Find X Rounded To One Decimal Place

Finding the value of 'x' and rounding it to one decimal place is a fundamental skill in mathematics with broad applications in various fields. Whether you're dealing with simple equations, complex algebraic expressions, or real-world problems, knowing how to isolate 'x' and present the solution accurately is crucial. This comprehensive guide will walk you through the methods, techniques, and considerations necessary to confidently find 'x' and round the result to one decimal place.

Understanding the Basics

Before diving into the methods, it’s essential to grasp the basic concepts:

• Equation: A statement that two expressions are equal. For example, 2x + 3 = 7.
• Variable: A symbol (usually a letter like 'x') that represents an unknown value.
• Constant: A fixed value that does not change. For example, in the equation 2x + 3 = 7, 3 and 7 are constants.
• Coefficient: A number multiplied by a variable. In the equation 2x + 3 = 7, 2 is the coefficient of 'x'.
• Rounding: Approximating a number to a specified number of decimal places.

Solving Linear Equations

Linear equations are the simplest form where 'x' appears to the first power. The goal is to isolate 'x' on one side of the equation.

Steps to Solve Linear Equations

1. Simplify both sides: Combine like terms and remove parentheses.
2. Isolate the term with 'x': Use addition or subtraction to move constants to the other side.
3. Solve for 'x': Divide by the coefficient of 'x'.
4. Round to one decimal place: If necessary, round the result to one decimal place.

Examples of Solving Linear Equations

Example 1: Solve 3x + 5 = 14 and round to one decimal place.

1. Subtract 5 from both sides: 3x + 5 - 5 = 14 - 5 3x = 9
2. Divide by 3: 3x / 3 = 9 / 3 x = 3
3. Since x is already a whole number, rounding to one decimal place is x = 3.0.

Example 2: Solve 4x - 7 = 2x + 3 and round to one decimal place.

1. Subtract 2x from both sides: 4x - 2x - 7 = 2x - 2x + 3 2x - 7 = 3
2. Add 7 to both sides: 2x - 7 + 7 = 3 + 7 2x = 10
3. Divide by 2: 2x / 2 = 10 / 2 x = 5
4. Since x is already a whole number, rounding to one decimal place is x = 5.0.

Example 3: Solve 5(x + 2) = 3x - 4 and round to one decimal place.

1. Expand the parentheses: 5x + 10 = 3x - 4
2. Subtract 3x from both sides: 5x - 3x + 10 = 3x - 3x - 4 2x + 10 = -4
3. Subtract 10 from both sides: 2x + 10 - 10 = -4 - 10 2x = -14
4. Divide by 2: 2x / 2 = -14 / 2 x = -7
5. Since x is already a whole number, rounding to one decimal place is x = -7.0.

Solving Quadratic Equations

Quadratic equations involve 'x' raised to the second power (x²). The standard form is ax² + bx + c = 0.

Methods to Solve Quadratic Equations

1. Factoring: Express the quadratic equation as a product of two binomials.
2. Completing the Square: Transform the equation into a perfect square trinomial.
3. Quadratic Formula: Use the formula x = (-b ± √(b² - 4ac)) / (2a) to find the solutions.

Factoring

Factoring involves expressing the quadratic equation as a product of two binomials. This method is effective when the quadratic equation can be easily factored.

Example: Solve x² - 5x + 6 = 0 and round to one decimal place.

1. Factor the quadratic equation: (x - 2)(x - 3) = 0
2. Set each factor equal to zero: x - 2 = 0 or x - 3 = 0
3. Solve for x: x = 2 or x = 3
4. Since both values are whole numbers, rounding to one decimal place gives x = 2.0 and x = 3.0.

Completing the Square

Completing the square involves transforming the equation into a perfect square trinomial. This method is useful when the quadratic equation is not easily factored.

Example: Solve x² + 6x - 7 = 0 and round to one decimal place.

1. Move the constant term to the other side: x² + 6x = 7
2. Add (b/2)² to both sides, where b is the coefficient of x (in this case, b = 6): (6/2)² = 3² = 9 x² + 6x + 9 = 7 + 9 x² + 6x + 9 = 16
3. Factor the left side as a perfect square: (x + 3)² = 16
4. Take the square root of both sides: x + 3 = ±√16 x + 3 = ±4
5. Solve for x: x = -3 + 4 or x = -3 - 4 x = 1 or x = -7
6. Since both values are whole numbers, rounding to one decimal place gives x = 1.0 and x = -7.0.

