Resposta :
temos f(x)=2x+4 e f(x)=10, então
[tex]\overbrace{f(x)}^{10}=2x+4\\\\10=2x+4\\\\2x=10-4\\\\2x=6\\\\x=\dfrac{6}{2}\\\\\boxed{\boxed{x=3}}[/tex]
temos f(x)=2x+4 e f(x)=10, então
[tex]\overbrace{f(x)}^{10}=2x+4\\\\10=2x+4\\\\2x=10-4\\\\2x=6\\\\x=\dfrac{6}{2}\\\\\boxed{\boxed{x=3}}[/tex]