Mathematics
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OpenStudy (anonymous):
In the system shown below, what is the sum of the x-coordinates of all solutions? pic attached
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OpenStudy (anonymous):
OpenStudy (raden):
from the 1st equation,
3y^2 - 5x^2 = 7 ----> 3y^2 = 5x^2 + 7
now, subtitute it into the 2nd equation
OpenStudy (anonymous):
k
OpenStudy (raden):
what u get ?
OpenStudy (anonymous):
idk how to do it, ik how to substitute but not this one
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OpenStudy (anonymous):
am i solving the second equation u wrote... the one the arrow is pointing too?
OpenStudy (raden):
well, i we substitute the value of 3y^2 = 5x^2 + 7 to 3y^2-x^2=23, gives
5x^2 + 7- x^2=23, right ?
OpenStudy (raden):
*if
OpenStudy (anonymous):
right
OpenStudy (raden):
then, just simplify it, what u get ?
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OpenStudy (anonymous):
5x^2 - x^2=16
OpenStudy (anonymous):
or 4x^2=16
OpenStudy (raden):
or 4x^2 - 16 = 0, right ?
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
x=-2 and x=2??
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OpenStudy (raden):
ok now without to find the roots, we can determine the sum of roots of equation kuadratic :
ax^2-bx+c=0,
just use formula x1 + x2 = -b/a
OpenStudy (anonymous):
which u get what once there plugged in?
OpenStudy (anonymous):
x1+x2=o/4?
OpenStudy (raden):
ax^2+bx+c=0... (sorry, trouble my keyboard)
yes.. 0/4 = 0
OpenStudy (anonymous):
so the sum of all x coordinates is 0?
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OpenStudy (raden):
yeps...
another way, from ur solution : x1=-2 and x2=2
so, x1+x2 = -2 + 2 = 0
OpenStudy (anonymous):
thankyou
OpenStudy (raden):
welcome