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 (12z + 15z) - 6 = 27z - 5 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 -6 = -5Identify contradiction: We see that -6 does not equal -5, which indicates that there is a contradiction. Therefore, there are no values of z that can satisfy this equation.

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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

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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Math Problems

Algebra 1

Find the number of solutions to a linear equation

Full solution

Q. 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

Combine like terms: Combine like terms on the left side of the equation. $\newline$ $12z + 15z - 6 = 27z - 5$ $\newline$ $(12z + 15z) - 6 = 27z - 5$ $\newline$ $27z - 6 = 27z - 5$
Subtract to isolate constants: Subtract $27z$ from both sides of the equation to isolate the constant terms. $\newline$ $27z - 6 - 27z = 27z - 5 - 27z$ $\newline$ $-6 = -5$
Identify contradiction: We see that $-6$ does not equal $-5$ , which indicates that there is a contradiction. Therefore, there are no values of $z$ that can satisfy this equation.