Problem 45 - Algebra Form and Function

Question

Decide for what value(s) of the constant \(A\) (if any) the equation has (a) The solution \(x=0\) (b) A positive solution in \(x\) (c) No solution in \(x\). $$ 3 x+5=A $$

Step-by-Step Solution

Answer

Question: Determine the values of the constant A for the following conditions regarding the solutions to the equation \(3x+5=A\): a) The solution is \(x=0\). b) The solution is a positive value of x. c) There is no solution for x. Answer: a) For \(x=0\), the value of A is \(A=5\). b) For a positive solution in x, the range of values for A is \(A > 5\). c) There is no value of A for which the equation has no solution in x.

Step 1: (a) Finding A when x=0 #

To find the value of A when x=0, substitute x with 0 in the given equation. $$ 3(0) + 5 = A \\ A = 5 $$ So the value of A that has a solution \(x=0\) is 5.

Step 2: (b) Finding A for a positive solution in x #

To find the values of A for which the equation has a positive solution in x, we need to isolate x: $$ 3x = A - 5 \\ x = \frac{A - 5}{3} $$ For x to be positive, the numerator (A - 5) must be greater than 0. $$ A - 5 > 0 \\ A > 5 $$ So, the equation has positive solutions for x when \(A > 5\).

Step 3: (c) Finding A for no solution in x #

Since the given equation has only one variable (x), there is no situation where there wouldn't be a solution for x. This can be explained by the equation being a simple linear equation which will always have a single unique solution for any value of A. Therefore, there's no value of the constant A for which the equation has no solution in x.

Key Concepts

Positive Solutions

Understanding when a value is a positive solution to an equation, like the linear equation given above, centers on determining the situations when the variable, here denoted as \(x\), is greater than zero. To achieve a positive solution, the equation must be rearranged to isolate \(x\), which allows us to clearly see the conditions under which \(x\) will be positive.

• The original equation: \(3x + 5 = A\)
• Rearranging gives: \(3x = A - 5\)
• Isolating \(x\): \(x = \frac{A - 5}{3}\)

For \(x\) to be positive, the expression \(A - 5\) must be greater than zero, which translates into \(A > 5\).

When the condition \(A > 5\) is met, the equation yields positive values for \(x\). This approach helps us understand how changing parameter \(A\) in the simple linear equation affects the solution of \(x\), emphasizing the importance of setting the right conditions for positive solutions in algebra.

Equation Substitution

Equation substitution is a powerful tool in solving linear equations. It involves substituting known values for specific variables to simplify the equation and find unknowns. Here, we apply substitution to find the solution to the equation:

• Substitute \(x = 0\) in the equation \(3x + 5 = A\)
• This gives: \(3(0) + 5 = A\)
• Resulting in \(A = 5\)

Substitution allows for a straightforward pathway to determine the constants that satisfy specific conditions, like \(x = 0\) in this case. The elegance of substitution lies in its simplicity, providing an easy-to-follow method to decompose and solve equations by transforming an equation's state.

This method also aids in verifying solutions, ensuring accuracy in equations and understanding complex algebraic relationships. By substituting values, we can quickly identify key values of parameters that yield special solutions to equations.

Unique Solution

Understanding unique solutions in the realm of linear equations is crucial. Linear equations are marked by having exactly one solution for any given value of its constants or variables, provided the equation is not compromised by divided by zero scenarios, which aren't applicable here.

• The given equation: \(3x + 5 = A\)
• Isolate \(x\): \(x = \frac{A - 5}{3}\)

This equation is linear in form, ensuring that for every value of \(A\), there exists a unique solution for \(x\).

Linear equations do not allow for configurations where there is "no solution," since they will always cross through every \(A\) value on a one-dimensional number line, meaning there will always be an \(x\) regardless of the \(A\) provided. In essence, linear equations guarantee a dependable relationship between parameters and variables, ensuring predictability and consistency in solutions.