2x 3 4 5x 3 July 8, 2019 16 32 64 128 2x 3y 4 3 5x 2y 7 by substitution method 2x 4 3x 3 5x 2 7x 11 0 1 2x 3 4 3 5x 2x 3 4 5x )= 4 5 3x 2x 4 5x 3 3x 2 2x 3 3x 4 5x 2 4x 8 4 2x 3 )+( 5x 3 2 3x 4 2x 3 5x 2 7 x 2 2x 2 3x 4 x 2 5x 3 3 4 2x 3 5x 3 1 2 2x^3 3x^2 5x - 4 divided by x^2 x 1 5x 3 4 2x 4 3 5 4 5x 3 )= 7 2x 3 2x 3 5x 4 x 3
