Solve for b. {:[(2)/(3)b+5=20-b],[b=◻]:}

Isolate variable terms: Isolate the variable terms on one side of the equation. We want to get all the terms with b on one side and the constant terms on the other side. (2/3)b + 5 = 20 - b First, we can add b to both sides to move the term from the right side to the left side. (2/3)b + b + 5 = 20 - b + bCombine like terms: Combine like terms. Now we need to combine the b terms on the left side. To combine (2/3)b and b, we need to express b as a fraction with the same denominator as (2/3)b. b is the same as (3/3)b, so we can add (2/3)b and (3/3)b. (2/3)b + (3/3)b = (5/3)b Now the equation looks like this: (5/3)b + 5 = 20Subtract to isolate term: Subtract 5 from both sides to isolate the term with b. (5/3)b + 5 - 5 = 20 - 5 (5/3)b = 15Multiply by reciprocal: Multiply both sides by the reciprocal of (5/3) to solve for b. The reciprocal of (5/3) is (3/5), so we multiply both sides by (3/5) to get b by itself. b = 15 * (3/5)Find value of b: Perform the multiplication to find the value of b. b = 15 * (3/5) b = (15 * 3) / 5 b = 45 / 5 b = 9

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Solve for
b.

{:[(2)/(3)b+5=20-b],[b=◻]:}

Solve for $b$ . $\newline$ $\begin{array}{l} \frac{2}{3} b+5=20-b \\ b=\square \end{array}$

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Math Problems

Algebra 1

Solve linear equations with variables on both sides

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Q. Solve for

b

\newline

\begin{array}{l} \frac{2}{3} b+5=20-b \\ b=\square \end{array}

Isolate variable terms: Isolate the variable terms on one side of the equation. $\newline$ We want to get all the terms with $b$ on one side and the constant terms on the other side. $\newline$ $\frac{2}{3}b + 5 = 20 - b$ $\newline$ First, we can add $b$ to both sides to move the term from the right side to the left side. $\newline$ $\frac{2}{3}b + b + 5 = 20 - b + b$
Combine like terms: Combine like terms. $\newline$ Now we need to combine the $b$ terms on the left side. $\newline$ To combine $\frac{2}{3}b$ and $b$ , we need to express $b$ as a fraction with the same denominator as $\frac{2}{3}b$ . $\newline$ $b$ is the same as $\frac{3}{3}b$ , so we can add $\frac{2}{3}b$ and $\frac{3}{3}b$ . $\newline$ $\frac{2}{3}b + \frac{3}{3}b = \frac{5}{3}b$ $\newline$ Now the equation looks like this: $\newline$ $\frac{2}{3}b$ $0$
Subtract to isolate term: Subtract $5$ from both sides to isolate the term with $b$ . $\newline$ $(\frac{5}{3})b + 5 - 5 = 20 - 5$ $\newline$ $(\frac{5}{3})b = 15$
Multiply by reciprocal: Multiply both sides by the reciprocal of $(\frac{5}{3})$ to solve for $b$ . The reciprocal of $(\frac{5}{3})$ is $(\frac{3}{5})$ , so we multiply both sides by $(\frac{3}{5})$ to get $b$ by itself. $b = 15 \times (\frac{3}{5})$
Find value of $b$ : Perform the multiplication to find the value of $b$ .
$b = 15 \times \left(\frac{3}{5}\right)$
$b = (15 \times 3) / 5$
$b = \frac{45}{5}$
$b = 9$