Please help. 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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Please help. 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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be unique y in second equation, then plug in to first equation... in seceon 8x - y = 10 -> y = 8x - 10 now plug in it to first equation and find x... y = x2 + 12x + 30 -> 8x - 10 = x2 +12x + 30 got it lady? ;)
hey @raebaby420 , sorry bout the "issues" last time anyway, as @mhmdrz91 said use the second equation to find what y equals (in terms of x) then sub in that result into the first equation (as per post above, cheers mhmd)

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so you'll get: 8x - 10 = x^2 +12x + 30 now combine "like terms" so 8x = x^2 + 12x + 30 +10 8x = x^2 + 12x + 40 continuing: 0 = x^2 + 12x -8x + 40 -->0 = x^2 + 4x + 40
0 = x^2 + 4x + 40 from here you've got two options: 1. the quadratic formula (google it, whay too long to type) or 2. factor above equation so: 0 = x^2 + 4x + 40 = (x + a ) ( x + b ) now solve for a and b
a * b = 40 ax + bx = 4x, so a + b = 4 now those 2 equations dont make any sense (as no 2 rational numbers both add to equal 4 and multiply to equal 40) so answer is: no real solution (d)
imagine for a second that the first equation was: y = x^2 + 18x + 30 and the 2nd equation was: y - 4x = -10 so rearrange eqn 2: y = 4x - 10 sub that into eqn 1 then you would get: y = x^2 + 18x + 30 4x - 10 = x^2 + 18x + 30 4x = x^2 + 18x + 30 +10 4x = x^2 + 18x + 40 0 = x^2 + 14x + 40 --> 0 = x^2 + 14x + 40 IS an equation that is solvable via factoring proof: 0 = x^2 + 14x + 40 0 = (x + a) ( x + b) now (from FOIL) a * b must equal 40 and a + b must equal 14 so the numbers that match that are 4 and 10 a * b = 40 = 4 *10 a + b = 14 = 4 + 10
therefore: 0 = x^2 + 14x + 40 0 = (x + a) ( x + b) 0 = (x + 4) (x + 10) so (x + 4) times (x + 10) = zero now the only thing that multiplies to equal zero is...ZERO so either (x + 4) = 0 [ as 0 times (x +10) = 0 ] which would mean that x = -4 (x + 4) = 0 (-4 + 4) = 0 0 = 0 proof OR (x + 10) = 0 [ as 0 times (x + 4) = 0 ] which would mean that x = -10 (x + 10) = 0 (-10 + 10) = 0 0 = 0 proof so there are really 2 answers to "what is x?" 1. if x = -4, (using our new 2nd equation) y - 4(x) = -10 y - 4 ( -4) = -10 y + (16) = -10 y = -10 -16 y = -26 giving coordinates of (-4, -26) [ as (x, y) ] or if x = -10 (also using the 2nd equation) y - 4(x) = -10 y - 4 (-10) = -10 y + 40 = -10 y = -50 giving coordinates of (-10, -50) [ as (x, y) ] hope this helps?

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