Problem 32 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation analytically. Check it analytically, and then support the solution graphically. $$1.1 x-2.5=0.3(x-2)$$

Step-by-Step Solution

Answer

The solution is \(x = 2.375\). Verified analytically and graphically.

Step 1: Expand the Right Side

First, expand the expression on the right side of the equation: \(0.3(x-2)\). Distribute the \(0.3\) to both terms inside the parentheses to get \(0.3x - 0.6\). The equation now becomes \(1.1x - 2.5 = 0.3x - 0.6\).

Step 2: Move All Terms Involving x to One Side

Subtract \(0.3x\) from both sides of the equation to get all terms involving \(x\) on one side: \(1.1x - 0.3x - 2.5 = -0.6\). This simplifies to \(0.8x - 2.5 = -0.6\).

Step 3: Isolate the Term Involving x

Add \(2.5\) to both sides of the equation to isolate the term involving \(x\): \(0.8x = 1.9\).

Step 4: Solve for x

Divide both sides of the equation by \(0.8\) to solve for \(x\): \(x = \frac{1.9}{0.8}\). Simplifying this gives \(x = 2.375\).

Step 5: Check the Solution Analytically

Substitute \(x = 2.375\) back into the original equation to verify the solution. Calculate both sides: Left side is \(1.1 \times 2.375 - 2.5 = 0.1125\) and the Right side is \(0.3(2.375 - 2) = 0.1125\). Both sides are equal, confirming \(x = 2.375\) is correct.

Step 6: Support the Solution Graphically

Graph the equations \(y = 1.1x - 2.5\) and \(y = 0.3(x-2)\). The intersection of the two lines occurs at the point \((2.375, 0.1125)\), verifying the solution \(x = 2.375\).

Key Concepts

Graphical Solutions

When solving linear equations, graphical solutions provide a visual representation of the problem. By graphing the two equations in the given problem, you can find the solution by identifying where the two graphs intersect. To graph the given equation, say we want to look at the equations \(y = 1.1x - 2.5\) and \(y = 0.3(x-2)\). Both of these lines can be plotted on a coordinate axis.

• The line from \(y = 1.1x - 2.5\) represents an incline that increases as \(x\) increases.
• The line \(y = 0.3(x-2)\) starts at \(y = -0.6\) when \(x = 0\) and also rises, but with a gentler slope.

Where these two lines meet is the solution to the equation. In this example, the intersection is at \((2.375, 0.1125)\). This confirms that \(x = 2.375\) is the solution, because at this \(x\)-value, both equations give the same \(y\)-value. Graphs are a nice way to support analytical solutions visually, adding another layer of comprehension to problem-solving.

Analytical Solutions

Analytical solutions involve manipulating algebraic expressions to find the value of variables. Here, we have used algebraic techniques to find \(x\) from the equation \(1.1x - 2.5 = 0.3(x-2)\). An essential first step is to simplify the equations. This often involves distributing expressions and combining like terms.

• First, distribute \(0.3\) across \(x-2\), giving \(0.3x - 0.6\).
• This converts the equation to \(1.1x - 2.5 = 0.3x - 0.6\).

By moving terms with \(x\) to one side, and constants on the other side, we simplify further to \(0.8x = 1.9\). Dividing both sides by \(0.8\) gives \(x = 2.375\). Analytical solutions follow systematic steps to ensure accuracy. Each step builds on the previous one until the value you need is isolated and found.

Solving Equations

Solving linear equations requires understanding both the properties of linearity and methods to isolate variables. Start by expanding any expressions and simplifying the equation through addition or subtraction of terms. Linear equations typically form a straight line when graphed, indicating they have constant rates of change.

• Arrange terms so that all variables are on one side and numbers on the other.
• Simplify the equation step by step, maintaining balance by performing equivalent operations on both sides.

For our exercise, we used these steps to find \(x = 2.375\). It's crucial to check your solutions by substituting back into the original equation. This ensures that no mistakes were made during simplification or when handling operations. By understanding these principles, you gain a solid foundation for tackling more complex algebraic problems.