Problem 15
Question
Solve and check each equation. Treat the constants in these equations as exact numbers. Leave your answers in fractional, rather than decimal, form. $$7 x-3=5 x+1$$
Step-by-Step Solution
Verified Answer
The solution to the equation \(7x - 3 = 5x + 1\) is \(x = 2/1\), which is confirmed by a check.
1Step 1: Isolate the variable on one side
Begin by moving all terms containing the variable to one side of the equation and constants to the other side. To do this, subtract 5x from both sides of the equation: \(7x - 5x - 3 = 5x - 5x + 1\). This simplifies to \(2x - 3 = 1\).
2Step 2: Isolate the variable
Now, to solve for x, you'll want to get x alone on one side of the equation. To isolate x, add 3 to both sides of the equation: \(2x - 3 + 3 = 1 + 3\). This simplifies to \(2x = 4\).
3Step 3: Solve for the variable
To find the value of x, divide both sides of the equation by 2: \(2x / 2 = 4 / 2\). This gives you \(x = 2\). Since the question asks to leave the answer in fractional form, the result is the same because \(2\) is an integer and can be expressed as a fraction with a denominator of \(1\): therefore, \(x = 2/1\).
4Step 4: Check the solution
To ensure \(x = 2/1\) is the correct solution, plug it back into the original equation: \(7(2/1) - 3 =? 5(2/1) + 1\). Simplifying both sides gives \(14 - 3 = 10 + 1\). This results in \(11 = 11\), which is a true statement, confirming our solution is correct.
Key Concepts
Isolate the VariableEquation Solving Step by StepChecking SolutionsFractional Answers
Isolate the Variable
In solving linear equations, the essential first step is to isolate the variable. This means that you want the variable on one side of the equation and all the constants on the other. This is crucial because it simplifies the equation, making it manageable to find the value of the unknown variable.
For instance, let's take the equation from our exercise \(7x - 3 = 5x + 1\). To isolate the variable \(x\), we look at moving terms containing \(x\) to one side. By subtracting \(5x\) from both sides, we get \(7x - 5x - 3 = 5x - 5x + 1\), which simplifies to \(2x - 3 = 1\). Our variable \(x\) is now clearer, and we're a step closer to finding its value.
Through this method, we effectively make our equation look simpler and set the stage to perform the next steps of solving the equation.
For instance, let's take the equation from our exercise \(7x - 3 = 5x + 1\). To isolate the variable \(x\), we look at moving terms containing \(x\) to one side. By subtracting \(5x\) from both sides, we get \(7x - 5x - 3 = 5x - 5x + 1\), which simplifies to \(2x - 3 = 1\). Our variable \(x\) is now clearer, and we're a step closer to finding its value.
Through this method, we effectively make our equation look simpler and set the stage to perform the next steps of solving the equation.
Equation Solving Step by Step
Having isolated the variable, we move on to solve the equation step by step. The goal is to perform the same operations on both sides of the equation until the variable is by itself.
Continuing with our example, we start with the simplified equation \(2x - 3 = 1\). To get \(x\) by itself, we add \(3\) to both sides leading to \(2x - 3 + 3 = 1 + 3\), resulting in \(2x = 4\). The variable \(x\) is almost isolated, but we need it to stand alone. Therefore, we divide both sides of the equation by \(2\) yielding \(x = 4/2\).
It's like putting together a puzzle – each piece must go in the right place, at the right time. If we 'jump ahead', or forget a piece, our solution won't fit the original problem.
Continuing with our example, we start with the simplified equation \(2x - 3 = 1\). To get \(x\) by itself, we add \(3\) to both sides leading to \(2x - 3 + 3 = 1 + 3\), resulting in \(2x = 4\). The variable \(x\) is almost isolated, but we need it to stand alone. Therefore, we divide both sides of the equation by \(2\) yielding \(x = 4/2\).
It's like putting together a puzzle – each piece must go in the right place, at the right time. If we 'jump ahead', or forget a piece, our solution won't fit the original problem.
Checking Solutions
After obtaining a potential solution for the variable, the final verification step is to check if this solution actually satisfies the original equation – a critical part of the problem-solving process.
In our current problem, we found that \(x = 2\), or in fractional form, \(x = 2/1\). To check, we substitute \(x\) back into the initial equation \(7x - 3\) and compare it with the other side \(5x + 1\). If both sides equal after the substitution, our solution is correct. With \(x = 2/1\), we get \(14 - 3\) on the left side and \(10 + 1\) on the right, which simplifies to \(11 = 11\), a true statement indicating our solution is indeed correct.
This crucial step ensures we've not made any mistakes along the way, and gives us confidence that our solution is the right fit for the original equation.
In our current problem, we found that \(x = 2\), or in fractional form, \(x = 2/1\). To check, we substitute \(x\) back into the initial equation \(7x - 3\) and compare it with the other side \(5x + 1\). If both sides equal after the substitution, our solution is correct. With \(x = 2/1\), we get \(14 - 3\) on the left side and \(10 + 1\) on the right, which simplifies to \(11 = 11\), a true statement indicating our solution is indeed correct.
This crucial step ensures we've not made any mistakes along the way, and gives us confidence that our solution is the right fit for the original equation.
Fractional Answers
Mathematical truth can be expressed in many forms, and it's important to present our answers in the form requested. When a problem asks for solutions in fractional form rather than decimal, it is typically to encourage an exact representation of the answer.
In problems like \(7x - 3 = 5x + 1\), when we solve for \(x\) and get an integer like \(2\), it's easy to overlook the request for a fractional answer. However, we must remember that any whole number can be expressed as a fraction with a denominator of \(1\), making our final answer for \(x\) as \(2/1\). This reinforces our understanding of fractions and integers and ensures we comply with the instructions provided.
Keeping answers in fractional form can also be beneficial when dealing with more complex equations, as it allows for precise calculations and reduces rounding errors that can occur with decimal approximations. By practicing this, students get used to working with fractions, an invaluable skill in higher mathematics.
In problems like \(7x - 3 = 5x + 1\), when we solve for \(x\) and get an integer like \(2\), it's easy to overlook the request for a fractional answer. However, we must remember that any whole number can be expressed as a fraction with a denominator of \(1\), making our final answer for \(x\) as \(2/1\). This reinforces our understanding of fractions and integers and ensures we comply with the instructions provided.
Keeping answers in fractional form can also be beneficial when dealing with more complex equations, as it allows for precise calculations and reduces rounding errors that can occur with decimal approximations. By practicing this, students get used to working with fractions, an invaluable skill in higher mathematics.
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