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:
\( ax + b - b = c - b \)
Simplifying, we get:
\( ax = c - b \)

Step 2: Solve for \( x \)

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

Now let's analyze each part (a to c, but the question here is about part c? Wait, the user's image shows part c: "If \( a \), \( b \), and \( c \) are rational numbers, \( x \) is a rational number. Choose the correct answer below:". Let's analyze the options:

• Option A: Says the statement is always true. But if \( a \), \( b \), \( c \) are rational, is \( x = \frac{c - b}{a} \) always rational? Wait, no—wait, rational numbers are closed under addition, subtraction, multiplication, and division (except by zero). So if \( a \), \( b \), \( c \) are rational and \( a

eq 0 \), then \( c - b \) is rational (since rationals are closed under subtraction), and dividing a rational by a non - zero rational gives a rational. Wait, but maybe the original problem has a different context? Wait, no—wait, the options:

Wait, the options:

• 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.

Wait, but in our solution, \( x=\frac{c - b}{a} \). If \( a \), \( b \), \( c \) are rational and \( a
eq0 \), then \( x \) is rational. But maybe the problem is considering cases where \( a = 0 \)? Wait, if \( a = 0 \), the equation is \( 0x + b = c \), or \( b = c \). If \( b
eq c \), no solution; if \( b = c \), infinite solutions. But the problem says "solve for \( x \)", so we assume \( a
eq0 \).

But the options: Let's re - examine. The key is: if \( a \), \( b \), \( c \) are rational, is \( x \) always rational?

Rational numbers are numbers that can be written as \( \frac{p}{q} \) where \( p,q \) are integers and \( q
eq0 \). If \( a=\frac{p_1}{q_1} \), \( b = \frac{p_2}{q_2} \), \( c=\frac{p_3}{q_3} \) (with \( q_1,q_2,q_3
eq0 \)), then \( c - b=\frac{p_3}{q_3}-\frac{p_2}{q_2}=\frac{p_3q_2 - p_2q_3}{q_2q_3} \), which is rational. Then \( \frac{c - b}{a}=\frac{\frac{p_3q_2 - p_2q_3}{q_2q_3}}{\frac{p_1}{q_1}}=\frac{(p_3q_2 - p_2q_3)q_1}{p_1q_2q_3} \), which is rational (since \( p_1
eq0 \) as \( a
eq0 \)). So the statement "If \( a \), \( b \), and \( c \) are rational numbers, \( x \) is a rational number" is always true. So the correct option is A.

1. Solve \( ax + b = c \): Subtract \( b \) ( \( ax=c - b \) ), then divide by \( a \) ( \( x=\frac{c - b}{a} \) ).
2. Analyze rationality: Rational numbers are closed under subtraction (\( c - b \) is rational if \( c,b \) are rational) and division (by non - zero \( a \), \( \frac{c - b}{a} \) is rational if \( c - b \) and \( a \) are rational). So if \( a,b,c \) are rational (\( a

eq0 \)), \( x \) is rational. Thus the statement is always true.

Answer:

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.