SOLUTION: 3x2+4y2=19 , x+2=1
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Question 159219
:
3x2+4y2=19 , x+2=1
Answer by
KnightOwlTutor(293)
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Solve for x x+2=1
subtract 2 from both sides
x=-1
Replace all x values with -1
3(-1)^2+4y^2=19
remember (-)(-)=(+)
3+4y^2=19
Subtract 3 form both sides
4y^2=16
Divide both sides by 4
y^2=4
y=2,-2