Let "a" and "b" be some number where:
a - b = 24
We want to find where a^2 + b^2 is a minimum. Instead of just logically figuring out that the answer is where a=b=12, I'll just use derivatives.
So we can first substitute for "a" where a = b+24
So we have (b+24)^2 + b^2 = b^2 +48b +576 + b^2
And that equals 2b^2 +48b +576
Then we take the derivative and set it equal to zero:
4b +48 = 0
4(b+12) = 0
b + 12 = 0
b = -12
Thus "a" must equal 12.
So:
a = 12
b = -12
And the sum of those two numbers squared is (12)^2 + (-12)^2 = 144 + 144 = 288.
The smallest sum is 288.
Answer:
x = 3
Step-by-step explanation:


Answer:
Step-by-step explanation:
Hello here is a solution :
Answer: Choice B
(-2, 5)
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Explanation:
The original system is

Multiply both sides of the second equation by 3. Doing so leads to this updated system of equations

Now add straight down
The x terms add to -4x+3x = -1x = -x
The y terms add to 3y+(-3y) = 0y = 0
The terms on the right hand sides add to 23+(-21) = 2
We end up with the equation -x = 2 which solves to x = -2
Now use this to find y. You can pick any equation with x,y in it
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-4x+3y = 23
-4(-2)+3y = 23
8+3y = 23
3y = 23-8
3y = 15
y = 15/3
y = 5
Or
x-y = -7
-2-y = -7
-y = -7+2
y = -5
y = 5
Either way, we get the same y value.
So that's why the solution is (x,y) = (-2, 5)