Quadratic Formula

The quadratic formula is a universal method for solving quadratic equations, regardless of whether they can be factored or easily transformed.

Formula: x = (-b ± √(b² - 4ac)) / (2a)

Example: Solve 2x² - 5x + 3 = 0 and round to one decimal place.

1. Identify a, b, and c: a = 2, b = -5, c = 3
2. Plug the values into the quadratic formula: x = (-(-5) ± √((-5)² - 4(2)(3))) / (2(2)) x = (5 ± √(25 - 24)) / 4 x = (5 ± √1) / 4 x = (5 ± 1) / 4
3. Solve for x: x = (5 + 1) / 4 or x = (5 - 1) / 4 x = 6 / 4 or x = 4 / 4 x = 1.5 or x = 1
4. Rounding to one decimal place gives x = 1.5 and x = 1.0.

Example: Solve x² + 4x + 2 = 0 and round to one decimal place.

1. Identify a, b, and c: a = 1, b = 4, c = 2
2. Plug the values into the quadratic formula: x = (-4 ± √(4² - 4(1)(2))) / (2(1)) x = (-4 ± √(16 - 8)) / 2 x = (-4 ± √8) / 2 x = (-4 ± 2√2) / 2 x = -2 ± √2
3. Solve for x: x = -2 + √2 or x = -2 - √2 x ≈ -2 + 1.414 or x ≈ -2 - 1.414 x ≈ -0.586 or x ≈ -3.414
4. Rounding to one decimal place gives x ≈ -0.6 and x ≈ -3.4.

Solving Equations with Radicals

Equations with radicals (square roots, cube roots, etc.) require isolating the radical and then raising both sides to the appropriate power to eliminate the radical.

Steps to Solve Equations with Radicals

1. Isolate the radical: Get the radical term alone on one side of the equation.
2. Raise both sides to the appropriate power: If it’s a square root, square both sides; if it’s a cube root, cube both sides, and so on.
3. Solve for 'x': Solve the resulting equation.
4. Check for extraneous solutions: Plug the solutions back into the original equation to make sure they are valid.
5. Round to one decimal place: If necessary, round the result to one decimal place.

Examples of Solving Equations with Radicals

Example 1: Solve √(2x + 5) = 3 and round to one decimal place.

1. The radical is already isolated.
2. Square both sides: (√(2x + 5))² = 3² 2x + 5 = 9
3. Solve for x: 2x = 9 - 5 2x = 4 x = 2
4. Check for extraneous solutions: √(2(2) + 5) = √(4 + 5) = √9 = 3 (valid)
5. Since x is already a whole number, rounding to one decimal place is x = 2.0.

Example 2: Solve √(3x - 2) + 4 = 7 and round to one decimal place.

1. Isolate the radical: √(3x - 2) = 7 - 4 √(3x - 2) = 3
2. Square both sides: (√(3x - 2))² = 3² 3x - 2 = 9
3. Solve for x: 3x = 9 + 2 3x = 11 x = 11 / 3 x ≈ 3.667
4. Check for extraneous solutions: √(3(11/3) - 2) + 4 = √(11 - 2) + 4 = √9 + 4 = 3 + 4 = 7 (valid)
5. Rounding to one decimal place gives x ≈ 3.7.

Example 3: Solve √(x + 1) = x - 1 and round to one decimal place.

1. The radical is already isolated.
2. Square both sides: (√(x + 1))² = (x - 1)² x + 1 = x² - 2x + 1
3. Rearrange the equation: x² - 2x + 1 - x - 1 = 0 x² - 3x = 0
4. Factor the equation: x(x - 3) = 0
5. Solve for x: x = 0 or x = 3
6. Check for extraneous solutions: For x = 0: √(0 + 1) = 0 - 1 → 1 = -1 (not valid) For x = 3: √(3 + 1) = 3 - 1 → √4 = 2 → 2 = 2 (valid)
7. Therefore, x = 3.
8. Since x is already a whole number, rounding to one decimal place is x = 3.0.

Solving Rational Equations

Rational equations involve fractions where 'x' appears in the denominator.

Steps to Solve Rational Equations

1. Find the least common denominator (LCD) of all fractions in the equation.
2. Multiply both sides of the equation by the LCD to eliminate the fractions.
3. Solve for 'x': Solve the resulting equation.
4. Check for extraneous solutions: Plug the solutions back into the original equation to make sure they are valid (i.e., they don't make any denominators zero).
5. Round to one decimal place: If necessary, round the result to one decimal place.

