Solve for k (5(k+2)-7)/6=(13-(4-k))/9

$\frac{5 (k + 2) - 7}{6} = \frac{13 - (4 - k)}{9}$

Simplify both sides.

Simplify the numerator.

Apply the distributive property.

$\frac{5 k + 5 \cdot 2 - 7}{6} = \frac{13 - (4 - k)}{9}$

Multiply $5$ by $2$ .

$\frac{5 k + 10 - 7}{6} = \frac{13 - (4 - k)}{9}$

Subtract $7$ from $10$ .

$\frac{5 k + 3}{6} = \frac{13 - (4 - k)}{9}$

Simplify the numerator.

Apply the distributive property.

$\frac{5 k + 3}{6} = \frac{13 - 1 \cdot 4 - - k}{9}$

Multiply $- 1$ by $4$ .

$\frac{5 k + 3}{6} = \frac{13 - 4 - - k}{9}$

Multiply $- - k$ .

Multiply $- 1$ by $- 1$ .

$\frac{5 k + 3}{6} = \frac{13 - 4 + 1 k}{9}$

Multiply $k$ by $1$ .

$\frac{5 k + 3}{6} = \frac{13 - 4 + k}{9}$

Subtract $4$ from $13$ .

$\frac{5 k + 3}{6} = \frac{9 + k}{9}$

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$(5 k + 3) \cdot 9 = 6 (9 + k)$

Solve the equation for $k$ .

Simplify $(5 k + 3) \cdot 9$ .

Start simplifying.

$0 + 0 + (5 k + 3) \cdot 9 = 6 (9 + k)$

Simplify by adding zeros.

$(5 k + 3) \cdot 9 = 6 (9 + k)$

Apply the distributive property.

$5 k \cdot 9 + 3 \cdot 9 = 6 (9 + k)$

Multiply.

Multiply $9$ by $5$ .

$45 k + 3 \cdot 9 = 6 (9 + k)$

Multiply $3$ by $9$ .

$45 k + 27 = 6 (9 + k)$

Simplify $6 (9 + k)$ .

Apply the distributive property.

$45 k + 27 = 6 \cdot 9 + 6 k$

Multiply $6$ by $9$ .

$45 k + 27 = 54 + 6 k$

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

Subtract $6 k$ from both sides of the equation.

$45 k + 27 - 6 k = 54$

Subtract $6 k$ from $45 k$ .

$39 k + 27 = 54$

Move all terms not containing $k$ to the right side of the equation.

Subtract $27$ from both sides of the equation.

$39 k = 54 - 27$

Subtract $27$ from $54$ .

$39 k = 27$

Divide each term in $39 k = 27$ by $39$ and simplify.

Divide each term in $39 k = 27$ by $39$ .

$\frac{39 k}{39} = \frac{27}{39}$

Simplify the left side.

Cancel the common factor of $39$ .

Cancel the common factor.

$\frac{39 k}{39} = \frac{27}{39}$

Divide $k$ by $1$ .

$k = \frac{27}{39}$

Simplify the right side.

Cancel the common factor of $27$ and $39$ .

Factor $3$ out of $27$ .

$k = \frac{3 (9)}{39}$

Cancel the common factors.

Factor $3$ out of $39$ .

$k = \frac{3 \cdot 9}{3 \cdot 13}$

Cancel the common factor.

$k = \frac{3 \cdot 9}{3 \cdot 13}$

Rewrite the expression.

$k = \frac{9}{13}$

The result can be shown in multiple forms.

Exact Form:

$k = \frac{9}{13}$

Decimal Form:

$k = 0.\overline{692307}$

Solve for k (5(k+2)-7)/6=(13-(4-k))/9

Related Questions

Support LCMGCF.com by disabling your adblocker.