To solve simultaneous equations where X + Y = 40 and X * Y = 80, express Y in terms of X as Y = 40 - X, substitute into the second equation to get X(40 - X) = 80, rearrange to form the quadratic equation X² - 40X + 80 = 0, then apply the quadratic formula x = [-b ± √(b² - 4ac)]/(2a) with a=1, b=-40, c=80 to find x = 20 ± 8√5, and finally solve for y using y = 40 - x to get the corresponding values y = 20 ∓ 8√5.
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Solve for x and y in this nice simultaneous equation | Math Olympiad Mathematics
Added:In this video, we want to solve for X and Y.
Given X + Y is 40 and X * Y is equal to 80.
Let us call this equation one.
Then this will be equation two.
Now, from equation one, I'm going to make Y the subject of the formula so that Y will now be equal to 40 - X.
Now, looking into equation two, in place of Y here, I'm going to put 40 - X.
Then I have X into 40 - X is equal to 80.
We need to carefully open up this bracket.
X * 40 will give us 40 X.
Then X * X here will give us X squared.
Then equal to 80.
We can rearrange this to give us negative X squared + 40 X.
Then bring this 80 to the left as negative 80.
Then equal to zero.
From here, I'm going to divide through by negative one on both sides.
So that this becomes positive X squared - 40 X + 80 is equal to zero.
This brings us to a quadratic equation, and to solve this, we first need to compare with the standard quadratic equation ax squared + bx + c = 0.
This is so that we can get value for a, b, and c.
a b being one which is the coefficient of x squared here.
b being -40 which is this, and then c being 80.
We're now going to apply the quadratic formula to solve for x, which will be -b + or - square root of b squared -4 * a * c then divided by 2 * a.
And since we have these parameters we're just going to plug it into this formula to get x is equal to -b which is -40 then + or - square root of -40 squared -4 * a a is one, and c is 80.
Square root of that then divided by 2 * a a is one.
This will give us x is equal to this will be positive 40 + or - square of this will be 40 squared minus 4 * 1 is 4. 4 * 80 here We can still leave this as 4 * 80.
Then divided by 2 * 1 is 2.
Let us do this as x is equal to 40 plus or minus square root of 40 squared is same thing as saying 40 * 40 then minus 4 * 80 here is 2 * 40.
Then divided by 2.
This will give us x is equal to 40 plus or minus square root of So, 40 is common here.
We can do 40 into What I'm left with 40 minus 4 * 2 is 8.
Then square root of that divided by 2.
Giving us x is equal to 40 plus or minus square root of This will be 40 * 32.
Then divided by 2.
This will give us x is equal to 40 plus or minus I'm going to write this as 40 here is 4 * 10, but I'll write it as 4 * 2 * 5 * 32 then / 2 Now X is going to be 40 + square here is 64.
So together with this we have 4 * 64 then * square root of 5 then divided by 2 This will give us X is equal to 40 + So square root of 4 is 2.
Square root of 64 is 8.
Then root 5 / 2 which would then give us X is equal to 40 + - 16 root 5 / 2 Therefore X is now equal to 40 / 2 + - 16 root 5 / 2 2 here is 1. 2 and 16 is 8.
2 here is 1. 2 and 20 and 40 is 20.
So X is equal to 20 + 8 root 5 And if we separate this we get X1 is equal to 20 plus 8 root 5.
And x2 equal to 20 minus 8 root 5.
So, this gives us two possible values for x and now we need to get two corresponding values for y.
Earlier in the video, we got y is equal to 40 minus x.
This would then imply that to get y1, we're going to do 40 minus x1. And similarly for y2, we're going to do 40 minus x2.
So, for x for y1, we're going to have y1 is equal to 40 minus 20 plus 8 root 5.
This will give us y1 is equal to 40 minus 20 is 20.
Minus times plus is minus.
Then 8 root 5.
So, y2 is going to be 20 plus 8 root 5.
Since the solutions appear to be symmetrical.
Given us all four possible solutions once to this problem.
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Bye.
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