Solve for b. {:[7b-15=5b-3],[b=]:}

Given equation: Start with the given equation 7b - 15 = 5b - 3. We want to isolate b, so we will move all the terms containing b to one side and the constant terms to the other side.Move terms: Subtract 5b from both sides to get the b terms on one side. 7b - 15 - 5b = 5b - 3 - 5b This simplifies to: 2b - 15 = -3Subtract 5b: Add 15 to both sides to isolate the term with b. 2b - 15 + 15 = -3 + 15 This simplifies to: 2b = 12Add 15: Divide both sides by 2 to solve for b. 2b / 2 = 12 / 2 This simplifies to: b = 6

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Solve for $b$ . $\newline$ $\begin{array}{l} 7 b-15=5 b-3 \\ b= \end{array}$

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Algebra 1

Solve linear equations with variables on both sides

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Q. Solve for

b

\newline

\begin{array}{l} 7 b-15=5 b-3 \\ b= \end{array}

Given equation: Start with the given equation $7b - 15 = 5b - 3$ . $\newline$ We want to isolate $b$ , so we will move all the terms containing $b$ to one side and the constant terms to the other side.
Move terms: Subtract $5b$ from both sides to get the $b$ terms on one side. $\newline$ $7b - 15 - 5b = 5b - 3 - 5b$ $\newline$ This simplifies to: $\newline$ $2b - 15 = -3$
Subtract $5b$ : Add $15$ to both sides to isolate the term with $b$ .
$2b - 15 + 15 = -3 + 15$
This simplifies to:
$2b = 12$
Add $15$ : Divide both sides by $2$ to solve for $b$ . $\frac{2b}{2} = \frac{12}{2}$ This simplifies to: $b = 6$