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If 
(x,y) is a solution to the system of equations shown, what is the product of the 
y-coordinates of the solutions?

x^(2)+y^(2)=9

x+y=3

If (x,y) (x, y) is a solution to the system of equations shown, what is the product of the y y -coordinates of the solutions?\newlinex2+y2=9 x^{2}+y^{2}=9 \newlinex+y=3 x+y=3

Full solution

Q. If (x,y) (x, y) is a solution to the system of equations shown, what is the product of the y y -coordinates of the solutions?\newlinex2+y2=9 x^{2}+y^{2}=9 \newlinex+y=3 x+y=3
  1. Given Equations: We are given the system of equations:\newline11. x2+y2=9 x^2 + y^2 = 9 \newline22. x+y=3 x + y = 3 \newlineWe need to find the y-coordinates of the solutions to these equations and then calculate their product. Let's start by expressing x in terms of y using the second equation.\newlinex=3y x = 3 - y
  2. Expressing x in terms of y: Now, we substitute x=3y x = 3 - y into the first equation to find y.\newline(3y)2+y2=9 (3 - y)^2 + y^2 = 9 \newlineExpanding the squared term, we get:\newline96y+y2+y2=9 9 - 6y + y^2 + y^2 = 9 \newlineSimplifying, we combine like terms:\newline2y26y=0 2y^2 - 6y = 0
  3. Substituting x into the first equation: We can factor out a y from the equation:\newliney(2y6)=0 y(2y - 6) = 0 \newlineThis gives us two possible solutions for y:\newline11. y=0 y = 0 \newline22. 2y6=0 2y - 6 = 0 which simplifies to y=3 y = 3
  4. Factoring out y: We have two y-coordinates for the solutions: y=0 y = 0 and y=3 y = 3 . To find the product of the y-coordinates, we multiply these two values together:\newlineProduct of y-coordinates = y1×y2=0×3=0 y_1 \times y_2 = 0 \times 3 = 0
  5. Finding the y-coordinates: The product of the y-coordinates of the solutions to the system of equations is 00.

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