HiSET - Rearrange formulas/equations to highlight a quantity of interest

Solve for :

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

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

$x = \frac{1}{3} y + \frac{4}{3}$

$x = \frac{2}{3} y -\frac{4}{3}$

$x = \frac{2}{3} y + \frac{1}{3}$

$x = \frac{4}{3} y + \frac{1}{3}$

Explanation

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

First, take the reciprocal of both sides:

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

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

Multiply both sides by :

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

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

Distribute on the right:

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

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

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

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

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

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

the correct response.

Solve for :

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

$x = y^{2}+ 2 y$

$x = y^{2}-2 y$

$x = y^{2}+ y$

$x = y^{2}- y$

$x = y^{2}$

Explanation

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

First, square both sides to eliminate the radical symbol:

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

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

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

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

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

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

Subtract 1 from both sides:

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

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

the correct response.

Solve for :

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

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

$x =40 c -\frac{3}{2} y$

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

Explanation

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

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

Multiply both sides by :

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

Subtract from both sides:

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

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

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

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

the correct response.

Solve for :

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

Assume is positive.

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

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

$x = 3 \sqrt {y - z}$

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

$x = \sqrt {3y - 3z}$

Explanation

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

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

Subtract from both sides:

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

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

Divide both sides by 9:

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

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

Take the square root of both sides:

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

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

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

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

the correct response.

Solve for :

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

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

$x= \frac{yz - 20}{y}$

Explanation

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

Subtract 20 from both sides:

Divide both sides by :

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

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

the correct response.

Solve for :

$32 -x^{2} = y$

You my assume is positive.

$x=\sqrt{ 32- y}$

$x=\sqrt{ 32}- y$

$x=\sqrt{ y - 32}$

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

$x=\sqrt{ y}- 32$

Explanation

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

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

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

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

$32 - y = x^{2}$

Take the positive square root of both sides:

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

the correct response.

Rearrange formulas/equations to highlight a quantity of interest

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