Solve in Terms of the Arbitrary Variable R T=(5R)/(S+U)

$T = \frac{5 R}{S + U}$

Step 1

Solve the equation for $S$ .

Rewrite the equation as $\frac{5 R}{S + U} = T$ .

$\frac{5 R}{S + U} = T$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$S + U, 1$

Remove parentheses.

$S + U, 1$

The LCM of one and any expression is the expression.

$S + U$

Multiply each term in $\frac{5 R}{S + U} = T$ by $S + U$ to eliminate the fractions.

Multiply each term in $\frac{5 R}{S + U} = T$ by $S + U$ .

$\frac{5 R}{S + U} (S + U) = T (S + U)$

Simplify the left side.

Cancel the common factor of $S + U$ .

Cancel the common factor.

$\frac{5 R}{S + U} (S + U) = T (S + U)$

Rewrite the expression.

$5 R = T (S + U)$

Simplify the right side.

Apply the distributive property.

$5 R = T S + T U$

Solve the equation.

Rewrite the equation as $T S + T U = 5 R$ .

$T S + T U = 5 R$

Subtract $T U$ from both sides of the equation.

$T S = 5 R - T U$

Divide each term in $T S = 5 R - T U$ by $T$ and simplify.

Divide each term in $T S = 5 R - T U$ by $T$ .

$\frac{T S}{T} = \frac{5 R}{T} + \frac{- T U}{T}$

Simplify the left side.

Cancel the common factor of $T$ .

Cancel the common factor.

$\frac{T S}{T} = \frac{5 R}{T} + \frac{- T U}{T}$

Divide $S$ by $1$ .

$S = \frac{5 R}{T} + \frac{- T U}{T}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $T$ .

Cancel the common factor.

$S = \frac{5 R}{T} + \frac{- T U}{T}$

Divide $- 1 U$ by $1$ .

$S = \frac{5 R}{T} - 1 U$

Rewrite $- 1 U$ as $- U$ .

$S = \frac{5 R}{T} - U$

Step 2

Solve the equation for $U$ .

Rewrite the equation as $\frac{5 R}{(\frac{5 R}{T} - U) + U} = T$ .

$\frac{5 R}{(\frac{5 R}{T} - U) + U} = T$

Factor each term.

Remove parentheses.

$\frac{5 R}{\frac{5 R}{T} - U + U} = T$

Add $- U$ and $U$ .

$\frac{5 R}{\frac{5 R}{T} + 0} = T$

Add $\frac{5 R}{T}$ and $0$ .

$\frac{5 R}{\frac{5 R}{T}} = T$

Multiply the numerator by the reciprocal of the denominator.

$5 R \frac{T}{5 R} = T$

Cancel the common factor of $5 R$ .

Cancel the common factor.

$5 R \frac{T}{5 R} = T$

Rewrite the expression.

$T = T$

Solve the equation.

Move all terms containing $T$ to the left side of the equation.

Subtract $T$ from both sides of the equation.

$T - T = 0$

Subtract $T$ from $T$ .

$0 = 0$

Since $0 = 0$ , the equation will always be true.

Always true

Solve in Terms of the Arbitrary Variable R T=(5R)/(S+U)

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