To solve for $\displaystyle x$ in a literal equation, use the properties of algebra to isolate $\displaystyle x$ on one side, just as if you were solving a regular equation. 

$\displaystyle \frac{2x+3y }{c} = 20$

Multiply both sides by $\displaystyle c$:

$\displaystyle \frac{2x+3y }{c} \cdot c = 20 \cdot c$

$\displaystyle 2x+3y = 20 c$

Subtract $\displaystyle 3y$ from both sides:

$\displaystyle 2x+3y - 3y = 20 c - 3y$

$\displaystyle 2x = 20 c - 3y$

Multiply both sides by $\displaystyle \frac{1}{2}$, distributing on the right:

$\displaystyle \frac{1}{2} \cdot 2x =\frac{1}{2} \cdot (20 c - 3y)$

$\displaystyle x =\frac{1}{2} \cdot 20 c -\frac{1}{2} \cdot 3y$

$\displaystyle x =10 c -\frac{3}{2} y$,

the correct response.

To solve for $\displaystyle x$ in a literal equation, use the properties of algebra to isolate $\displaystyle x$ on one side, just as if you were solving a regular equation. 

$\displaystyle 9x^{2}+z = y$

Subtract $\displaystyle z$ from both sides:

$\displaystyle 9x^{2}+z - z = y - z$

$\displaystyle 9x^{2} = y - z$

Divide both sides by 9:

$\displaystyle \frac{9x^{2}}{9} = \frac{y - z}{9}$

$\displaystyle x^{2} = \frac{y - z}{9}$

Take the square root of both sides:

$\displaystyle x =\sqrt{ \frac{y - z}{9}}$

Simplify the expression on the right by splitting it, and taking the square root of numerator and denominator:

$\displaystyle x = \frac{\sqrt {y - z}}{\sqrt {9}}$

$\displaystyle x = \frac{\sqrt {y - z}}{3}$,

the correct response.

To solve for $\displaystyle x$ in a literal equation, use the properties of algebra to isolate $\displaystyle x$ on one side, just as if you were solving a regular equation. 

First, take the reciprocal of both sides:

$\displaystyle \frac{y+1}{x+1} = 3$

$\displaystyle \frac{x+1}{y+1} = \frac{1}{3}$

Multiply both sides by $\displaystyle y+1$:

$\displaystyle \frac{x+1}{y+1} \cdot (y+1)= \frac{1}{3}(y+1)$

$\displaystyle x+1= \frac{1}{3}(y+1)$

Distribute on the right:

$\displaystyle x+1= \frac{1}{3} \cdot y+ \frac{1}{3} \cdot 1$

$\displaystyle x+1= \frac{1}{3} y+ \frac{1}{3}$

Subtract 1 from both sides, rewriting 1 as $\displaystyle \frac{3}{3}$ to facilitate subtraction:

$\displaystyle x+1 - 1 = \frac{1}{3} y+ \frac{1}{3} - 1$

$\displaystyle x = \frac{1}{3} y+ \frac{1}{3} - \frac{3}{3}$

$\displaystyle x = \frac{1}{3} y - \frac{2}{3}$,

the correct response.

To solve for $\displaystyle x$ in a literal equation, use the properties of algebra to isolate $\displaystyle x$ on one side, just as if you were solving a regular equation. 

$\displaystyle xy+20 = z$

Subtract 20 from both sides:

$\displaystyle xy+20- 20 = z - 20$

$\displaystyle xy = z - 20$

Divide both sides by $\displaystyle y$:

$\displaystyle \frac{xy}{y} = \frac{z - 20}{y}$

$\displaystyle x= \frac{z - 20}{y}$,

the correct response.

To solve for $\displaystyle x$ in a literal equation, use the properties of algebra to isolate $\displaystyle x$ on one side, just as if you were solving a regular equation. 

First, add $\displaystyle x^{2} - y$ to both sides:

$\displaystyle 32 -x^{2} +(x^{2} - y) = y +(x^{2} - y)$

$\displaystyle 32 -x^{2} +x^{2} - y = x^{2} +y - y$

$\displaystyle 32 - y = x^{2}$

Take the positive square root of both sides:

$\displaystyle x=\sqrt{ 32- y}$,

the correct response.

To solve for $\displaystyle x$ in a literal equation, use the properties of algebra to isolate $\displaystyle x$ on one side, just as if you were solving a regular equation. 

First, square both sides to eliminate the radical symbol:

$\displaystyle \sqrt{x+1}= y+1$

$\displaystyle (\sqrt{x+1})^{2}=( y+1)^{2}$

$\displaystyle x+1=( y+1)^{2}$

Rewrite the expression on the right using the square of a binomial pattern:

$\displaystyle x+1= y^{2}+ 2 (y)(1)+1^{2}$

$\displaystyle x+1= y^{2}+ 2 y+1$

Subtract 1 from both sides:

$\displaystyle x+1-1 = y^{2}+ 2 y+1 - 1$

$\displaystyle x = y^{2}+ 2 y$,

the correct response.