Question

Not sure if I plug 43 in

Answer 1

`(x^2-2x)+(x+4)+(x+3)+(3x-4)~~ = ~~43 \\\\\\ x^2+3x+3=43\implies x^2+3x-40=0 \\\\\\ (x+8)(x-5)=0\implies x= \begin{cases} -8&\bigotimes\\\\ 5&\checkmark \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{x+3}{8}\hspace{5em} \stackrel{x+4}{9}\hspace{5em} \stackrel{x^2+2x}{35}\hspace{5em} \stackrel{3x-4}{11}`

again, let's notice, we can't use x = -8, because whatever "x" may be, it has to be greater than 0.