How do you solve #a + b = 4# and #2a + 3b = 11# using substitution?

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Answer 1 · 2016-02-29T14:35:17

#a,b=1,3#

#color(blue)(a+b=4),color(red)(2a+3b=11)#

Solve for first equation

#rarra+b=4#

#rarrb=4-a#

Substitute the value of #b# to the second equation

#rarr2(4-a)+3b=11#

#rarr8-2b+3b=11#

#rarr8+b=11#

#rarrb=11-8#

#color(green)(rArrb=3#

So substitute the value to the first equation

#rarra+3=4#

#rarra=4-3#

#color(green)(rArra=1#