Question

lesson 6 practice problems
1 solve each of these equations. explain or show your reasoning.
2b + 8 - 5b + 3 = -13 + 8b - 5
2x + 7 - 5x + 8 = 3(5 + 6x) - 12x
2c - 3 = 2(6 - c) + 7c

Explanation:

First Equation: \(2b + 8 - 5b + 3 = -13 + 8b - 5\)

Step1: Combine like terms on both sides

Left side: \(2b - 5b + 8 + 3 = -3b + 11\)
Right side: \(-13 - 5 + 8b = 8b - 18\)
So the equation becomes \(-3b + 11 = 8b - 18\)

Step2: Add \(3b\) to both sides

\(-3b + 3b + 11 = 8b + 3b - 18\)
\(11 = 11b - 18\)

Step3: Add 18 to both sides

\(11 + 18 = 11b - 18 + 18\)
\(29 = 11b\)

Step4: Divide both sides by 11

\(b = \frac{29}{11}\)

Second Equation: \(2x + 7 - 5x + 8 = 3(5 + 6x) - 12x\)

Step1: Combine like terms on left, expand right

Left side: \(2x - 5x + 7 + 8 = -3x + 15\)
Right side: \(15 + 18x - 12x = 15 + 6x\)
Equation: \(-3x + 15 = 6x + 15\)

Step2: Add \(3x\) to both sides

\(-3x + 3x + 15 = 6x + 3x + 15\)
\(15 = 9x + 15\)

Step3: Subtract 15 from both sides

\(15 - 15 = 9x + 15 - 15\)
\(0 = 9x\)

Step4: Divide by 9

\(x = 0\)

Third Equation: \(2c - 3 = 2(6 - c) + 7c\)

Step1: Expand the right side

\(2(6 - c) + 7c = 12 - 2c + 7c = 12 + 5c\)
Equation: \(2c - 3 = 5c + 12\)

Step2: Subtract \(2c\) from both sides

\(2c - 2c - 3 = 5c - 2c + 12\)
\(-3 = 3c + 12\)

Step3: Subtract 12 from both sides

\(-3 - 12 = 3c + 12 - 12\)
\(-15 = 3c\)

Step4: Divide by 3

\(c = -5\)

Answer:

For \(2b + 8 - 5b + 3 = -13 + 8b - 5\), \(b = \boldsymbol{\frac{29}{11}}\)
For \(2x + 7 - 5x + 8 = 3(5 + 6x) - 12x\), \(x = \boldsymbol{0}\)
For \(2c - 3 = 2(6 - c) + 7c\), \(c = \boldsymbol{-5}\)