How to solve 2(3k-1)-(2k+1=1 2

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 2(3k-1)-(2k+1=1 2 | Equatix

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Question

$$\frac { 2 ( 3 k - 1 ) - ( 2 k + 1 ) } { 7 k - 2 } = \frac { 1 } { 2 }$$

Solve for k

$k=4$

Steps for Solving Linear Equation

Variable $k$ cannot be equal to $\frac{2}{7}$ since division by zero is not defined. Multiply both sides of the equation by $2\left(7k-2\right)$, the least common multiple of $7k-2,2$.

$$2\left(2\left(3k-1\right)-\left(2k+1\right)\right)=7k-2$$

Use the distributive property to multiply $2$ by $3k-1$.

$$2\left(6k-2-\left(2k+1\right)\right)=7k-2$$

To find the opposite of $2k+1$, find the opposite of each term.

$$2\left(6k-2-2k-1\right)=7k-2$$

Combine $6k$ and $-2k$ to get $4k$.

$$2\left(4k-2-1\right)=7k-2$$

Subtract $1$ from $-2$ to get $-3$.

$$2\left(4k-3\right)=7k-2$$

Use the distributive property to multiply $2$ by $4k-3$.

$$8k-6=7k-2$$

Subtract $7k$ from both sides.

$$8k-6-7k=-2$$

Combine $8k$ and $-7k$ to get $k$.

$$k-6=-2$$

Add $6$ to both sides.

$$k=-2+6$$

Add $-2$ and $6$ to get $4$.

$$k=4$$

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