Problem 51 - Intermediate Algebra

Question

Solve each formula for the specified variable. \(\frac{E}{e}=\frac{R+r}{r}\) for \(r\) (from engineering)

Step-by-Step Solution

Answer

The solution is \( r = \frac{eR}{E-e} \).

Step 1: Cross-multiply

Cross-multiply to get rid of the fractions. Multiply both sides by the denominators to get: \( E imes r = e imes (R + r) \).

Step 2: Distribute

Distribute \( e \) on the right side of the equation: \( Er = eR + er \).

Step 3: Isolate terms with 'r'

Move all terms containing \( r \) to one side of the equation by subtracting \( er \): \( Er - er = eR \).

Step 4: Factor 'r' out

Factor out \( r \) from the left side: \( r(E - e) = eR \).

Step 5: Solve for 'r'

Divide both sides by \( E-e \) to solve for \( r \): \( r = \frac{eR}{E-e} \).

Key Concepts

Cross-Multiplication

Cross-multiplication is a method used to eliminate fractions from an equation, making it simpler and more straightforward to solve. When you see an equation involving two fractions set equal to one another, like in our exercise, cross-multiplication can be applied to eliminate the denominators and work with a simpler equation.
To cross-multiply, multiply each numerator by the opposite fraction's denominator. For example, if you have \( \frac{a}{b} = \frac{c}{d} \), you would perform cross-multiplication as follows:

• Multiply \( a \) by \( d \)
• Multiply \( c \) by \( b \)

This results in the equation \( ad = bc \), which has no fractions and is often easier to manipulate further. This step was crucial in our problem to break down the fractions into a form that can be further simplified.

Isolation of Variables

The next step in solving our formula is isolating the variable we are interested in—in this case, the variable \( r \). Isolation of a variable means rearranging the equation so that the variable appears on one side of the equation by itself, with all other terms on the opposite side.
This usually involves several algebraic operations, including addition, subtraction, multiplication, and division, to "move" terms from one side of the equation to the other.
In our exercise, the variable \( r \) needs to be gathered on one side. After distributing and arranging terms:\( Er = eR + er \), the subtraction is applied to isolate all \( r \) terms on the same side like: \( Er - er = eR \).
By isolating \( r \), we create a stepping stone for the next step: factorization. It's an essential method used frequently not only in mathematics but also in solving real-world problems, making it an invaluable skill to develop.

Factorization

In algebra, factorization refers to breaking down an expression into a product of its factors. When isolating a variable like \( r \) leads to a set of terms grouped together, factorization becomes handy.
In our problem, after grouping all terms involving \( r \), we see the equation in the form \( Er - er = eR \).
Notice how both terms on the left contain \( r \). By factoring \( r \) out of these terms, the equation simplifies to:

• \( r(E - e) \)

This results in \( r \) being multiplied by \( E - e \), leaving a concise form that prepares the problem for the final operation needed to solve for \( r \).
Understanding factorization helps simplify problems and is central to solving not just algebraic expressions but also complex engineering equations efficiently.

Engineering Equations

Engineering equations often come across as complex due to their technical nature, but the fundamental mathematical methods we apply to them are very similar to basic algebraic processes. In engineering, precision and clarity are crucial.
This is why being proficient in manipulating equations is essential.
For instance, our example equation, \( \frac{E}{e} = \frac{R+r}{r} \), originates from engineering, where it could represent relationships between specific forces, resistances, or other variables in a system.

• Cross-multiplication aids in adjusting these equations into manageable forms.
• Isolating variables helps identify specific relationships.
• Factorization reduces complicated expressions into simpler forms.

This transformation process exemplifies how mathematical techniques are tailored to solve practical problems in engineering, ensuring solutions are not only accurate but also applicable.Engineers routinely solve such equations to derive parameters critical to designing and optimizing systems, making these skills indispensable.Understanding the background and meaning of these equations can provide deeper insight into how theoretical math applies to tangible engineering challenges.