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3x+4y=23 and 5x+3y=31. Solve by elimination.

Sagot :

edmc96
let [tex]3x+4y=23 [/tex] be equation 1
let [tex]5x+3y=31[/tex]  be equation 2

do [tex](equ1)*3[/tex] and [tex](equ2)*4[/tex]

now [tex]9x+12y=69 [/tex]
and [tex]20x+12y=124[/tex]
now eliminate the [tex]y[/tex] by doing [tex](equ2)-(equ1)[/tex]
[tex]11x=55[/tex]
[tex]x=5[/tex]

now substitute [tex]x=5[/tex] back into equation 1 or 2
[tex]3*(5)+4y=23[/tex]
[tex]4y=23-15[/tex]
[tex]y=2[/tex]
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