How do you solve 2/3(x-7)=4/5?

Jul 5, 2016

The solution is $x = \frac{41}{5}$.

Explanation:

We can first apply the multiplication on the left side of the $=$

$\frac{2}{3} \left(x - 7\right) = \frac{4}{5}$

$\frac{2}{3} \cdot x - \frac{2}{3} \cdot 7 = \frac{4}{5}$

We can continue calculating $\frac{2}{3} \cdot 7 = \frac{14}{3}$

$\frac{2}{3} x - \frac{2}{3} \cdot 7 = \frac{4}{5}$

$\frac{2}{3} x - \frac{14}{3} = \frac{4}{5}$

Now we want to isolate the term with the $x$ and to do this we can add on both sides $\frac{14}{3}$ in this way we can cancel the $- \frac{14}{3}$ from the side where there is the term with the $x$

$\frac{2}{3} x - \cancel{\frac{14}{3}} + \cancel{\frac{14}{3}} = \frac{4}{5} + \frac{14}{3}$

$\frac{2}{3} x = \frac{4}{5} + \frac{14}{3}$

On the right side we can calculate $\frac{4}{5} + \frac{14}{3}$

$\frac{2}{3} x = \frac{4}{5} + \frac{14}{3}$

$\frac{2}{3} x = \frac{3 \cdot 4 + 14 \cdot 5}{3 \cdot 5}$

$\frac{2}{3} x = \frac{12 + 70}{15}$

$\frac{2}{3} x = \frac{82}{15}$.

Finally we want to remove the coefficient in front of the $x$. We have that $x$ is multiplied by $2$ and divided by $3$, so we will divide by $2$ and multiply by $3$ both side of the equation

$\frac{2}{3} x = \frac{82}{15}$

$\cancel{\frac{2}{3}} \cdot \cancel{\frac{3}{2}} x = \frac{82}{15} \cdot \frac{3}{2}$

$x = \frac{82}{15} \cdot \frac{3}{2}$

we can simplify $\frac{82}{2} = 41$ and $\frac{3}{15} = \frac{1}{5}$

$x = \frac{82}{15} \cdot \frac{3}{2} = \frac{41}{5}$.

Sep 4, 2016

$x = \frac{41}{5}$

Explanation:

We have: $\frac{2}{3} \left(x - 7\right) = \frac{4}{5}$

Let's cross-multiply:

$\implies 10 \left(x - 7\right) = 12$

Then, let's divide both sides of the equation by $2$:

$\implies 5 \left(x - 7\right) = 6$

$\implies 5 x - 35 = 6$

Adding $35$ to both sides:

$\implies 5 x = 41$

Finally, let's solve for $x$ by dividing both sides by $5$:

$\implies x = \frac{41}{5}$