Find the Value of x and Round to the Nearest Tenth: A Step‑by‑Step Guide
When solving algebraic problems, you often need to determine the unknown variable x and then present your answer with a specific level of precision. Worth adding: one common requirement is to find the value of x round the nearest tenth. This instruction appears in textbooks, standardized tests, and real‑world applications where measurements are reported to one decimal place. Because of that, mastering this skill not only improves accuracy but also builds confidence when interpreting results. Below, you will find a thorough explanation of the concept, a clear procedure, worked‑out examples, common pitfalls, and practice opportunities to reinforce your understanding.
Understanding Rounding to the Nearest Tenth
Before diving into solving for x, it helps to revisit what “round to the nearest tenth” means.
- The tenths place is the first digit to the right of the decimal point (e.g., in 3. 4 7, the 4 is in the tenths place).
- To round a number to the nearest tenth, look at the digit in the hundredths place (the second digit right of the decimal).
- If that digit is 0‑4, keep the tenths digit unchanged and drop all following digits.
- If that digit is 5‑9, increase the tenths digit by one and drop all following digits.
Example: 5. 63 → hundredths digit is 3 (<5) → round to 5.6.
Example: 7. 89 → hundredths digit is 9 (≥5) → round to 7.9.
Rounding is essential when the exact solution contains many decimal places, but the context only justifies one‑decimal precision (e.Which means g. , length measurements, financial figures, or scientific data).
General Procedure to Find x and Round the Nearest Tenth
Regardless of the equation type, follow these steps:
-
Isolate the variable
Use inverse operations (addition, subtraction, multiplication, division, factoring, etc.) to get x by itself on one side of the equation It's one of those things that adds up. Worth knowing.. -
Solve for the exact value
Perform the necessary arithmetic. If the solution involves radicals, fractions, or trigonometric functions, keep the exact form for now. -
Convert to a decimal (if needed)
Use a calculator or long division to obtain a decimal representation with at least three decimal places. This ensures you can see the hundredths digit clearly for rounding. -
Apply rounding rules
Examine the hundredths digit and round the tenths digit accordingly. -
State the final answer
Write the rounded value with the appropriate units or context, if any.
Worked‑Out Examples
Example 1: Linear Equation
Problem: Solve (3x - 7 = 11) and round x to the nearest tenth Worth keeping that in mind..
Solution
- Add 7 to both sides: (3x = 18).
- Divide by 3: (x = 6).
- As a decimal, (x = 6.0).
- Hundredths digit is 0 → no change.
Answer: (x = 6.0).
Example 2: Quadratic Equation (Irrational Root)
Problem: Solve (x^2 - 5x + 6 = 0) and round the larger root to the nearest tenth.
Solution
- Factor: ((x-2)(x-3)=0) → roots (x=2) and (x=3).
- Larger root is (x=3).
- Decimal form: (3.0).
- Hundredths digit = 0 → stays 3.0.
Answer: (x = 3.0).
Note: If the quadratic did not factor nicely, you would use the quadratic formula and then round Surprisingly effective..
Example 3: Quadratic Equation Requiring the Formula
Problem: Solve (2x^2 + 4x - 1 = 0) and round the positive root to the nearest tenth.
Solution
- Identify (a=2), (b=4), (c=-1).
- Quadratic formula: (x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}).
- Discriminant: (b^2-4ac = 16 - (-8) = 24).
- (\sqrt{24} \approx 4.898979).
- Positive root:
[ x = \frac{-4 + 4.898979}{4} = \frac{0.898979}{4} \approx 0.22474475. ] - Look at hundredths digit: 2 (in 0.2247…) → less than 5, so tenths digit stays 2.
- Rounded value: (0.2).
Answer: (x \approx 0.2).
Example 4: Trigonometric Equation
Problem: Solve (\sin(x) = 0.6) for (0^\circ \le x \le 180^\circ) and round x to the nearest tenth of a degree.
Solution
- Use inverse sine: (x = \sin^{-1}(0.6)).
- Calculator gives (x \approx 36.86989765^\circ).
- Hundredths digit is 6 (≥5) → increase tenths digit from 8 to 9.
- Rounded: (36.9^\circ).
Answer: (x \approx 36.9^\circ).
Example 5: Word Problem (Geometry)
Problem: A rectangular garden has a length that is 3 meters more than twice its width. If the perimeter is 54 meters, find the width and round to the nearest tenth of a meter.
Solution
- Let width = (w). Then length = (2w + 3).
- Perimeter formula: (2(\text{length} + \text{width}) = 54).
- Substitute: (2((2w+3) + w) = 54) → (2(3w+3) = 54).
- Simplify: (6w + 6 = 54) → (6w = 48) → (w = 8).
- Decimal: (8.0). Hundredths digit = 0 → stays 8.0.
Answer: Width = (8.0) m.