Question

analyzing solution sets to linear equations with the variable on both sides - item 31009
use the drop - down menus to complete each equation so the statement about its solution is true.
no solutions
$3x + 9 + 4x + x = \square x + \square$
one solution
$3x + 9 + 4x + x = \square x + \square$
infinitely many solutions
$3x + 9 + 4x + x = \square x + \square$

Explanation:

for No Solutions:

Step1: Simplify left side

Combine like terms: \(3x + 4x + x=8x\), so left side is \(8x + 9\).
For no solutions, coefficients of \(x\) must be equal, and constants different. So first box: \(8\), second box: (any number ≠9, e.g., 5)

Step2: Verify

Equation: \(8x + 9 = 8x + 5\). Subtract \(8x\): \(9 = 5\) (false), no solutions.

for One Solution:

Step1: Simplify left side

Left side: \(8x + 9\) (as above).
For one solution, coefficients of \(x\) must be different. Let first box be \(7\) (≠8), second box any (e.g., 9).

Step2: Verify

Equation: \(8x + 9 = 7x + 9\). Subtract \(7x\) and 9: \(x = 0\) (one solution).

for Infinitely Many Solutions:

Step1: Simplify left side

Left side: \(8x + 9\).
For infinitely many solutions, both sides must be identical. So first box: \(8\), second box: \(9\).

Step2: Verify

Equation: \(8x + 9 = 8x + 9\). Subtract \(8x + 9\): \(0 = 0\) (true), infinitely many solutions.

Answer:

s:

• No Solutions: \(3x + 9 + 4x + x = \boldsymbol{8}x + \boldsymbol{5}\) (second box can be any ≠9)
• One Solution: \(3x + 9 + 4x + x = \boldsymbol{7}x + \boldsymbol{9}\) (first box ≠8, second box any)
• Infinitely Many Solutions: \(3x + 9 + 4x + x = \boldsymbol{8}x + \boldsymbol{9}\)