Question

(7+x)в квадрате-(5+x)в квадрате=65 (2-y)в квадрате-(6-y)в квадрате =4x

Answer 1

$(7+x)^2-(5+x)^2=65 \\ 49+14x+x^2-25-10x-x^2=65 \\ 4x+24=65 \\ 4x=65 -24 \\ 4x=41 \\ x=41/4 \\ x=10,25$

$(2-y)^2-(6-y)^2=4x \\ 4-4y+y^2-36+12y-y^2=4x \\ 8y-32=4x\\ 8y-4x=32$