Answer:
[tex]x=3[/tex]
Step-by-step explanation:
we have
[tex]\frac{3}{3x}+\frac{1}{x+4}=\frac{10}{7x}[/tex]
The domain of the function is all real numbers except the zero, because the denominator can not be zero
Multiply by [tex][3x(x+4)7x][/tex] both sides
[tex]\frac{3[3x(x+4)7x]}{3x}+\frac{[3x(x+4)7x]}{x+4}=\frac{10[3x(x+4)7x]}{7x}[/tex]
[tex]\frac{3[3x(x+4)7x]}{3x}+\frac{[3x(x+4)7x]}{x+4}=\frac{10[3x(x+4)7x]}{7x}[/tex]
[tex]3[(x+4)7x]+[3x(7x)]=10[3x(x+4)][/tex]
[tex]21x^{2}+84x+21x^{2}=30x^{2}+120x[/tex]
[tex]42x^{2}+84x=30x^{2}+120x[/tex]
[tex]42x^{2}-30x^{2}+84x-120x=0[/tex]
[tex]12x^{2}-36x=0[/tex]
Divide by [tex]12[/tex] both sides
[tex]x^{2}-3x=0[/tex]
we know that
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2}-3x=0[/tex]
so
[tex]a=1\\b=-3\\c=0[/tex]
substitute in the formula
[tex]x=\frac{3(+/-)\sqrt{(-3)^{2}-4(1)(0)}} {2(1)}[/tex]
[tex]x=\frac{3(+/-)\sqrt{9}} {2}[/tex]
[tex]x=\frac{3(+/-)3} {2}[/tex]
[tex]x=\frac{3+3} {2}=3[/tex]
[tex]x=\frac{3-3} {2}=0[/tex]
remember that
The domain of the function is all real numbers except the zero
so
the solution is
[tex]x=3[/tex]
