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y^(2)+25=x

10 y=x
If 
(x,y) is a solution to the system of equations shown, which of the following is an 
x-coordinate of the solution?
Choose 1 answer:
(A) 50
(B) -50
(C) -5
(D) 5

y2+25=x y^{2}+25=x \newline10y=x 10 y=x \newlineIf (x,y) (x, y) is a solution to the system of equations shown, which of the following is an x x -coordinate of the solution?\newlineChoose 11 answer:\newline(A) 5050\newline(B) 50-50\newline(C) 5-5\newline(D) 55

Full solution

Q. y2+25=x y^{2}+25=x \newline10y=x 10 y=x \newlineIf (x,y) (x, y) is a solution to the system of equations shown, which of the following is an x x -coordinate of the solution?\newlineChoose 11 answer:\newline(A) 5050\newline(B) 50-50\newline(C) 5-5\newline(D) 55
  1. Substitute xx for 10y10y: Let's substitute 10y10y for xx in the first equation, since 10y=x10y = x from the second equation.\newliney2+25=10yy^2 + 25 = 10y
  2. Rearrange equation and solve for y: Now, let's rearrange the equation to set it to zero and solve for y. \newliney210y+25=0y^2 - 10y + 25 = 0
  3. Factor the equation: We notice that the equation y210y+25y^2 - 10y + 25 is a perfect square trinomial, which factors into (y5)2(y - 5)^2.\newline(y5)2=0(y - 5)^2 = 0
  4. Take the square root to find yy: To find the value of yy, we take the square root of both sides.\newliney5=0y - 5 = 0\newliney=5y = 5
  5. Substitute yy back into the equation to find xx: Now that we have the value of yy, we can substitute it back into either of the original equations to find xx. Let's use the second equation 10y=x10y = x.
    10(5)=x10(5) = x
    x=50x = 50

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