In Exercises 29–42, solve each system by the method of your choic...

Question

In Exercises 29–42, solve each system by the method of your choice. x^2+(y−2)^2=4, x^2−2y=0

Accepted Answer

Welcome back. I am so glad you're here. We are given a system of equations. The first equation is x minus squared plus Y squared equals 256. The second equation is Y squared minus X equals zero. And we are asked to solve the system of equations. Well, I am going to solve it using substitution. I am going to solve equation two for y squared. So I'm going to add 16 X to both sides. So now equation two, redes Y squared equals 16 X. And in equation one where I have a Y squared, I can substitute this 16 X. So now equation one reads parentheses X minus squared plus instead of y squared. Now, 16 X equals 256. And now I can solve for X. First I am going to foil out this x minus times x minus 16 Plus the x equals 256. So first X times X is X squared, Outsides X times negative 16 is minus 16 X. Insides likewise minus 16 X and lasts negative 16 times negative 16 is plus 256. Bring down the plus 16 X equals 256. Now, I'm going to do two things. I'm going to bring this 256 from the right hand side and subtract it from both sides. Bring it to the left so that the right hand side will be equal to zero. So that hopefully we can factor this quadratic and solve for X. And then I'm going to combine like terms with these exes so we still have X squared and then negative 16, X minus 16 X plus 16 X leaves us with just negative 16 X 256 minus 256 to 0. So X squared minus 16, X equals zero. And now I can factor an X out in front because both of those terms have an X. And that will leave us with x minus 16 in parentheses equal zero. I can set both of these factors equal to zero, X equals zero and X minus 16 equals zero. For the second one, I can add 16 to both sides. And then I get a solution of X equals 16. So I have two X values X equals zero and X equals 16. So I can set up a solution set and I have two X values. The first one is zero and the second one is 16. And now I can plug those back in to one of our equations for X and then solve for Y. I'm going to use equation to which will be y squared minus 16, X equal zero. And since I have two X values, I'm going to set up two equations and the first one plugging zero in for X. So instead of 16 times X it's 16 times zero. Well 16 times zero is zero. So that will just be Y squared equals zero to get that Y by itself take the square to both sides and we get y equals plus or minus zero, but negative zero is just zero. So it's Y equals zero. So for the first ordered pair, when X equals zero, Y equals zero. Now let's see what Y equals. When X equals 16. Plugging 16 in up here for X. So that'll be a Y squared minus 16 times 16 is so minus 256. We can add 256 to both sides of the equation and we get a y squared Equals 256 to get that Y by itself, take the square root of both sides and we get Y equals plus or minus 16 squared of 256 is 16. So when X equals 16, Y actually has two different values. So when X equals 16, 1 value for Y is a negative 16 and when X equals 16, another value for Y is a positive 16. So this is our solution set. We look at our answer choices and this matches with answer choice. A well done. We'll catch you on the next one

In Exercises 29–42, solve each system by the method of your choice. x^2+(y−2)^2=4, x^2−2y=0

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