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Please Help c: What are the solutions to the following system of equations? y = x2 + 12x + 30 8x − y = 10 (−4, −2) and (2, 5) (−2, −4) and (2, 5) (−2, −4) and (5, 2) No Real Solutions

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What is the solution to the following system of equations? y = x2 + 10x + 11 y = x2 + x − 7 (−2, −5) (2, −5) (−2, 5)
rewrite the 2nd equation with y as the subject y = 8x - 10 now equate it to the 1st equation and you get \[8x - 10 = x^2 + 12x + 10\] now you have 1 equation in 1 unknown to solve
y = x2 + 12x + 30 8x − y = 10 express y from second equation: \(y=8x-10\) substitute in the 1º equation: \(8x-10=x^2 + 12x + 30\) rearange terms: \(x^2+4x+40=0\) solve this. You will get 2 solutions, which later substitute in the 2º equation to get corresponding y value

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Other answers:

oops should be ' \[8x - 10 = x^2 + 12x + 30\] just solve for x
Mm.. ill try. One sec, ill let you know what I get
:O maths
and can you check your equations as they may be a typo in them...
you can get the ans by by making the quadratic x
we will put all the x on one side
The what? LOL
because based on the information there are any solutions that are real numbers
oops... which is an option
Ok.. so its D.. cool
maths is confusing
But idk how to get D
*takes out the gun shoots maths*
well do you know about the discriminant...? thats how you check...
No.. but I got 8x-10=x2+12x+30 so far..
What do I do from there?
subtract 8x from both sides... then add 10 to both sides of the equation...
Oh.. I knew that. For some reason I thought I needed to bring my X from the right to the left.
Ok.. so I got x=x2+4x+40 Is that right?
ya you got it right
and the discriminant is for a quadratic \[ax^2 + bx + c = 0\] thew discriminant is \[b^2 - 4ac\] you have a = 1, b = 4 and c = 40 it will be less than zero... meaning complex roots
Sooo its D.
Thanks everyone! c:

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