Question

which equation has infinitely many solutions? a 9x - 3 = 3x + 6x + 2 + 3 b 12(x + 8) = 11x - 5 c 11x - 2x + 15 = 8 + 7 + 9x d 5x - 8 = 11 - 7x + 12x

Explanation:

Step1: Simplify Option A

Simplify the right - hand side of $9x - 3=3x + 6x+2 + 3$. Combine like terms: $3x+6x + 2+3=9x + 5$. The equation becomes $9x-3=9x + 5$, subtract $9x$ from both sides: $-3
eq5$, no solution.

Step2: Simplify Option B

Expand the left - hand side of $12(x + 8)=11x-5$. Using the distributive property, $12x+96=11x - 5$. Subtract $11x$ from both sides: $12x-11x+96=11x-11x - 5$, $x=-101$, one solution.

Step3: Simplify Option C

Simplify the left - hand side of $11x-2x + 15=8 + 7+9x$. Combine like terms: $9x+15$. Simplify the right - hand side: $8 + 7+9x=15+9x$. The equation $9x + 15=9x+15$ is an identity. Subtract $9x$ from both sides, $15 = 15$. This equation has infinitely many solutions.

Step4: Simplify Option D

Simplify the right - hand side of $5x-8=11-7x + 12x$. Combine like terms: $11-7x + 12x=11 + 5x$. The equation becomes $5x-8=5x + 11$. Subtract $5x$ from both sides: $-8
eq11$, no solution.

Answer:

C. $11x - 2x+15=8 + 7+9x$