Question

given the equation ( ax + b = c ), solve for ( x ). describe each statement in parts a to c below as always, sometimes, or never true. explain your answer. c. if a, b, and c are rational numbers, x is a rational number. choose the correct answer below a. the statement is always true. the solution for x is an expression involving addition, subtraction, multiplication, and/or division, and these operations always give rational numbers when a, b, and c are rational numbers. b. the statement is sometimes true. the solution for x is an expression that may involve radicals that cannot be simplified, and these are not rational numbers. c. the statement is sometimes true because the solution could also be a fraction. d. the statement is never true because the solution involves division, meaning, x will be a fraction.

Explanation:

First, let's solve the equation \( ax + b = c \) for \( x \):

Step 1: Isolate the term with \( x \)

Subtract \( b \) from both sides of the equation:
\( ax + b - b = c - b \)
Simplifies to: \( ax = c - b \)

Step 2: Solve for \( x \)

Divide both sides by \( a \) (assuming \( a
eq 0 \)):
\( x = \frac{c - b}{a} \)

Now, analyze each statement:

Part c: "If \( a \), \( b \), and \( c \) are rational numbers, \( x \) is a rational number."

We need to determine if this is always, sometimes, or never true.

• Rational numbers are closed under addition, subtraction, multiplication, and division (except division by zero).
• If \( a \), \( b \), \( c \) are rational, then \( c - b \) is rational (closure of subtraction), and \( \frac{c - b}{a} \) is rational (closure of division, since \( a

eq 0 \) and \( a \) is rational).

Thus, the solution \( x = \frac{c - b}{a} \) will always be a rational number when \( a \), \( b \), \( c \) are rational (and \( a
eq 0 \)).

Answer:

for the equation:
\( x = \frac{c - b}{a} \) (for \( a
eq 0 \))