Problem 12
Question
Determine whether each value of \(x\) is a solution of the equation. Equation $$ 3+\frac{1}{x+2}=4 $$ Values (a) \(x=-1\) (b) \(x=-2\) (c) \(x=0\) (d) \(x=5\)
Step-by-Step Solution
Verified Answer
The solution is that only \(x = -1\) makes the equation true.
1Step 1: Substituting \(x = -1\)
The original equation is \(3 + \frac{1}{x+2} = 4\). Substituting \(x = -1\) into the equation gives \(3 + \frac{1}{-1+2} = 4\). Simplifying this results in \(3 + 1 = 4\), which is true. Therefore, \(x = -1\) is a solution to the equation.
2Step 2: Substituting \(x = -2\)
Substitute \(x = -2\) into the equation. This results in \(3 + \frac{1}{-2 +2} = 4\). Simplifying this, an undefined operation (division by zero) arises, so \(x = -2\) is not a solution of the equation.
3Step 3: Substituting \(x = 0\)
Substitute \(x = 0\) into the equation. This gives \(3 + \frac{1}{0+2} = 4\), which simplifies to \(3 + \frac{1}{2} = 4\). Simplifying further gives \(3.5 = 4\), which is not true. Therefore, \(x = 0\) is not a solution.
4Step 4: Substituting \(x = 5\)
Substitute \(x = 5\) into the equation. This results in \(3 + \frac{1}{5+2} = 4\), which simplifies to \(3 + \frac{1}{7} = 4\). Simplifying further gives \(3.14\... \neq 4\), which is not true. Therefore, \(x = 5\) is not a solution.
Key Concepts
Substitution MethodSolutions of EquationsUndefined OperationsSimplifying Expressions
Substitution Method
The substitution method is a cornerstone technique in algebra that allows you to determine if a particular value is a solution to an equation. In practice, this method involves replacing the variable in the equation with a given numerical value and simplifying the expression to see if the equation holds true.
For instance, consider the equation \(3 + \frac{1}{x+2} = 4\). To investigate whether \(x = -1\) is a solution, you substitute \(x\) with -1 which transforms the equation into \(3 + \frac{1}{-1+2} = 4\). Simplifying the right side gives you \(3 + 1 = 4\), which verifies that \(x = -1\) is indeed a solution. The strength of the substitution method lies in its straightforward nature – it provides a clear and direct way to validate solutions of equations.
For instance, consider the equation \(3 + \frac{1}{x+2} = 4\). To investigate whether \(x = -1\) is a solution, you substitute \(x\) with -1 which transforms the equation into \(3 + \frac{1}{-1+2} = 4\). Simplifying the right side gives you \(3 + 1 = 4\), which verifies that \(x = -1\) is indeed a solution. The strength of the substitution method lies in its straightforward nature – it provides a clear and direct way to validate solutions of equations.
Solutions of Equations
Finding the solutions of equations is central to algebra. A solution is a value (or set of values) that makes an equation true when substituted for the unknown variable. It is essential to remember that not every value will satisfy a given equation, and an equation might have one, multiple, or no solutions at all.
For the equation \(3 + \frac{1}{x+2} = 4\), we look for values of \(x\) that make the equation accurate. Evaluating each option separately, we found that \(x = -1\) is a valid solution while \(x = -2\), \(x = 0\), and \(x = 5\) are not. This could be due to the equation not being balanced with these values or because they result in undefined operations.
For the equation \(3 + \frac{1}{x+2} = 4\), we look for values of \(x\) that make the equation accurate. Evaluating each option separately, we found that \(x = -1\) is a valid solution while \(x = -2\), \(x = 0\), and \(x = 5\) are not. This could be due to the equation not being balanced with these values or because they result in undefined operations.
Undefined Operations
In algebra, certain operations are undefined. For example, division by zero is one such operation that has no meaning in the realm of arithmetic and is thus considered undefined. If the substitution yields a situation where you are asked to divide by zero, the operation is not valid, and consequently, the value is not a solution to your equation.
Referring to the original equation \(3 + \frac{1}{x+2} = 4\) and substituting \(x = -2\), we're led to a division by zero: \(3 + \frac{1}{-2+2} = 4\). In this case, the fraction becomes \(\frac{1}{0}\) which is not defined, meaning that \(x = -2\) does not provide a legitimate solution. Being aware of undefined operations is crucial when determining the validity of potential solutions.
Referring to the original equation \(3 + \frac{1}{x+2} = 4\) and substituting \(x = -2\), we're led to a division by zero: \(3 + \frac{1}{-2+2} = 4\). In this case, the fraction becomes \(\frac{1}{0}\) which is not defined, meaning that \(x = -2\) does not provide a legitimate solution. Being aware of undefined operations is crucial when determining the validity of potential solutions.
Simplifying Expressions
Simplifying expressions is a key process in algebra. It involves combining like terms and performing arithmetic operations to rewrite expressions in their most basic form. Simplification helps in understanding the structure of an expression and in discovering its properties.
With our initial equation, after substituting variables like \(x = 0\) and \(x = 5\), we get expressions like \(3 + \frac{1}{2}\) and \(3 + \frac{1}{7}\), respectively. Simplifying these expressions gives \(3.5\) and approximately \(3.14\), clearly neither of which equals to 4. Hence, neither \(x = 0\) nor \(x = 5\) are solutions. Simplifying expressions is a fundamental skill that confirms whether the algebraic manipulations carried out when applying the substitution method meet the conditions of the given equation.
With our initial equation, after substituting variables like \(x = 0\) and \(x = 5\), we get expressions like \(3 + \frac{1}{2}\) and \(3 + \frac{1}{7}\), respectively. Simplifying these expressions gives \(3.5\) and approximately \(3.14\), clearly neither of which equals to 4. Hence, neither \(x = 0\) nor \(x = 5\) are solutions. Simplifying expressions is a fundamental skill that confirms whether the algebraic manipulations carried out when applying the substitution method meet the conditions of the given equation.
Other exercises in this chapter
Problem 12
Solve the quadratic equation by factoring. $$ x^{2}-10 x+9=0 $$
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Write an equation that represents the statement. The difference of \(n\) and 7 is 4 .
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Solve the inequality. Then graph the solution set on the real number line. \(x^{2}+4 x+4 \geq 9\)
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Find the real solution(s) of the polynomial equation. Check your solutions. \(x^{4}-12 x^{2}+11=0\)
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