How many solutions does the following equation have? 12 z-6+15 z=27 z-5 Choose 1 answer: (A) No solutions (B) Exactly one solution (C) Infinitely many solutions

Combine like terms: Combine like terms on the left side of the equation. 12z + 15z - 6 = 27z - 5 This simplifies to: 27z - 6 = 27z - 5Subtract to isolate constants: Subtract 27z from both sides of the equation to isolate the constant terms. 27z - 6 - 27z = 27z - 5 - 27z This simplifies to: -6 = -5Identify contradiction: We see that -6 does not equal -5, which indicates that there is a contradiction. Since the variable z has been eliminated and we are left with a false statement, this means there are no solutions to the equation.

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How many solutions does the following equation have? $\newline$ $12z-6+15z=27z-5$ $\newline$ Choose $1$ answer: $\newline$ (A) No solutions $\newline$ (B) Exactly one solution $\newline$ (C) Infinitely many solutions

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Q. How many solutions does the following equation have?

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12z-6+15z=27z-5

\newline

Choose

1

answer:

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(A) No solutions

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(B) Exactly one solution

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Combine like terms: Combine like terms on the left side of the equation. $\newline$ $12z + 15z - 6 = 27z - 5$ $\newline$ This simplifies to: $\newline$ $27z - 6 = 27z - 5$
Subtract to isolate constants: Subtract $27z$ from both sides of the equation to isolate the constant terms. $\newline$ $27z - 6 - 27z = 27z - 5 - 27z$ $\newline$ This simplifies to: $\newline$ $-6 = -5$
Identify contradiction: We see that $-6$ does not equal $-5$ , which indicates that there is a contradiction. Since the variable $z$ has been eliminated and we are left with a false statement, this means there are no solutions to the equation.