To determine which equation is equivalent to \(4x + 6 = 1\), we can start by solving the original equation for \(x\):
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Subtract 6 from both sides: \[ 4x + 6 - 6 = 1 - 6 \implies 4x = -5 \]
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Divide by 4: \[ x = -\frac{5}{4} \]
Now, we will check each option by substituting \(x = -\frac{5}{4}\) into the equations and checking if the left-hand side equals 1.
Option A: \[ 5(x+2)-x = 1 \]
Substituting \(x = -\frac{5}{4}\): \[ 5\left(-\frac{5}{4}+2\right) - \left(-\frac{5}{4}\right) = 5\left(-\frac{5}{4} + \frac{8}{4}\right) + \frac{5}{4} \] \[ = 5\left(\frac{3}{4}\right) + \frac{5}{4} = \frac{15}{4} + \frac{5}{4} = \frac{20}{4} = 5 \quad (\text{not } 1) \]
Option B: \[ 18 + 6(x-2) - 2x = 1 \]
Substituting \(x = -\frac{5}{4}\): \[ 18 + 6\left(-\frac{5}{4} - 2\right) - 2\left(-\frac{5}{4}\right) \] \[ = 18 + 6\left(-\frac{5}{4} - \frac{8}{4}\right) + \frac{10}{4} \] \[ = 18 + 6\left(-\frac{13}{4}\right) + \frac{10}{4} \] \[ = 18 - \frac{78}{4} + \frac{10}{4} = 18 - \frac{68}{4} = 18 - 17 = 1 \quad (\text{is } 1) \]
Option C: \[ 9x - 5(-1) - 1 = 1 \]
Substituting \(x = -\frac{5}{4}\): \[ 9\left(-\frac{5}{4}\right) + 5 - 1 \] \[ = -\frac{45}{4} + 5 - 1 = -\frac{45}{4} + \frac{20}{4} - \frac{4}{4} = -\frac{45}{4} + \frac{16}{4} = -\frac{29}{4} \quad (\text{not } 1) \]
Option D: \[ 1 + 3(x + 2) = 1 \]
Substituting \(x = -\frac{5}{4}\): \[ 1 + 3\left(-\frac{5}{4} + 2\right) = 1 + 3\left(-\frac{5}{4} + \frac{8}{4}\right) \] \[ = 1 + 3\left(\frac{3}{4}\right) = 1 + \frac{9}{4} \] \[ = \frac{4}{4} + \frac{9}{4} = \frac{13}{4} \quad (\text{not } 1) \]
The only equation that is equivalent to \(4x + 6 = 1\) is Option B:
\[ \boxed{B} \]
1 answer