Problem 55 - Elementary and Intermediate Algebra

Question

Solve for the specified variable or expression. $$ 2 g=c h+d h \text { for } h $$

Step-by-Step Solution

Answer

h = \frac{2g}{c + d}

Step 1: Isolate the Expression Involving h

Start by grouping the terms involving \(h\) on one side of the equation. The equation is \(2g = ch + dh\). You can factor \(h\) from the right side of the equation: \(2g = h(c + d)\).

Step 2: Solve for h

To solve for \(h\), divide both sides of the equation by \((c + d)\) to isolate \(h\): \(h = \frac{2g}{c + d}\).

Step 3: Verify the Solution

To ensure the solution is correct, plug \(h = \frac{2g}{c + d}\) back into the original equation and verify both sides are equal. Substitute \(h\) in the expression \(ch + dh\), which simplifies to \( (c+d) \times \frac{2g}{c+d} = 2g\). This confirms that both sides are equal and the solution is correct.

Key Concepts

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations to make solving them easier. In this exercise, we began with the equation \( 2g = ch + dh \). The goal was to solve for \( h \), which means we needed to isolate \( h \) on one side of the equation. To do so, we noticed that both \( ch \) and \( dh \) share the common variable \( h \). This allowed us to factor out \( h \), giving us the equation \( 2g = h(c + d) \). This step is a classic example of using algebraic manipulation to simplify the problem. By factoring, we're combining terms with a shared factor to make the equation easier to solve. Here are some general techniques used in algebraic manipulation:

• Combining like terms to simplify expressions.
• Factoring to reveal common elements, as seen in the problem.
• Expanding expressions or distributing terms when necessary.
• Rearranging terms to isolate variables or expressions of interest.

These techniques not only speed up the solving process but also make it easier to understand complex mathematical relationships.

Isolating Variables

Isolating a variable means getting the variable alone on one side of the equation, leading to a direct solution. In our exercise, after factoring out \( h \), we had the equation \( 2g = h(c + d) \). The next logical step to isolate \( h \) is to divide both sides of the equation by \( (c + d) \). This operation effectively undoes the multiplication of \( h \) by \( (c + d) \), resulting in\[ h = \frac{2g}{c + d} \]. Isolating the variable helps to express the unknown directly without surrounding distractions. This is particularly helpful in algebra where solving for one variable helps in solving multi-variable systems. When isolating variables:

• Perform operations that will leave the variable alone on one side.
• Ensure that any arithmetic operations you apply are reversible (e.g., addition and subtraction, multiplication and division).
• Maintain balance by applying operations equally to both sides of the equation.

Once isolated, the variable provides a clear view of its relationship with other elements in the equation.

Verifying Solutions

Verifying solutions is a crucial step in solving equations. It involves substituting the solution back into the original equation to ensure it satisfies the equation completely. In our example, the solution found was \( h = \frac{2g}{c + d} \). By substituting \( h \) back into the original equation \( 2g = ch + dh \), we replace \( h \) with \( \frac{2g}{c + d} \), making the expression \((c + d) \times \frac{2g}{c + d} \). Simplifying this expression gives us back \( 2g \), showing both sides are equal. This confirms the validity of our solution. Verification is vital because it:

• Ensures your solution is accurate and reliable.
• Identifies errors in earlier steps if the solution doesn't check out.
• Strengthens understanding and confidence in the method used to solve.

Always remember, a solution isn't fully complete until it has been verified to satisfy the original equation!