How do you solve the system of equations $3m - 2n = - 3$ and $9m - 7n= - 12$ ?

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Answer 1 · 2016-11-22T21:07:39

$n = 3$ and $m = 1$

Step 1) Solve the first equation for $m$ :

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

$3m = 2n - 3$

$(3m)/3 = (2n - 3)/3$

$m = (2n)/3 - 3/3$

$m = (2n)/3 - 1$

Step 2) Substitute $(2n)/3 - 1$ for $m$ in the second equation and solve for $n$ :

$9((2n)/3 - 1) - 7n = -12$

$(18n/3) - 9 - 7n = -12$

$6n - 9 - 7n + 9 = -12 + 9$

$-1n = -3$

$(-1n)/-1 = (-3)/(-1)$

$n = 3$

Step 3) Substitute $3$ for $n$ in the solution for the first equation and solve for $m$ :

$m = (2 * 3)/3 - 1$

$m = 2 - 1$

$m = 1$

How do you solve the system of equations 3m - 2n = - 33m - 2n = - 3 and 9m - 7n= - 129m - 7n= - 12?