Solve 3u-2/4+2u/1=8/3div2u+3/3

$$\frac { 3 u - 2 } { 4 } + \frac { 2 u } { 1 } = \frac { 8 } { 3 } \div \frac { 2 u + 3 } { 3 }$$

Solve for u

$$u=\frac{3\sqrt{465}-29}{44}\approx 0.811172181$$

$$u=\frac{-3\sqrt{465}-29}{44}\approx -2.129353999$$

Solution Steps

Multiply both sides of the equation by 4.

$$3u-2+4\times 2u=12\left(2u+3\right)^{-1}\times \frac{8}{3}$$

Multiply 4 and 2 to get 8.

$$3u-2+8u=12\left(2u+3\right)^{-1}\times \frac{8}{3}$$

Combine 3u and 8u to get 11u.

$$11u-2=12\left(2u+3\right)^{-1}\times \frac{8}{3}$$

Multiply 12 and \frac{8}{3} to get 32.

$$11u-2=32\left(2u+3\right)^{-1}$$

Subtract 32\left(2u+3\right)^{-1} from both sides.

$$11u-2-32\left(2u+3\right)^{-1}=0$$

Reorder the terms.

$$11u-2-32\times \frac{1}{2u+3}=0$$

Variable u cannot be equal to -\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2u+3.

$$\left(2u+3\right)\left(11u-2\right)-32\times \frac{1}{2u+3}\left(2u+3\right)=0$$

Use the distributive property to multiply 2u+3 by 11u-2 and combine like terms.

$$22u^{2}+29u-6-32\times \frac{1}{2u+3}\left(2u+3\right)=0$$

Express -32\times \frac{1}{2u+3} as a single fraction.

$$22u^{2}+29u-6+\frac{-32}{2u+3}\left(2u+3\right)=0$$

Express \frac{-32}{2u+3}\left(2u+3\right) as a single fraction.

$$22u^{2}+29u-6+\frac{-32\left(2u+3\right)}{2u+3}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 22u^{2}+29u-6 times \frac{2u+3}{2u+3}.

$$\frac{\left(22u^{2}+29u-6\right)\left(2u+3\right)}{2u+3}+\frac{-32\left(2u+3\right)}{2u+3}=0$$

Since \frac{\left(22u^{2}+29u-6\right)\left(2u+3\right)}{2u+3} and \frac{-32\left(2u+3\right)}{2u+3} have the same denominator, add them by adding their numerators.

$$\frac{\left(22u^{2}+29u-6\right)\left(2u+3\right)-32\left(2u+3\right)}{2u+3}=0$$

Do the multiplications in \left(22u^{2}+29u-6\right)\left(2u+3\right)-32\left(2u+3\right).

$$\frac{44u^{3}+66u^{2}+58u^{2}+87u-12u-18-64u-96}{2u+3}=0$$

Combine like terms in 44u^{3}+66u^{2}+58u^{2}+87u-12u-18-64u-96.

$$\frac{44u^{3}+124u^{2}+11u-114}{2u+3}=0$$

Variable u cannot be equal to -\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2u+3.

$$44u^{3}+124u^{2}+11u-114=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -114 and q divides the leading coefficient 44. List all candidates \frac{p}{q}.

$$±\frac{57}{22},±\frac{57}{11},±\frac{114}{11},±\frac{57}{2},±57,±114,±\frac{57}{44},±\frac{57}{4},±\frac{19}{22},±\frac{19}{11},±\frac{38}{11},±\frac{19}{2},±19,±38,±\frac{19}{44},±\frac{19}{4},±\frac{3}{22},±\frac{3}{11},±\frac{6}{11},±\frac{3}{2},±3,±6,±\frac{3}{44},±\frac{3}{4},±\frac{1}{22},±\frac{1}{11},±\frac{2}{11},±\frac{1}{2},±1,±2,±\frac{1}{44},±\frac{1}{4}$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$u=-\frac{3}{2}$$

By Factor theorem, u-k is a factor of the polynomial for each root k. Divide 44u^{3}+124u^{2}+11u-114 by 2\left(u+\frac{3}{2}\right)=2u+3 to get 22u^{2}+29u-38. Solve the equation where the result equals to 0.

$$22u^{2}+29u-38=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 22 for a, 29 for b, and -38 for c in the quadratic formula.

$$u=\frac{-29±\sqrt{29^{2}-4\times 22\left(-38\right)}}{2\times 22}$$

Do the calculations.

$$u=\frac{-29±3\sqrt{465}}{44}$$

Solve the equation 22u^{2}+29u-38=0 when ± is plus and when ± is minus.

$$u=\frac{-3\sqrt{465}-29}{44} u=\frac{3\sqrt{465}-29}{44}$$

Remove the values that the variable cannot be equal to.

$$u\in \emptyset$$

List all found solutions.

$$u=-\frac{3}{2} u=\frac{-3\sqrt{465}-29}{44} u=\frac{3\sqrt{465}-29}{44}$$

Variable u cannot be equal to -\frac{3}{2}.

$$u=\frac{3\sqrt{465}-29}{44} u=\frac{-3\sqrt{465}-29}{44}$$