Problem 58
Question
Classify each equation as a contradiction, an identity, or a conditional equation. Give the solution set. Use a graph or table to support your answer. $$-3[-5-(-9+2 x)]=2(3 x-1)$$
Step-by-Step Solution
Verified Answer
The equation is a contradiction with no solution (empty set).
1Step 1: Simplify Both Sides of the Equation
The given equation is \[-3[-5-(-9+2x)]=2(3x-1)\].First, simplify the left side of the equation by distributing the -3 inside the brackets:Inside the brackets: \[-5-(-9+2x) \] simplifies to \[-5+9-2x\] which equals \[4-2x\].Now, distribute the -3: \[-3(4-2x) = -12+6x\].The right side is simplified by distributing 2: \[2(3x-1) = 6x-2\].So, the equation becomes \[ -12+6x = 6x - 2 \].
2Step 2: Evaluate the Equation
To evaluate, compare the simplified expressions:From the previous step, the equation is \[-12 + 6x = 6x - 2\].Cancel out the \(6x\) terms on both sides:\(-12 = -2\).Since this statement is false, it indicates that there is no possible value for \(x\) that can satisfy the equation.
3Step 3: Classify the Equation
Since the result \(-12 = -2\) is a false statement, the equation has no solutions no matter what value x takes. This makes the equation a contradiction, as it is impossible for any x to satisfy the equation. The solution set is therefore the empty set.
4Step 4: Graphical or Tabular Support
We can attempt to graph the expressions of both sides of the equation to show there are no values of \(x\) satisfying both simultaneously.On graphing, \(-12 + 6x\) and \(6x - 2\) are parallel lines meaning they never intersect; hence, there is no solution.As a table of values for both expressions would also never match, this further confirms that the solution set is empty.
Key Concepts
ContradictionIdentityConditional EquationSolution Set
Contradiction
A contradiction in the context of equations is an assertion in which it results in a statement that is always false, regardless of the values substituted into the variable(s). For instance, when simplifying the equation \(-12 + 6x = 6x - 2\), eliminating identical terms from both sides led to \(-12 = -2\), which is a false statement. This false result indicates an impossibility, making the equation contradictory.
In arithmetic and algebra, contradictions are not solvable as no value of the variable will ever satisfy the equation. Due to this, the solution set of a contradictory equation is termed as the empty set (denoted by \(\varnothing\)), reflecting the fact there are no solutions.
In arithmetic and algebra, contradictions are not solvable as no value of the variable will ever satisfy the equation. Due to this, the solution set of a contradictory equation is termed as the empty set (denoted by \(\varnothing\)), reflecting the fact there are no solutions.
- Contradictory equations do not share any common values between expressions.
- Graphically, these are represented by parallel lines that never meet.
- Understanding contradictions helps identify equations that can't be solved with real numbers.
Identity
An identity is a type of equation that holds true for all possible values of the variable involved. In other words, no matter what number you substitute for the variable, the equation is always true. An example would be an equation like \(2x + 3 = 2x + 3\), where any value for \(x\) makes both sides equal.
Identities are useful because they highlight intrinsic truths in algebraic expressions.
Identities are useful because they highlight intrinsic truths in algebraic expressions.
- They illustrate fundamental principles or facts, applicable universally.
- Identities have an infinite solution set since any real number suits the equation.
- These are valuable in simplifying complex algebraic expressions and formulas.
Conditional Equation
A conditional equation is a statement that is only true for specific value(s) of the variable. Unlike identities, which are always true, and contradictions, which are never true, conditional equations rely on particular solutions to be valid.
For example, the equation \(x + 2 = 5\) is a conditional equation because it only holds when \(x\) equals 3. Solving conditional equations involves finding this specific value or set of values.
For example, the equation \(x + 2 = 5\) is a conditional equation because it only holds when \(x\) equals 3. Solving conditional equations involves finding this specific value or set of values.
- The solution set contains the specific values making the equation true.
- These solutions can often be found using algebraic manipulation.
- Conditional equations help in modeling real-world scenarios where conditions must be met.
Solution Set
The solution set of an equation represents all the values of the variable that satisfy the equation. It encapsulates all the possible solutions, highlighting the specific conditions under which the equation holds true.
In the case of the exercise equation where \-12 = -2\ showed a false statement, the solution set was thus empty, such that \(\varnothing\). For a conditional equation like \x + 2 = 5\, the solution set would just be \(\{3\}\), since only \(x = 3\) satisfies the equation.
In the case of the exercise equation where \-12 = -2\ showed a false statement, the solution set was thus empty, such that \(\varnothing\). For a conditional equation like \x + 2 = 5\, the solution set would just be \(\{3\}\), since only \(x = 3\) satisfies the equation.
- The solution set is comprehensive, showing either a single value, multiple values, or no values at all (empty set).
- In graphical terms, it represents where two expressions intersect.
- Understanding solution sets is crucial to solving equations accurately and knowing which values apply.
Other exercises in this chapter
Problem 57
Find \(f(x)\) at the indicated value of \(x\). $$f(x)=-x^{2}+x+2 ; x=4$$
View solution Problem 57
Find a decimal approximation of each root or power Round answers to the nearest thousandth. $$\sqrt[3]{33}$$
View solution Problem 58
Find the equation of the line described, giving it in slope-intercept form if possible. Find the equation of the line that is the perpendicular bisector of the
View solution Problem 58
Solve each problem. Speeding Fines Suppose that speeding fines are determined by \(y=10(x-65)+50, x>65,\) where \(y\) is the cost in dollars of the fine if a pe
View solution