What is the intersection point of 2x+19y=3 and 1x-3=7y

[tex]2x+19y=3\\19y=-2x+3\\y=-\frac{2}{19}x+\frac{3}{19}\\\\x-3=7y\\y=\frac{1}{7}x-\frac{3}{7}\\\\-\frac{2}{19}x+\frac{3}{19}=\frac{1}{7}x-\frac{3}{7}\\-\frac{14}{133}x+\frac{21}{133}=\frac{19}{133}x-\frac{57}{133}\\\frac{78}{133}=\frac{33}{133}x\\\frac{78}{33}=x\\x=\frac{26}{11}\\\\(\frac{26}{11})-3=7y\\(\frac{26}{11})-\frac{33}{11}=7y\\\frac{-7}{11}=7y\\-11=y[/tex]

(26/11, -11)

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Panoyin Panoyin · Answer 2 · 2015-03-19T20:39:48-04:00

2x + 19y = 3 ⇒ 2x + 19y = 3 ⇒ 2x + 19y = 3
1x - 3 = 7y    ⇒ 1x -   7y = 3 ⇒ 2x + 14y = 6
      5y = -3
   5     5
   y = -3/5
      2x + 19(-3/5) = 3
   2x - 11 2/5 = 3
      +11 2/5 +11 2/5
   2x = 14 2/5
      2 2
       x = 7 1/5
   (x, y) = (7 1/5, -3/5)