Does 7(m+2)−5=13(21m+9)−6

have one solution, no solution, or infinitely many solutions?

1 answer

To determine whether the equation \( 7(m+2)−5=13(21m+9)−6 \) has one solution, no solution, or infinitely many solutions, we first simplify both sides of the equation.

Starting with the left side: \[ 7(m + 2) - 5 = 7m + 14 - 5 = 7m + 9 \]

Now simplifying the right side: \[ 13(21m + 9) - 6 = 273m + 117 - 6 = 273m + 111 \]

Now we have the equation: \[ 7m + 9 = 273m + 111 \]

Next, we will move all terms involving \( m \) to one side and constant terms to the other side. Subtract \( 7m \) from both sides: \[ 9 = 273m - 7m + 111 \] \[ 9 = 266m + 111 \]

Now, subtract 111 from both sides: \[ 9 - 111 = 266m \] \[ -102 = 266m \]

Now, divide both sides by 266: \[ m = \frac{-102}{266} \]

Simplifying \(-102/266\): \[ m = \frac{-51}{133} \]

Thus, the equation has one solution, which is \( m = \frac{-51}{133} \).

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