Question

(x+1)(x+2)-(x-3)(x+4)=6
(3x-1)(2x+7)-(x+1)(6x-5)=7​

Answer 1

$(x+1)(x+2)-(x-3)(x+4)=6\x^2+2x+x+2-(x^2+4x-3x-12)=6\x^2+2x+x+2-(x^2+x-12)=6\x^2+2x+x+2-x^2-x+12=6\2x+14=6\2x=6-14\2x=-8\x=-4$

$(3x-1)(2x+7)-(x+1)(6x-5)=7\6x^2+21x-2x-7-(6x^2-5x+6x-5)=7\6x^2+21x-2x-7-(6x^2+x-5)=7\6x^2+21x-2x-7-6x^2-x+5=7\18x-2=7\18x=7+2\18x=9\x=frac{1}{2}$