An interesting question!
Instead of solving each set of non-linear system of equations 6 times, I will try a simpler way, which is more adapted to multiple choice questions. However, please be warned that this method will not improve immensely math skills, but will help with reasoning, and possibly a broader understand in system of equations.
For simplicity, I will denote the sets of system of non-linear equations by S={A,B,C,D,E,F} from left to right.
1. given solutions {(-2,3),(7,-6)}
We first check each member of S, i.e. A,B,C,D,E,F for the linear conditions.
A: x+y=-2+3=1 ≠ 3 so no
B: x-y=-2-3=-5 ≠ 1 so no
C: 2x+y=-2(2)+3=-1 ≠ 1 so no
D: x+2y=-2+3(2)=4 ≠ 2 so no
E: -x+y=2+3=5 ≠ 1 so no
F: x+y=-2+3=1 ≠ 3 so YES now check the other solution 7-6=-1 OK
Now check the non-linear condition for F:
S1: (-2,3)
y-15=3-15=-12
-x^2+4x=-(-2)^2+4*(-2)=-12 good
S2:(7,-6)
y-15=-6-15=-21
-x^2+4x=-(7^2)+4(7)=-49+28=-21 also good,
So Solution set (1) matches tile F
(2) Given solutions {(-5,8),(3,0)}
We proceed in a similar way, to find that
-5+8=3 & 3+0=3 (matches A) and nothing else.
Check non-linear conditions (optionally, I transposed the terms to make comparison easier)
A. x^2+x-y=(-5)^2+(-5)-8=25-5-8=12 (looks good)
x^2+x-y=(3)^2+(3)-0=9+3=12 (looks even better)
So we determined that {(-5,8),(3,0)} is the solution for tile A.
(3) For solution set {(-2,5),(3,-5)}
Check linear conditions:
x+y=3 and -3, so does not satisfy A,& F.
x-y=-7 and 8, so does not satisfy B
2x+y=5 and 1 so satisfies C
x+2y=8 & -7 so does not satisfy D
-x+y=7 & -8 so does not satisfy E
Thus C is our only set. NOTE: if two distinct points satisfy one linear condition (say C), then both points cannot satisfy another non-equivalent linear condition, i.e. we didn't really have to check points D & E (which are not equivalent linear conditions as C) once we have found C.
Now check non-linear conditions:
x^2-3x-y=(-2)^2-3(-2)-5=5 ok
x^2-3x-y=(3)^2-3(3)-(-5)=5 ok
So solution set (3) matches tile C.
@officiallyqueenz, for your benefit, I will leave one for you as exercise. Please proceed to find the system for solution set (4). Post a your work as comments if you get stuck, or your answer for verification if you wish.
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Answer:
12. 7xy ∛(xy^2)
13. 3x^2|y| sqrt(7x)
14. 5 x^4 y ^4 y ∛(y^2)
Step-by-step explanation:
12. (343 x^4y^5) ^ 1/3
separate
343^1/3 x^4 ^ 1/3 y^5 ^ 1/3
using the power of power we can multiply the powers
343 ^ 1/3 x^ 4/3 y ^ 5/3
simplify
7 x^ 4/3 y ^ 5/3
if the power is greater than 1 we can split it
7 x x^ 1/3 y y ^ 2/3
7xy ( xy^2 ) ^ 1/3
7xy ∛(xy^2)
13. (189 x^5y^6)^ 1/2/(3y^4)^ 1/2
combine
( (189 x^5y^6)/(3y^4))^ 1/2
(189/3 * x^5 * y^6/y^4) ^ 1/2
when we divide exponents with the same base, we subtract the exponent
(63 * x^5 * y^2) ^ 1/2
split
63 ^ 1/2 x^ 5 ^ 1/2 y ^ 2 ^ 1/2
sqrt (63) * sqrt * (x^5) * sqrt(y^2)
sqrt(9*7) sqrt( x^4 *x) * sqrt(y^2)
sqrt(9) * sqrt(7) sqrt(x^4) sqrt(x) sqrt(y^2)
we need to make sure to take the positive value of sqrt(y^2)
3sqrt(7) x^2 sqrt(x) | y|
3x^2|y| sqrt(7x)
14. (625 x^17 y^16) ^1/3 / (5 x^5 y^2) ^ 1/3
combine
(625 x^17 y^16 / 5 x^5 y^2) ^ 1/3
(625/5 * x^17/x^5 * y^16/y^2) ^ 1/3
when we divide exponents with the same base, we subtract the exponent
(125 x^(17-5) y^(16-2)) ^ 1/3
(125 x^12 y ^ 14) ^ 1/3
split
125 ^ 1/3 x^ 12 ^ 1/3 y^ 14 ^ 1/3
power to the power ( multiply the power)
125 ^ 1/3 x^ 4 y ^ 14/3
5 x^4 y ^ 14/3
if the power is greater than 1 we can split it
5 x^4 y ^4 y ^2/3
5 x^4 y ^4 y ∛(y^2)
