To solve a system of equations where x + y = 100 and x * y = 100, express y in terms of x from the first equation (y = 100 - x), substitute into the second equation to form a quadratic equation (x(100 - x) = 100), which simplifies to x² - 100x + 100 = 0. Apply the quadratic formula x = [-b ± √(b² - 4ac)] / (2a) with a = 1, b = -100, and c = 100 to find x = 50 ± 20√6, then solve for y using y = 100 - x to get y = 50 ∓ 20√6. The solution pairs are (50 + 20√6, 50 - 20√6) and (50 - 20√6, 50 + 20√6).
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Solve for x and y in this nice Algebra equation | Math Olympiad MathematicsAdded:
In this video, let us solve for x and y.
Given x + y is = 100 and x * y is also = 100.
Let us call this equation one and call this equation two.
Then from equation one if we solve for y we get 100 - x.
So we have y is = 100 - x. I'm going to call that equation three.
Now if we go back into equation two and make a substitution for y using 100 - x then we get x into 100 - x then equal to 100.
Next we open up this bracket. X * 100 will give us 100 X.
Then - X * X. This will give us X^2 is = 100.
Or we can rearrange this to give us -x^ 2 + 100x is equal to 100.
I'm going to move this to the left so that we have -x^2 + 100x - 100 is = 0.
Now because the coefficient of x squ here is negative I can multiply through by -1.
So negative * negative here will give give us positive. So we have x^2 this becomes negative 100x and this becomes positive 100 then equal to z.
And now we have a quadratic equation.
Let us solve for x.
To begin, I'll first compare this with a x^2 + b x + c is equal to zero. This will allow me to get the coefficient a and b and the constant stem c which will be 1 and 100 respectively.
Then to solve for x we use the quadratic formula.
x is = tob + or minus square<unk> of b ^ 2 - 4 a c over 2 a.
So all we need to do is just plug in this values into this formula.
This will then give us x = minus b itself is negative. So we have - 100 plus or minus square<unk> of -00^ 2 - 4 * a is 1 and c is 100.
Then all of that divided by two because a is one.
This will then give us x is = positive 100 plus or minus square root of we have 100^ 2 - 4 * 100 then divided by 2 * 1 is 2 giving us x is = 100 plus or minus square root of 100 is common here. So let's do 100 into brackets 100 - 4 then / 2 x is = 100 + or minus square<unk> of 100 * 100 - 4 is 96. six.
So we get x is = to 100 plus or minus. If we separate this we get square<unk> of 100 *<unk> of 96 then divided by two.
So that this gives us x is = 100 plus or minus square root of 100 is 10.
If we look at 96 this is 16 * 6 then divide by two.
So we get x is = 100 + or - 10 *<unk> 16 *<unk> 6 then / 2 giving us x is = 100 + or minus square root of 16 is 4 4 * 10 is 40 then <unk>6 / 2.
I'm going to separate this division to give us x is = 100 / 2 plus or - 40 <unk>6 / 2. 2 here is 1. 2 here 50 2 here 1 2 here 20.
So this will give us x is = 50 plus or minus 20 20<unk> 6 or we can separate this into x1 is = 50 + 20 <unk>6 or x2 2 is equal to 50 - 20 <unk>6.
And with these two values of x, we can then get the corresponding values of y.
Recall from equation three, we had y is = 100 - x.
Therefore, y1 is going to be 100 - x1. And we know x1 is 50 + 20<unk> 6.
This will imply that y1 is = 100 minus 50 + 20 <unk>6 which will give us y1 is equal to 100 - 50 here is 50 - * + is - 20 <unk>6 six.
Similarly, we're going to have y2 = 100 - this will be 50 - 20<unk> 6 and then we get y2 is equal to 100 - 50 here will give us 50 - * - is plus. So we have 50 + 20 <unk>6 giving us every possible values for this problem. Now let us do a quick check to confirm that these values are correct.
To check, we need to substitute each of this into the given problem, which was x + y = 100 and xy also equal to 100.
We can use either of these solution sets to check for correctness. I'm going to use the first one. Now starting with the first equation, x + y is = 100. This will then imply 50 + 20<unk> 6 + 50 - 20<unk> 6 to give us 100.
So this here cancels this and then we have 50 + 50 that will give us 100. So we have 100 is equal to 100. And because the left hand side balances the right hand side confirms that this solution is correct.
But we need to also check with the second equation. Putting this values.
Therefore, xy is = 100.
This will then imply 50 plus 06 * 50 - 2006 to give us 100.
50 here * 50 here will give us 2500.
Then we do 50 here times this. This will give us - 1,6.
And this here times this will give us plus 1, <unk>6.
And finally this here times this here because the signs are different. So we have 20 <unk>6 raised to power 2 to give us 100.
This is 2500.
This and this take care of each other cuz they have opposing signs. Then we have - 20<unk> 6 raised to power 2 to give us 100.
Here this is 2500 minus by law of indices this will give us 20 raised to power 2 * <unk>6 raised to power 2 to give us 100.
This takes care of this.
So we have 2 500 minus this is 400 * 6 to give us 100.
So we have 2 500 minus this will be 2 400 to give us 100 and this is 100.
So 100 is equal to 100. Again the left hand side balances the right hand side.
Also confirming that these solutions are perfectly correct.
Thanks for watching. Please like and share and also remember to subscribe to my channel and I'll see you in my next video. Bye.
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