Solve for k (4k+3)/6+(4k-8)/9=(5k-4)/3-(k-3)/2

$\frac{4 k + 3}{6} + \frac{4 k - 8}{9} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Step 1

Simplify $\frac{4 k + 3}{6} + \frac{4 k - 8}{9}$ .

Factor $4$ out of $4 k - 8$ .

Factor $4$ out of $4 k$ .

$\frac{4 k + 3}{6} + \frac{4 (k) - 8}{9} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Factor $4$ out of $- 8$ .

$\frac{4 k + 3}{6} + \frac{4 k + 4 \cdot - 2}{9} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Factor $4$ out of $4 k + 4 \cdot - 2$ .

$\frac{4 k + 3}{6} + \frac{4 (k - 2)}{9} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

To write $\frac{4 k + 3}{6}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{4 k + 3}{6} \cdot \frac{3}{3} + \frac{4 (k - 2)}{9} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

To write $\frac{4 (k - 2)}{9}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{4 k + 3}{6} \cdot \frac{3}{3} + \frac{4 (k - 2)}{9} \cdot \frac{2}{2} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Write each expression with a common denominator of $18$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{4 k + 3}{6}$ by $\frac{3}{3}$ .

$\frac{(4 k + 3) \cdot 3}{6 \cdot 3} + \frac{4 (k - 2)}{9} \cdot \frac{2}{2} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Multiply $6$ by $3$ .

$\frac{(4 k + 3) \cdot 3}{18} + \frac{4 (k - 2)}{9} \cdot \frac{2}{2} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Multiply $\frac{4 (k - 2)}{9}$ by $\frac{2}{2}$ .

$\frac{(4 k + 3) \cdot 3}{18} + \frac{4 (k - 2) \cdot 2}{9 \cdot 2} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Multiply $9$ by $2$ .

$\frac{(4 k + 3) \cdot 3}{18} + \frac{4 (k - 2) \cdot 2}{18} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Combine the numerators over the common denominator.

$\frac{(4 k + 3) \cdot 3 + 4 (k - 2) \cdot 2}{18} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Simplify the numerator.

Apply the distributive property.

$\frac{4 k \cdot 3 + 3 \cdot 3 + 4 (k - 2) \cdot 2}{18} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Multiply $3$ by $4$ .

$\frac{12 k + 3 \cdot 3 + 4 (k - 2) \cdot 2}{18} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Multiply $3$ by $3$ .

$\frac{12 k + 9 + 4 (k - 2) \cdot 2}{18} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Apply the distributive property.

$\frac{12 k + 9 + (4 k + 4 \cdot - 2) \cdot 2}{18} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Multiply $4$ by $- 2$ .

$\frac{12 k + 9 + (4 k - 8) \cdot 2}{18} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Apply the distributive property.

$\frac{12 k + 9 + 4 k \cdot 2 - 8 \cdot 2}{18} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Multiply $2$ by $4$ .

$\frac{12 k + 9 + 8 k - 8 \cdot 2}{18} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Multiply $- 8$ by $2$ .

$\frac{12 k + 9 + 8 k - 16}{18} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Add $12 k$ and $8 k$ .

$\frac{20 k + 9 - 16}{18} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Subtract $16$ from $9$ .

$\frac{20 k - 7}{18} = \frac{5 k - 4}{3} - \frac{k - 3}{2}$

Step 2

Simplify $\frac{5 k - 4}{3} - \frac{k - 3}{2}$ .

To write $\frac{5 k - 4}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{20 k - 7}{18} = \frac{5 k - 4}{3} \cdot \frac{2}{2} - \frac{k - 3}{2}$

To write $- \frac{k - 3}{2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{20 k - 7}{18} = \frac{5 k - 4}{3} \cdot \frac{2}{2} - \frac{k - 3}{2} \cdot \frac{3}{3}$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{5 k - 4}{3}$ by $\frac{2}{2}$ .

$\frac{20 k - 7}{18} = \frac{(5 k - 4) \cdot 2}{3 \cdot 2} - \frac{k - 3}{2} \cdot \frac{3}{3}$

Multiply $3$ by $2$ .

$\frac{20 k - 7}{18} = \frac{(5 k - 4) \cdot 2}{6} - \frac{k - 3}{2} \cdot \frac{3}{3}$

Multiply $\frac{k - 3}{2}$ by $\frac{3}{3}$ .

$\frac{20 k - 7}{18} = \frac{(5 k - 4) \cdot 2}{6} - \frac{(k - 3) \cdot 3}{2 \cdot 3}$

Multiply $2$ by $3$ .

