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

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Question

Solve for :

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

Tap to reveal answer

Answer

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.

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Question

Solve for :

Assume is positive.

Tap to reveal answer

Answer

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 from both sides:

Divide both sides by 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.

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Question

Solve for :

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

Solve for :

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

Tap to reveal answer

Answer

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}$= y+1$

$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.

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Question

Solve for :

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

You my assume is positive.

Tap to reveal answer

Answer

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 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.

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Question

Solve for :

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

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

Solve for :

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

Solve for :

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

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

Solve for :

Assume is positive.

Tap to reveal answer

Answer

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 from both sides:

Divide both sides by 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.

← Didn't Know|Knew It →

Question

Solve for :

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

Tap to reveal answer

Answer

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}$= y+1$

$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.

← Didn't Know|Knew It →

Question

Solve for :

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

You my assume is positive.

Tap to reveal answer

Answer

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 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.

← Didn't Know|Knew It →

Question

Solve for :

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

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

Solve for :

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

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

Solve for :

Assume is positive.

Tap to reveal answer

Answer

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 from both sides:

Divide both sides by 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.

← Didn't Know|Knew It →

Question

Solve for :

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

Solve for :

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

Tap to reveal answer

Answer

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}$= y+1$

$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.

← Didn't Know|Knew It →

Question

Solve for :

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

You my assume is positive.

Tap to reveal answer

Answer

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 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.

← Didn't Know|Knew It →

Question

Solve for :

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

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

Solve for :

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

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

Solve for :

Assume is positive.

Tap to reveal answer

Answer

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 from both sides:

Divide both sides by 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.

← Didn't Know|Knew It →

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