Solve 7/3x+1/3x=3x+40/3+7/3x

$$\frac { 7 } { 3 } x + \frac { 1 } { 3 } x = 3 x + \frac { 40 } { 3 } + \frac { 7 } { 3 } x$$

Solve for x

$$x=-5$$

Steps for Solving Linear Equation

Combine \frac{7}{3}x and \frac{1}{3}x to get \frac{8}{3}x.

$$\frac{8}{3}x=3x+\frac{40}{3}+\frac{7}{3}x$$

Combine 3x and \frac{7}{3}x to get \frac{16}{3}x.

$$\frac{8}{3}x=\frac{16}{3}x+\frac{40}{3}$$

Subtract \frac{16}{3}x from both sides.

$$\frac{8}{3}x-\frac{16}{3}x=\frac{40}{3}$$

Combine \frac{8}{3}x and -\frac{16}{3}x to get -\frac{8}{3}x.

$$-\frac{8}{3}x=\frac{40}{3}$$

Multiply both sides by -\frac{3}{8}, the reciprocal of -\frac{8}{3}.

$$x=\frac{40}{3}\left(-\frac{3}{8}\right)$$

Multiply \frac{40}{3} times -\frac{3}{8} by multiplying numerator times numerator and denominator times denominator.

$$x=\frac{40\left(-3\right)}{3\times 8}$$

Do the multiplications in the fraction \frac{40\left(-3\right)}{3\times 8}.

$$x=\frac{-120}{24}$$

Divide -120 by 24 to get -5.

$$x=-5$$