so we can get 2 equations here, please ask if u want to know how i got these.
\(4a + 4b = 16\)
\(a^2 + 2ab + b^2 = 16\)
so now we combine them
\(4a + 4b = a^2 + 2ab + b^2\)
\(4a + 4b - 2ab - 2b^2 = (a+b)(a-b)\)
more on how that works if you ask me.
\(4(a+b) - 2b(a+b) = (a+b)(a-b)\)
\(4-2b = a-b\)
i think thats correct..
\(a+b = 4\)
thats depressing oofoof
okok for b) its just 4, literally
and for c i don't think any integers work therfore no real numbers work becuase the max number for a or b is 1, becuase 2^4 is already 16 and negatives^4 are always postive for i think c is none.
and for a) when i did the failed equations u see a 2ab when i squared the first equation,try to single that out and u will get the answer. for a
HINT. for a, do \(2ab = a^4+b^4-a^2 - b^2\)
that will be equal to \(2ab = (a^2)(a^2 - 1) + (b^2)(b^2 - 1)\)
then use the second equation to substiture the a^2 and b^2 and u will get somewhere...
Ok lets see...
alright lets make these look better.
alright more simplifying:
Oh yay! A quadratic equation! Now time to solve!
Substitute this into xy = 20 and you will get
so x is NOT 4±2i but it is -4\pm2i so yeah!
If anyone can check my work that would be great
Next time the image would be greatly appreciated, lets see what i can do..
What do you mean by finding the equation of the medians? I think it is how to find the medians
ok so find the midpoint of the line segments with this:
for part b thou, they intersect at \(\left(2.66666666666,2.3333333333\right)\).
Divide the first congruence by 7, remembering to divide 14 by as well. We find that the first congruence is equivalent to . Subtracting 13 from both sides and multiplying both sides by 5 (which is the modular inverse of 2, modulo 9) gives for the second congruence. Finally, adding to both sides in the third congruence and multiplying by 17 (which is the modular inverse of 3, modulo 25) gives . So we want to solveLet's first find a simultaneous solution for the second and third congruences. We begin checking numbers which are 17 more than a multiple of 25, and we quickly find that 42 is congruent to 17 (mod 25) and 6 (mod 9). Since 42 does not satisfy the first congruence we look to the next solution . Now we have found a solution for the system, so we can appeal to the Chinese Remainder Theorem to conclude that the general solution of the system is , where 450 is obtained by taking the least common multiple of 2, 9, and 25. So the least four-digit solution is .