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Mathematics 21 Online
OpenStudy (hunter535):

What is the solution set for this linear-quadratic system of equations? y = x2 − x − 12 y − x − 3 = 0 A:{(-3, 0), (0, 3)} B:{(-3, 0), (4, 0)} C:{(-3, 0), (5, 8)} D:{(4, 0), (0, 3)}

OpenStudy (sshayer):

y=x+3 \[y=x^2-x-12\] \[x^2-x-12=x+3,x^2-2x-15=0\] solve this quadratic ,then find x then corresponding y

OpenStudy (triciaal):

one approach factor the first equation to get when y = 0 for the second equation use expression for y = 0 compare and note that x value

OpenStudy (triciaal):

|dw:1467866003568:dw|

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