Examples of Solving Rational Equations

Example 1: Solve (2 / x) + (3 / 2) = 2 and round to one decimal place.

1. Find the LCD: The LCD of x and 2 is 2x.
2. Multiply both sides by the LCD: 2x * ((2 / x) + (3 / 2)) = 2x * 2 2x * (2 / x) + 2x * (3 / 2) = 4x 4 + 3x = 4x
3. Solve for x: 4 = 4x - 3x 4 = x x = 4
4. Check for extraneous solutions: (2 / 4) + (3 / 2) = 0.5 + 1.5 = 2 (valid)
5. Since x is already a whole number, rounding to one decimal place is x = 4.0.

Example 2: Solve (1 / (x - 1)) = (2 / (x + 1)) and round to one decimal place.

1. Find the LCD: The LCD of (x - 1) and (x + 1) is (x - 1)(x + 1).
2. Multiply both sides by the LCD: (x - 1)(x + 1) * (1 / (x - 1)) = (x - 1)(x + 1) * (2 / (x + 1)) (x + 1) = 2(x - 1)
3. Solve for x: x + 1 = 2x - 2 1 + 2 = 2x - x 3 = x x = 3
4. Check for extraneous solutions: (1 / (3 - 1)) = (2 / (3 + 1)) (1 / 2) = (2 / 4) 0.5 = 0.5 (valid)
5. Since x is already a whole number, rounding to one decimal place is x = 3.0.

Example 3: Solve (x / (x + 2)) + (2 / x) = 1 and round to one decimal place.

1. Find the LCD: The LCD of (x + 2) and x is x(x + 2).
2. Multiply both sides by the LCD: x(x + 2) * ((x / (x + 2)) + (2 / x)) = x(x + 2) * 1 x(x + 2) * (x / (x + 2)) + x(x + 2) * (2 / x) = x(x + 2) x² + 2(x + 2) = x² + 2x x² + 2x + 4 = x² + 2x
3. Solve for x: x² + 2x + 4 - x² - 2x = 0 4 = 0 (This is a contradiction, so there is no solution.)
4. Since there is no solution, no rounding is needed.

Solving Systems of Equations

Systems of equations involve two or more equations with two or more variables. The goal is to find the values of the variables that satisfy all equations simultaneously.

Methods to Solve Systems of Equations

1. Substitution: Solve one equation for one variable and substitute that expression into the other equation.
2. Elimination: Add or subtract the equations to eliminate one variable.

Substitution Method

Example: Solve the system of equations:

• y = 2x + 1
• 3x + y = 11

and find the value of x, rounded to one decimal place.

1. Substitute the first equation into the second equation: 3x + (2x + 1) = 11
2. Solve for x: 5x + 1 = 11 5x = 10 x = 2
3. Since x is already a whole number, rounding to one decimal place is x = 2.0.

Elimination Method

Example: Solve the system of equations:

• 2x + 3y = 8
• x - y = 1

and find the value of x, rounded to one decimal place.

1. Multiply the second equation by 3 to make the coefficients of y opposites: 3(x - y) = 3(1) 3x - 3y = 3
2. Add the modified second equation to the first equation: (2x + 3y) + (3x - 3y) = 8 + 3 5x = 11
3. Solve for x: x = 11 / 5 x = 2.2
4. Rounding to one decimal place gives x = 2.2.

Rounding to One Decimal Place

Rounding to one decimal place involves looking at the digit in the second decimal place. If this digit is 5 or greater, round the first decimal place up. If it is less than 5, leave the first decimal place as it is.

Examples of Rounding

• 3.14 rounded to one decimal place is 3.1.
• 3.15 rounded to one decimal place is 3.2.
• -2.78 rounded to one decimal place is -2.8.
• -2.72 rounded to one decimal place is -2.7.
• 5.00 rounded to one decimal place is 5.0.
• 5 rounded to one decimal place is 5.0.

Conclusion

Finding the value of 'x' and rounding it to one decimal place is a fundamental mathematical skill with various applications. By understanding the different types of equations and the appropriate methods to solve them, you can confidently tackle any problem. Whether it's linear, quadratic, radical, or rational equations, the key is to isolate 'x' and then apply the rules of rounding accurately. Always remember to check for extraneous solutions in radical and rational equations to ensure the validity of your results. With practice and a clear understanding of these concepts, you'll master the art of finding 'x' and presenting your solutions with precision.