How do you solve #(x+2)/3+(x-3)/4=1#?

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Answer 1 · 2018-03-22T19:09:07

#x = 13/"7"#

#(x +2)/"3"# + #(x - 3)/"4"# = 1

#((x +2) (4) )/"3 (4)"# + #((x - 3) (3))/"4 (3)"# = 1/1

#((x + 2) 4 + (x - 3) 3)/"12"# = 1

#(4x + 8 +3x - 9)/"12"# = 1

#(7x - 1)/"12"# = 1 (multiply both sides with 12)

7x - 1 = 12

7x = 13

x = #13/"7"#