Question

Solve for x: 3 over 3 x plus 1 over quantity x plus 4 equals 10 over 7x .

x = −3
x = 3
x = −3 and x = 3
No solution

Answer 1

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From thus the answer would be

X = 3

:)

Answer 2

`x=3`

Explanation:

we have

`(3)/(3x)+(1)/(x+4)=(10)/(7x)`

The domain of the function is all real numbers except the zero, because the denominator can not be zero

Multiply by
`[3x(x+4)7x]` both sides

`(3[3x(x+4)7x])/(3x)+([3x(x+4)7x])/(x+4)=(10[3x(x+4)7x])/(7x)`

`(3[3x(x+4)7x])/(3x)+([3x(x+4)7x])/(x+4)=(10[3x(x+4)7x])/(7x)`

`3[(x+4)7x]+[3x(7x)]=10[3x(x+4)]`

`21x^(2)+84x+21x^(2)=30x^(2)+120x`

`42x^(2)+84x=30x^(2)+120x`

`42x^(2)-30x^(2)+84x-120x=0`

`12x^(2)-36x=0`

Divide by
`12` both sides

`x^(2)-3x=0`

we know that

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`x^(2)-3x=0`

so

`a=1\\b=-3\\c=0`

substitute in the formula

`x=\frac{3(+/-)\sqrt{(-3)^(2)-4(1)(0)}} {2(1)}`

`x=\frac{3(+/-)√(9)} {2}`

`x=\frac{3(+/-)3} {2}`

`x=\frac{3+3} {2}=3`

`x=\frac{3-3} {2}=0`

remember that

The domain of the function is all real numbers except the zero

so

the solution is

`x=3`