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To solve the linear-quadratic system y=x^2+10 and y=-7x-2 using the elimination method, we can substitute the value of y from one equation into the other and solve for x.
We have the equation y=x^2+10.
Substituting y from the second equation into the first equation, we get:
-7x-2 = x^2 + 10
Rearranging this equation, we have:
x^2 + 7x + 12 = 0
Now we can factor this quadratic equation:
(x + 3)(x + 4) = 0
Setting each factor equal to zero, we have:
x + 3 = 0 or x + 4 = 0
Solving for x, we find:
x = -3 or x = -4
Now we substitute these values of x into one of the original equations to solve for y.
For x = -3:
y = (-3)^2 + 10
y = 9 + 10
y = 19
For x = -4:
y = (-4)^2 + 10
y = 16 + 10
y = 26
Therefore, the solutions to the linear-quadratic system are (-3, 19) and (-4, 26).