Home \( > \) CC2 \( > \) Chapter \( 6> \) Lesson \( 6.2 .3> \) Problem 6-81 Solve each equation below for \( x \). Check your final answer. \( \begin{array}{ll}\text { a. } 4 x=6 x-14 & \text { b. } 3 x+5=50 \\ \boxtimes \square \operatorname{step} 1 \text { (a): }\end{array} \)
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Bonus Knowledge
Let's solve the first equation step by step and then tackle the second one! For equation (a) \( 4x = 6x - 14 \): Step 1: Start by isolating \( x \). Subtract \( 6x \) from both sides: \[ 4x - 6x = -14 \] This simplifies to: \[ -2x = -14 \] Step 2: Now, divide both sides by -2: \[ x = 7 \] To check, substitute \( x = 7 \) back into the original equation: \[ 4(7) = 6(7) - 14 \] This gives: \[ 28 = 42 - 14 \] Which simplifies to: \[ 28 = 28 \] (True!) Now, for equation (b) \( 3x + 5 = 50 \): Step 1: Isolate \( 3x \) by subtracting 5 from both sides: \[ 3x = 50 - 5 \] This simplifies to: \[ 3x = 45 \] Step 2: Now, divide both sides by 3: \[ x = 15 \] Check this by substituting \( x = 15 \): \[ 3(15) + 5 = 50 \] This gives: \[ 45 + 5 = 50 \] Which simplifies to: \[ 50 = 50 \] (True!) So, the solutions are \( x = 7 \) for (a) and \( x = 15 \) for (b). Happy solving!