Solve for x: 7 over 8 minus 1 over x equals 3 over 4. x = 1
x = 2
x = 4
x = 8

Solve for x: 2 over quantity x plus 2 plus 1 over 5 equals 6 over quantity x plus 5.

x = 0
x = −13
x = 13
x = 0 and x = 13

Classify the solutions of 3 over x plus 5, plus one fifth, equals 2 over x plus 5 as extraneous or non-extraneous.

x = −5; extraneous
x = −5; non-extraneous
x = −10; extraneous
x = −10; non-extraneous