How do you solve the following system?: $x + 3 - 2 y = 7, y = 2 x + 9$ $x + 3-2y=7 , y=2x+9$

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Answer 1 · 2016-03-09T13:53:03

$(x, y) = (- \frac{22}{3}, - \frac{17}{3})$ $(x,y)=(-22/3,-17/3)$

$1) x + 3 - 2 y = 7$ $1)x+3-2y=7$

$2) y = 2 x + 9$ $2)y=2x+9$

Substitute the value of $y$ $y$ to the first equation

$\to x + 3 - 2 (2 x + 9) = 7$ $rarrx+3-2(2x+9)=7$

$\to x + 3 - 4 x - 8 = 7$ $rarrx+3-4x-8=7$

$\to x + 3 - 4 x = 7 + 18$ $rarrx+3-4x=7+18$

$\to x + 3 - 4 x = 25$ $rarrx+3-4x=25$

$\to x - 4 x = 25 - 3$ $rarrx-4x=25-3$

$\to x - 4 x = 22$ $rarrx-4x=22$

$\to - 3 x = 22$ $rarr-3x=22$

$\Rightarrow x = - \frac{22}{3}$ $rArrx=-22/3$

Now,Substitute the value of $x$ $x$ to the second equation

$\to y = 2 (- \frac{22}{3}) + 9$ $rarry=2(-22/3)+9$

$\to y = - \frac{44}{3} + 9$ $rarry=-44/3+9$

$\Rightarrow y = - \frac{17}{3}$ $rArry=-17/3$

How do you solve the following system?: x+3−2y=7,y=2x+9 x + 3-2y=7 , y=2x+9