What number needs to be added to 5 and 3 so that the ratio of the first number to the second
1 answer:
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2 ways: Easy and hard
Hard=A
Easy=B
A: 1/2x+4 work from there so we do fun stuff with it make something that can be simplified so 1/2x+4 times (2/2)=x+8 now square the whole thing and put the result in a square root thingie (x+8)^2=x^2+16x+64
multiply the whole thing by 4/4 and put ![\sqrt{16} [\tex] on top so then [tex] \sqrt{x^2+16x+64}](https://tex.z-dn.net/?f=%20%5Csqrt%7B16%7D%20%5B%5Ctex%5D%20on%20top%20so%20then%20%0A%5Btex%5D%20%5Csqrt%7Bx%5E2%2B16x%2B64%7D%20)
times 
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to solve it, factor out the 16 in the square root and then square root 16 to get 4
then it will be (4 times square root of equation)/4=square root of equatio
factor square root of equation and square root it and get x+8
divide by 2 to get 1/2x+4
B: 1/2x+4
put stuff that cancels out
1/2x+3x-3x+4+56-56
move them around
3 and 1/2x-3x+60-56
or
2x-3x+1 and 1/2x+30-20+30-36
then just add like terms to solve
Hard=A
Easy=B
A: 1/2x+4 work from there so we do fun stuff with it make something that can be simplified so 1/2x+4 times (2/2)=x+8 now square the whole thing and put the result in a square root thingie (x+8)^2=x^2+16x+64
B: 1/2x+4
put stuff that cancels out
1/2x+3x-3x+4+56-56
move them around
3 and 1/2x-3x+60-56
or
2x-3x+1 and 1/2x+30-20+30-36
then just add like terms to solve