Question

Match the description of the one variable equation with the number of solutions it will have.

ax+c1=bx+c2

Answer 1

To determine the number of solutions for the equation \( ax + c_1 = bx + c_2 \), let's rearrange it:

1. Subtract \( bx \) from both sides:
\[ ax - bx + c_1 = c_2 \]
\[ (a - b)x + c_1 = c_2 \]

2. Now, isolate \( x \):
\[ (a - b)x = c_2 - c_1 \]
\[ x = \frac{c_2 - c_1}{a - b} \]

Now we can analyze the number of solutions based on the values of \( a \) and \( b \):

1. **If \( a - b \neq 0 \)**:
- This means \( a \neq b \), and the equation will have one unique solution:
\[
x = \frac{c_2 - c_1}{a - b}
\]

2. **If \( a - b = 0 \)** (or \( a = b \)):
- The equation simplifies to \( c_1 = c_2 \):
- If \( c_1 = c_2 \), then the equation is true for all \( x \) (infinitely many solutions).
- If \( c_1 \neq c_2 \), then the equation is false for all \( x \) (no solutions).

In summary:
- One solution: \( a \neq b \)
- Infinitely many solutions: \( a = b \) and \( c_1 = c_2 \)
- No solutions: \( a = b \) and \( c_1 \neq c_2 \)