Problem 40

Question

Solve \(4.5-7.2 x=3.4 x-49.5 .\) Round to the nearest tenth. You can multiply an equation with decimal coefficients by a power of ten to get an equivalent equation with integer coefficients. Multiply each side of this equation by 10 to rewrite the equation without decimals. $$ \begin{aligned} 4.5-7.2 x &=3.4 x-49.5 \\ 10(4.5-7.2 x) &=10(3.4 x-49.5) \\ 45-72 x &=34 x-495 \\ 45 &=106 x-495 \\ 540 &=106 x \\ \frac{540}{106} &=x \\ 5.094339623 & \approx x \\ 5.1 & \approx x \end{aligned} $$ The solution is approximately 5.1. Check this in the original equation. Solve the equation. Round to the nearest tenth. $$ 1.1 x+3.2=0.2 x-1.4 $$

Step-by-Step Solution

Verified
Answer
The solution to the equation \(4.5-7.2x = 3.4x - 49.5\) is \(x \approx 5.1\), rounded to the nearest tenth.
1Step 1: Rewrite The Equation
First step is to remove the decimal numbers by multiplying both sides of the equation by 10. This leads to the new equation: \(45 - 72x = 34x - 495\)
2Step 2: Rearrange The Equation
Next step is to isolate the variable 'x' by moving all 'x' terms to one side and constant terms to the other side. This results in: \(45 + 495 = 72x + 34x\), which simplifies to \(540 = 106x\)
3Step 3: Solve For 'x'
Solve for the variable 'x' by dividing both sides of the equation by 106: \(x = 540 / 106\)
4Step 4: Round to the Nearest Tenth
Finally, round the number to the nearest tenth. Hence, \(x \approx 5.1\)

Key Concepts

Understanding Decimal CoefficientsWorking with Integer CoefficientsIsolating Variables
Understanding Decimal Coefficients
When solving equations, decimal coefficients can make calculations more complex. Decimals can be unwieldy, but there's a neat trick to simplify them. By multiplying both sides of an equation by 10 (or a higher power of 10), you convert these decimals into integers.
This gets rid of the decimal places and makes the math easier to handle. For example, in the equation \(4.5 - 7.2x = 3.4x - 49.5\), any decimal can be eliminated by multiplying each term by 10.
  • This transforms the equation into \(45 - 72x = 34x - 495\), now free from decimals.
  • The multiplication does not change the equation's solutions; it only simplifies the working process.
  • This method is particularly useful when dealing with large systems of equations, as it reduces errors and simplifies calculations.
Deciding a multiplication factor depends on the number of decimal places you need to clear. Every decimal digit indicates another zero for your multiplying factor.
Working with Integer Coefficients
Once you've eliminated decimals, you work with integer coefficients. This simplifies calculations significantly, making addition, subtraction, multiplication, and division much easier to manage.
In our transformed equation, \(45 - 72x = 34x - 495\), we see integer coefficients in action. Working with integers:
  • Makes manual calculations faster, especially useful if you're working without a calculator.
  • Improves accuracy since operations on whole numbers are less prone to rounding errors compared to decimals.
  • Facilitates easier identification of potential errors in your steps.
In this state, you can rearrange, combine, or factor the terms as necessary, leading us smoothly towards isolating the variable.
Isolating Variables
Isolating the variable in an equation involves separating the variable from all other numbers. It's like dissecting the equation to extract the unknown part you need to solve.
After converting to integer coefficients, isolate \(x\) by following these steps:
  • Combine all like terms involving \(x\) on one side, and constants on the other. For the equation \(45 - 72x + 495 = 34x\), reorganize to \(540 = 106x\).
  • Notice how constants and variables are segregated. This makes the next step of division straightforward.
  • Divide both sides by 106, the coefficient of \(x\), to solve for \(x\). This finds the value of \(x\) as \(x = \frac{540}{106}\), approximately \(5.1\).
Isolating variables is crucial because it simplifies down to a simple arithmetic operation, uncovering the variable's value effectively and efficiently. Remember to check your solution with the original equation to ensure everything balances!