How do you solve #4x - 3y = 17#, #5x + 4y = 60#?

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Answer 1 · 2015-11-13T18:33:29

#x=8#, #y=5#.

From the first equation, you know that

#x=(17+3y)/4#

Substitute this value in the second equation:

#5((17+3y)/4)+4y=60 -> (85+15y)/4+4y=60#

We can multiply everything by #4# to get rid of the denominator:

#85+15y+16y = 240#

Isolating the #y# terms:

#31y = 155 -> y=155/31 = 5#

Now that we knoe that #y=5#, we can obtain #x# by substitution:

#4x-3y=17 -> 4x-15=17 -> 4x=32#

and thus #x=8#