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Question
- santa claims that a linear equation can have exactly one solution, no solution, or infinitely many solutions. for each value of a, determine how many solutions would result for the linear equation 3x + 8 = a
| exactly one solution | no solution | infinitely many solutions |
|---|---|---|
| a = 6x | ||
| a = 3x + 7 | ||
| a = 3x + 8 |
Step 1: Analyze \(a = 0\)
The equation is \(3x + 8=0\). Solve for \(x\):
Subtract 8 from both sides: \(3x= - 8\)
Divide by 3: \(x =-\frac{8}{3}\). This is a unique solution, so exactly one solution.
Step 2: Analyze \(a = 6x\)
The equation is \(3x + 8=6x\)
Subtract \(3x\) from both sides: \(8 = 3x\)
Divide by 3: \(x=\frac{8}{3}\). This is a unique solution, so exactly one solution. Wait, there was a mistake earlier. Let's re - analyze:
Case 1: \(a = 0\)
Equation: \(3x+8 = 0\)
Subtract 8: \(3x=0 - 8=-8\)
Divide by 3: \(x =-\frac{8}{3}\). So exactly one solution.
Case 2: \(a = 6x\)
Equation: \(3x + 8=6x\)
Subtract \(3x\) from both sides: \(8=6x - 3x=3x\)
Divide by 3: \(x=\frac{8}{3}\). Exactly one solution.
Case 3: \(a = 3x + 7\)
Equation: \(3x+8=3x + 7\)
Subtract \(3x\) from both sides: \(8 = 7\). This is a false statement. So no solution.
Case 4: \(a=3x + 8\)
Equation: \(3x + 8=3x + 8\)
Subtract \(3x\) from both sides: \(8 = 8\). This is a true statement for all values of \(x\). So infinitely many solutions.
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- \(a = 0\): Exactly one solution.
- \(a = 6x\): Exactly one solution.
- \(a=3x + 7\): No solution.
- \(a = 3x+8\): Infinitely many solutions.
So the completed table is:
| Exactly one solution | No solution | Infinitely many solutions | |
|---|---|---|---|
| \(a = 6x\) | \(\checkmark\) | ||
| \(a=3x + 7\) | \(\checkmark\) | ||
| \(a = 3x+8\) | \(\checkmark\) |