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6y + 12 = 3y - 3
Subtract 3y on both sides
3y + 12 = -3
Subtract 12 on both sides
3y = -15
Divide by 3 on both sides
y = -5
Answer:f
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Step-by-step explanation:
so we have the points of (0,-7),(7,-14),(-3,-19), let's plug those in the y = ax² + bx + c form, since we have three points, we'll plug each one once, thus a system of three variables, and then we'll solve it by substitution.

well, from the 1st equation, we know what "c" is already, so let's just plug that in the 2nd equation and solve for "b".

well, now let's plug that "b" into our 3rd equation and solve for "a".
![\bf -19=9a-3b-7\implies -12=9a-3b\implies -12=9a-3(-1-7a) \\\\\\ -12=9a+3+21a\implies -15=9a+21a\implies -15=30a \\\\\\ -\cfrac{15}{30}=a\implies \blacktriangleright -\cfrac{1}{2}=a \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{and since we know that}}{-1-7a=b}\implies -1-7\left( -\cfrac{1}{2} \right)=b\implies -1+\cfrac{7}{2}=b\implies \blacktriangleright \cfrac{5}{2}=b \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill y=-\cfrac{1}{2}x^2+\cfrac{5}{2}x-7~\hfill](https://tex.z-dn.net/?f=%5Cbf%20-19%3D9a-3b-7%5Cimplies%20-12%3D9a-3b%5Cimplies%20-12%3D9a-3%28-1-7a%29%20%5C%5C%5C%5C%5C%5C%20-12%3D9a%2B3%2B21a%5Cimplies%20-15%3D9a%2B21a%5Cimplies%20-15%3D30a%20%5C%5C%5C%5C%5C%5C%20-%5Ccfrac%7B15%7D%7B30%7D%3Da%5Cimplies%20%5Cblacktriangleright%20-%5Ccfrac%7B1%7D%7B2%7D%3Da%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Band%20since%20we%20know%20that%7D%7D%7B-1-7a%3Db%7D%5Cimplies%20-1-7%5Cleft%28%20-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%3Db%5Cimplies%20-1%2B%5Ccfrac%7B7%7D%7B2%7D%3Db%5Cimplies%20%5Cblacktriangleright%20%5Ccfrac%7B5%7D%7B2%7D%3Db%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20y%3D-%5Ccfrac%7B1%7D%7B2%7Dx%5E2%2B%5Ccfrac%7B5%7D%7B2%7Dx-7~%5Chfill)