Problem 67
Question
If the equations \((a+1)^{3} x+(a+2)^{3} y=(a+3)^{3},(a+1) x+(a+2) y\) \(=a+3, x+y=1\) are consistent then \(a\) is equal to (A) 1 (B) \(-1\) (C) 2 (D) \(-2\)
Step-by-Step Solution
Verified Answer
The value of \(a\) is \(-2\).
1Step 1: Write down the equations
We begin by writing down the given system of equations: \((a+1)^{3} x + (a+2)^{3} y = (a+3)^{3}\), \((a+1) x + (a+2) y = a+3\), \(x+y = 1\).
2Step 2: Express in terms of variables
From equation \(x+y=1\), replace \(y\) with \(1-x\) in the other two equations. This simplifies the system to two variables: \((a+1)^{3} x + (a+2)^{3} (1-x) = (a+3)^{3}\), and \((a+1) x + (a+2)(1-x) = a+3\).
3Step 3: Simplify the equations
Substitute \(y = 1-x\) into the equations:1. \((a+1)^{3} x + (a+2)^{3} - (a+2)^{3} x = (a+3)^{3}\)2. \((a+1) x + (a+2) - (a+2) x = a+3\).
4Step 4: Solve the second equation
Simplify the second equation: \((a+1)x + (a+2) - (a+2)x = a+3\) which simplifies to \(x = 1\).
5Step 5: Substitute x into the first equation
Substituting \(x = 1\) into the first simplified equation results in:\((a+1)^{3} - (a+2)^{3} + (a+2)^{3} = (a+3)^{3}\).
6Step 6: Check consistency by solving the algebraic equation
After simplifying, check whether this equality holds:\((a+1)^{3} = (a+3)^{3}\).This implies solving \(a+1 = a+3\), which is impossible unless \(a = -2\) allows consistent solution without contradictions.
Key Concepts
Consistency of EquationsSubstitution MethodAlgebraic Manipulation
Consistency of Equations
In a system of equations, consistency is crucial. It indicates that there is at least one solution that satisfies all equations in the system. In this case, to determine the value of \( a \) that ensures consistency, we need each equation to align perfectly without any contradictions. When equations are consistent:
- They intersect at common points.
- There is at least one solution.
- All expressions are logically coherent within the context of the equations.
Substitution Method
The substitution method is a useful technique to solve a system of equations. It involves expressing one variable in terms of another and substituting this expression into the other equations. This step leads to equations with fewer variables, simplifying the process of finding solutions. Here's how it works in this scenario:
- From the simple equation \( x+y=1 \), we can express \( y \) as \( 1-x \).
- Substitute \( y = 1-x \) into the remaining equations.
- This results in equations predominantly in terms of \( x \), simplifying the solving process.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying equations to find solutions. It is vital in solving the system of equations and checking their consistency. The general steps include:
- Simplifying expressions by combining like terms and performing operations that maintain equality.
- Substituting values, like \( x = 1 \), into equations to test their validity.
- Breaking complex expressions into simpler components for easier handling.
Other exercises in this chapter
Problem 65
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View solution Problem 68
If the system of equations \(x \sin \alpha+y \sin \beta+z \sin \gamma=0, x \cos \alpha+y \cos \beta+z \cos \gamma\) \(=0, x+y+z=0\), where \(\alpha, \beta, \gam
View solution Problem 69
If \(x_{1} \neq 0, x_{2} \neq 0, x_{3} \neq 0\), then the determinant \(\left|\begin{array}{ccc}x_{1}+a_{1} b_{1} & a_{1} b_{2} & a_{1} b_{3} \\\ a_{2} b_{1} &
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