2x 1 7x 5 July 8, 2019 16 32 64 128 2x 1 )( 7x 5 )= 0 2x 3 1 7x 5 3 X 2 2x 1 7x 5 Lim x 1 2x 2 7x 5 7x 3y 1 2x 5y 12 3x 1 2x 3 7x 5 2x 1 )( 3x2 7x 5 2x 3 1 7x 50 3 2x 1 7x 2 3 5 Y 2x 1 5y 7x 11 substitution 2x 1 5 7x 1 13 4 2x 1 )-( 5 7x 3x 2x 1 7x 3 5x x 24