$\frac{20 k - 7}{18} = \frac{(5 k - 4) \cdot 2}{6} - \frac{(k - 3) \cdot 3}{6}$

Combine the numerators over the common denominator.

$\frac{20 k - 7}{18} = \frac{(5 k - 4) \cdot 2 - (k - 3) \cdot 3}{6}$

Simplify the numerator.

Apply the distributive property.

$\frac{20 k - 7}{18} = \frac{5 k \cdot 2 - 4 \cdot 2 - (k - 3) \cdot 3}{6}$

Multiply $2$ by $5$ .

$\frac{20 k - 7}{18} = \frac{10 k - 4 \cdot 2 - (k - 3) \cdot 3}{6}$

Multiply $- 4$ by $2$ .

$\frac{20 k - 7}{18} = \frac{10 k - 8 - (k - 3) \cdot 3}{6}$

Apply the distributive property.

$\frac{20 k - 7}{18} = \frac{10 k - 8 + (- k - - 3) \cdot 3}{6}$

Multiply $- 1$ by $- 3$ .

$\frac{20 k - 7}{18} = \frac{10 k - 8 + (- k + 3) \cdot 3}{6}$

Apply the distributive property.

$\frac{20 k - 7}{18} = \frac{10 k - 8 - k \cdot 3 + 3 \cdot 3}{6}$

Multiply $3$ by $- 1$ .

$\frac{20 k - 7}{18} = \frac{10 k - 8 - 3 k + 3 \cdot 3}{6}$

Multiply $3$ by $3$ .

$\frac{20 k - 7}{18} = \frac{10 k - 8 - 3 k + 9}{6}$

Subtract $3 k$ from $10 k$ .

$\frac{20 k - 7}{18} = \frac{7 k - 8 + 9}{6}$

Add $- 8$ and $9$ .

$\frac{20 k - 7}{18} = \frac{7 k + 1}{6}$

Step 3

Move all terms containing $k$ to the left side of the equation.

Subtract $\frac{7 k + 1}{6}$ from both sides of the equation.

$\frac{20 k - 7}{18} - \frac{7 k + 1}{6} = 0$

To write $- \frac{7 k + 1}{6}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{20 k - 7}{18} - \frac{7 k + 1}{6} \cdot \frac{3}{3} = 0$

Write each expression with a common denominator of $18$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{7 k + 1}{6}$ by $\frac{3}{3}$ .

$\frac{20 k - 7}{18} - \frac{(7 k + 1) \cdot 3}{6 \cdot 3} = 0$

Multiply $6$ by $3$ .

$\frac{20 k - 7}{18} - \frac{(7 k + 1) \cdot 3}{18} = 0$

Combine the numerators over the common denominator.

$\frac{20 k - 7 - (7 k + 1) \cdot 3}{18} = 0$

Simplify the numerator.

Apply the distributive property.

$\frac{20 k - 7 + (- (7 k) - 1 \cdot 1) \cdot 3}{18} = 0$

Multiply $7$ by $- 1$ .

$\frac{20 k - 7 + (- 7 k - 1 \cdot 1) \cdot 3}{18} = 0$

Multiply $- 1$ by $1$ .

$\frac{20 k - 7 + (- 7 k - 1) \cdot 3}{18} = 0$

Apply the distributive property.

$\frac{20 k - 7 - 7 k \cdot 3 - 1 \cdot 3}{18} = 0$

Multiply $3$ by $- 7$ .

$\frac{20 k - 7 - 21 k - 1 \cdot 3}{18} = 0$

Multiply $- 1$ by $3$ .

$\frac{20 k - 7 - 21 k - 3}{18} = 0$

Subtract $21 k$ from $20 k$ .

$\frac{- k - 7 - 3}{18} = 0$

Subtract $3$ from $- 7$ .

$\frac{- k - 10}{18} = 0$

Factor $- 1$ out of $- k$ .

$\frac{- (k) - 10}{18} = 0$

Rewrite $- 10$ as $- 1 (10)$ .

$\frac{- (k) - 1 (10)}{18} = 0$

Factor $- 1$ out of $- (k) - 1 (10)$ .

$\frac{- (k + 10)}{18} = 0$

Rewrite $- (k + 10)$ as $- 1 (k + 10)$ .

$\frac{- 1 (k + 10)}{18} = 0$

Move the negative in front of the fraction.

$- \frac{k + 10}{18} = 0$

Step 4

Set the numerator equal to zero.

$k + 10 = 0$

Step 5

Subtract $10$ from both sides of the equation.

$k = - 10$

Solve for k (4k+3)/6+(4k-8)/9=(5k-4)/3-(k-3)/2

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