How to solve y-2=(1)/(5)(3y-1)

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 y-2=(1)/(5)(3y-1) | Equatix

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Question

$$y-2= \frac{ 1 }{ 5 } (3y-1)$$

Solve for y

$y = \frac{9}{2} = 4\frac{1}{2} = 4.5$

Steps for Solving Linear Equation

Use the distributive property to multiply $\frac{1}{5}$ by $3y-1$.

$$y-2=\frac{1}{5}\times 3y+\frac{1}{5}\left(-1\right)$$

Multiply $\frac{1}{5}$ and $3$ to get $\frac{3}{5}$.

$$y-2=\frac{3}{5}y+\frac{1}{5}\left(-1\right)$$

Multiply $\frac{1}{5}$ and $-1$ to get $-\frac{1}{5}$.

$$y-2=\frac{3}{5}y-\frac{1}{5}$$

Subtract $\frac{3}{5}y$ from both sides.

$$y-2-\frac{3}{5}y=-\frac{1}{5}$$

Combine $y$ and $-\frac{3}{5}y$ to get $\frac{2}{5}y$.

$$\frac{2}{5}y-2=-\frac{1}{5}$$

Add $2$ to both sides.

$$\frac{2}{5}y=-\frac{1}{5}+2$$

Convert $2$ to fraction $\frac{10}{5}$.

$$\frac{2}{5}y=-\frac{1}{5}+\frac{10}{5}$$

Since $-\frac{1}{5}$ and $\frac{10}{5}$ have the same denominator, add them by adding their numerators.

$$\frac{2}{5}y=\frac{-1+10}{5}$$

Add $-1$ and $10$ to get $9$.

$$\frac{2}{5}y=\frac{9}{5}$$

Multiply both sides by $\frac{5}{2}$, the reciprocal of $\frac{2}{5}$.

$$y=\frac{9}{5}\times \frac{5}{2}$$

Multiply $\frac{9}{5}$ times $\frac{5}{2}$ by multiplying numerator times numerator and denominator times denominator.

$$y=\frac{9\times 5}{5\times 2}$$

Cancel out $5$ in both numerator and denominator.

$$y=\frac{9}{2}$$

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