Solve ((m-7)(3m+10)/(2m+3)(3n+10)=-2/3

$$\frac { ( ( m - 7 ) ( 3 m + 10 ) } { ( 2 m + 3 ) ( 3 n + 10 ) } = - \frac { 2 } { 3 }$$

Solve for n

$$n=-\frac{9m^{2}+7m-150}{6\left(2m+3\right)}$$

$$m\neq -\frac{3}{2}\text{ and }m\neq 7\text{ and }m\neq -\frac{10}{3}$$

Steps for Solving Linear Equation

Variable n cannot be equal to -\frac{10}{3} since division by zero is not defined. Multiply both sides of the equation by 3\left(2m+3\right)\left(3n+10\right), the least common multiple of \left(2m+3\right)\left(3n+10\right),3.

$$3\left(m-7\right)\left(3m+10\right)=-2\left(2m+3\right)\left(3n+10\right)$$

Use the distributive property to multiply 3 by m-7.

$$\left(3m-21\right)\left(3m+10\right)=-2\left(2m+3\right)\left(3n+10\right)$$

Use the distributive property to multiply 3m-21 by 3m+10 and combine like terms.

$$9m^{2}-33m-210=-2\left(2m+3\right)\left(3n+10\right)$$

Use the distributive property to multiply -2 by 2m+3.

$$9m^{2}-33m-210=\left(-4m-6\right)\left(3n+10\right)$$

Use the distributive property to multiply -4m-6 by 3n+10.

$$9m^{2}-33m-210=-12mn-40m-18n-60$$

Swap sides so that all variable terms are on the left hand side.

$$-12mn-40m-18n-60=9m^{2}-33m-210$$

Add 40m to both sides.

$$-12mn-18n-60=9m^{2}-33m-210+40m$$

Combine -33m and 40m to get 7m.

$$-12mn-18n-60=9m^{2}+7m-210$$

Add 60 to both sides.

$$-12mn-18n=9m^{2}+7m-210+60$$

Add -210 and 60 to get -150.

$$-12mn-18n=9m^{2}+7m-150$$

Combine all terms containing n.

$$\left(-12m-18\right)n=9m^{2}+7m-150$$

Divide both sides by -12m-18.

$$\frac{\left(-12m-18\right)n}{-12m-18}=\frac{9m^{2}+7m-150}{-12m-18}$$

Dividing by -12m-18 undoes the multiplication by -12m-18.

$$n=\frac{9m^{2}+7m-150}{-12m-18}$$

Divide 7m+9m^{2}-150 by -12m-18.

$$n=-\frac{9m^{2}+7m-150}{6\left(2m+3\right)}$$

Variable n cannot be equal to -\frac{10}{3}.

$$n=-\frac{9m^{2}+7m-150}{6\left(2m+3\right)}\text{, }n\neq -\frac{10}{3}$$

Solve for m

$$m=\frac{\sqrt{144n^{2}-480n+5449}}{18}-\frac{2n}{3}-\frac{7}{18}$$

$$m=-\frac{\sqrt{144n^{2}-480n+5449}}{18}-\frac{2n}{3}-\frac{7}{18}\text{, }n\neq -\frac{10}{3}$$