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

Question

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 \( c_1 \neq c_2 \)