can someone please help me im stumped on these two.
use analytical methods to solve the equation.
1+(1/x)=(42/x²)
(7/x-4)=1+(9/x+4)
1 + 1/x = 42/x^2.
Multiply both sides by x^2 and get:
x^2 + x = 42,
x^2 + x - 42 = 0,
(x+7)(x-6) = 0.
x+7 = 0,
x = -7.
x-6 = 0,
x = 6.
Solution Set: x=-7, and 6.
7/(x-4) = 1 + 9/(x+4).
7/(x-4) - 9/(x+4) = 1,
LCM = (x-4)(x+4),
(7(x+4) - 9(x-4)) / (x-4)(x+4) = 1,
(7x+28-9x+36) / (x-4)(x+4) = 1,
(-2x+64) / (x--4)(x+4) = 1,
Cross Multiply:
(x-4)(x+4) = -2x + 64,
x^2 - 16 = -2x+64,
x^2 + 2x - 80 = 0,
(x-8)(x+10) = 0.
x-8 = 0,
x = 8.
x+10 = 0,
x = -10.
Solution Set: x = 8, and -10.