How do you solve the system #2x - 3y = -11# and #3x + 2y = 29#?

1 Answer
Jun 12, 2015

Answer:

Let #M = ((2,3),(-3,2))#. Then #det(M) = 2^2+3^2=13#

#1/13((2,3),(-3,2))((-11),(29)) = 1/13((65),(91)) = ((5),(7))#

#x=5# and #y=7#

Explanation:

Since the second equation swaps the coefficients of #x# and #y# with exactly one sign change, we can tell that the second equation is of a line perpendicular to that described by the first.

Hence all we need to do to get #x# and #y# is to multiply #((-11),(29))# by a rotation matrix constructed from the coefficients of #x# and #y#.

graph{(2x-3y+11)(3x+2y-29) = 0 [-16.18, 23.82, -4.16, 15.84]}