Problem 19 - Prealgebra

Question

Solve each equation. Check your solution. $$\frac{h}{-7}=20$$

Step-by-Step Solution

Answer

The solution is \( h = -140 \).

Step 1: Isolate the Variable

Start by isolating the variable \( h \) in the equation \( \frac{h}{-7} = 20 \). To do this, multiply both sides of the equation by \(-7\):\[\frac{h}{-7} \times (-7) = 20 \times (-7)\]This simplifies to: \[h = -140\]

Step 2: Check the Solution

Now, substitute \( h = -140 \) back into the original equation to ensure it satisfies \( \frac{h}{-7} = 20 \).\[\frac{-140}{-7} = 20\]Calculate the left-hand side:\[\frac{-140}{-7} = 20\]Since both sides are equal, \( h = -140 \) is indeed the correct solution.

Key Concepts

Isolate the Variable

When solving linear equations, your main goal is to "isolate the variable." This means you want the variable you are solving for to be on one side of the equation all by itself. Imagine it as untangling a knot. You want the variable alone to clearly see its value.
In the equation \( \frac{h}{-7} = 20 \), our task is to isolate \( h \). To do this, consider what operation is currently affecting \( h \). Here, \( h \) is divided by \(-7\). To undo this operation, perform the opposite action – in this case, multiply both sides of the equation by \(-7\).
This step would look like:\[\frac{h}{-7} \times (-7) = 20 \times (-7)\]By multiplying, the \(-7\) on the left side cancels out and isolates \( h \):\[h = -140\]
Thus, isolating the variable allows us to directly determine its value.

Check the Solution

Once you have found a potential solution, it is crucial to "check the solution" to ensure it is correct. This process involves plugging your found value back into the original equation to verify your work. Think of this as proofreading your own writing; you're ensuring everything makes sense and there are no errors.
In our problem, we found \( h = -140 \). Plug \( h = -140 \) back into the original equation \( \frac{h}{-7} = 20 \):\[\frac{-140}{-7} = 20\]
Perform the division on the left-hand side:\[20 = 20\]
Both sides are equal, confirming that \( h = -140 \) is indeed the solution. Checking the solution serves as a confirmation step, catching any potential mistakes before they become problems later on.

Multiplying Equations

Sometimes, when solving for a variable, you may need to use "multiplying equations" to simplify the steps. This involves multiplying both sides of an equation by the same number, either to eliminate fractions or to isolate the variable, as we saw earlier.
In the equation \( \frac{h}{-7} = 20 \), multiplying each side by \(-7\) was necessary to remove the fraction and bring \( h \) out of the denominator. It's important to multiply *both* sides to maintain the balance of the equation, much like balancing a seesaw.
Performing the multiplication for this problem yields:\[\frac{h}{-7} \times (-7) = 20 \times (-7)\]
This step helps simplify the equation to an easier, more workable form. Remember, the operation used must "cancel" or "undo" the undesired parts of the equation, bringing you closer to the solution.