Problem 50
Question
Solve the equation. Round your answer to two decimal places. $$7.2 x-4.7=62.3$$
Step-by-Step Solution
Verified Answer
The solution to the equation \(7.2 x-4.7=62.3\) (rounded to two decimal places) is approximately \(x = 9.31\).
1Step 1: Move the constant to the right-hand side
Firstly, add 4.7 to both sides of the equation to isolate the x term on one side. This will result in \(7.2x = 62.3 + 4.7\).
2Step 2: Simplify the right-hand side
Then simplify the right-hand side by adding 62.3 and 4.7, which gives us \(7.2x = 67.0\).
3Step 3: Isolate x
Next, divide both sides of the equation by 7.2 to get \(x\). This leaves us with \(x = 67.0 / 7.2\).
4Step 4: Solve for x
When you divide 67.0 by 7.2, rounding to two decimal places, you reach the solution for \(x\).
Key Concepts
Algebraic EquationsIsolate VariablesRounding Decimal Places
Algebraic Equations
Algebraic equations are mathematical statements that express the equality of two algebraic expressions. These expressions can include variables, coefficients, and constants. The goal in solving an algebraic equation is to find the value or values of the variables that make the equation true.
Take the equation from our exercise, for instance: \(7.2x - 4.7 = 62.3\). Here, \(x\) is the variable, \(7.2\) and \( -4.7\) are coefficients, and \(62.3\) is a constant. Solving the equation means finding the value of \(x\) that, when substituted into the equation, satisfies the equality. Administrators and educators, who develop curricula for mathematics education, emphasize the importance of understanding algebraic equations because they form the foundation for more advanced topics in mathematics and are widely used in various applications, such as science, engineering, and economics.
Take the equation from our exercise, for instance: \(7.2x - 4.7 = 62.3\). Here, \(x\) is the variable, \(7.2\) and \( -4.7\) are coefficients, and \(62.3\) is a constant. Solving the equation means finding the value of \(x\) that, when substituted into the equation, satisfies the equality. Administrators and educators, who develop curricula for mathematics education, emphasize the importance of understanding algebraic equations because they form the foundation for more advanced topics in mathematics and are widely used in various applications, such as science, engineering, and economics.
Isolate Variables
To isolate a variable means to manipulate an equation such that the variable we want to solve for is on one side by itself. This is typically done by using operations that reverse the effect of whatever is being done to the variable. For instance, if a variable is being multiplied by a number, we can isolate it by dividing by that number.
In our exercise, we used two steps to isolate \(x\).
In our exercise, we used two steps to isolate \(x\).
Adding to Both Sides
We first added \(4.7\) to both sides, which effectively moved the constant to the other side, simplifying the equation to \(7.2x = 67.0\).Dividing Both Sides
Next, we divided both sides by \(7.2\), which is the coefficient of \(x\), eventually leaving \(x\) isolated on one side of the equation. The ability to isolate variables effectively is fundamental for problem-solving across all STEM (science, technology, engineering, mathematics) fields.Rounding Decimal Places
Rounding decimal places is the process of adjusting a number to the nearest value based on the number of decimal places specified. When solving problems in mathematics, physics, or finance, precise values are important, but sometimes, we approximate these values for simplicity or because the situation allows for a margin of error.
In our step-by-step solution, once we isolated \(x\), we needed to divide \(67.0\) by \(7.2\). Instead of working with an overly precise and lengthy decimal, we round the result to two decimal places for a more practical answer. For example, if the raw division gave us \(9.305555\), we would look at the third decimal place, which is \(5\). Since \(5\) is equal to or greater than \(5\), we round up, giving \(x = 9.31\). If the third decimal place had been less than \(5\), we would round down, and our answer would remain \(x = 9.30\). Learning to round efficiently is important in real-world applications where exact values are not always necessary or feasible, such as engineering tolerances, statistical analysis, or financial forecasting.
In our step-by-step solution, once we isolated \(x\), we needed to divide \(67.0\) by \(7.2\). Instead of working with an overly precise and lengthy decimal, we round the result to two decimal places for a more practical answer. For example, if the raw division gave us \(9.305555\), we would look at the third decimal place, which is \(5\). Since \(5\) is equal to or greater than \(5\), we round up, giving \(x = 9.31\). If the third decimal place had been less than \(5\), we would round down, and our answer would remain \(x = 9.30\). Learning to round efficiently is important in real-world applications where exact values are not always necessary or feasible, such as engineering tolerances, statistical analysis, or financial forecasting.
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Problem 50
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