6 n = 3 + \frac{300}{7}$ $[1] => 6n = 3 + 300/7$

$[1] \Rightarrow 6 n = \frac{21 + 300}{7}$ $[1] => 6n = (21 + 300) /7$

$[1] \Rightarrow 6 n = \frac{321}{7}$ $[1] => 6n = 321 / 7$

$[1] \Rightarrow n = \frac{321}{7 \cdot 6}$ $[1] => n = 321 /(7 * 6)$

$[1] \Rightarrow n = \frac{107 \cdot 3}{7 \cdot 2 \cdot 3}$ $[1] => n = (107 * 3) / (7 * 2 * 3)$

$[1] \Rightarrow n = \frac{107}{14}$ $[1] => n = 107 / 14$

How do you solve using elimination of 5m + 3n = 1.55m+3n=1.55m + 3n = 1.5 and -8m - 2n = 20−8m−2n=20-8m - 2n = 20?