How do you solve #3a - b = 11# and #2a + 3b = 0# using substitution?

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Answer 1 · 2018-03-06T14:14:08

#color(green)(a = 3, b = -2#

#3a - b = 11# Eqn (1)

#b = 3a - 11#

#2a + 3b = 0# Eqn (2)

Substituting value of b in terms of a in Eqn (2),

#2a + 9a - 33 = 0#

#11a = 33# or #color(green)(a = 3#

Substituting value of a in Eqn (1),

#(3 * 3) - b = 11#

#b = 9 - 11 #, #color(green)( b = - 2#