Q63. - Algebra 2

Question

Solve the equation and check the solution.

 

$\frac{5}{7}y-3=\frac{3}{7}y+1$

Step-by-Step Solution

Answer

The solution is $y=14$.

Step 1- Consider the equation.

Consider the equation $\frac{5}{7}y-3=\frac{3}{7}y+1$.

 

Multiply both sides of the equation by 7.

$\begin{array}{c}\frac{5}{7}y-3=\frac{3}{7}y+1\\ 7\times \frac{5}{7}y-3\left(7\right)=7\times \frac{3}{7}y+1\left(7\right)\\ 5y-21=3y+7\end{array}$

Step 2- Step description.

Now, simplify the equation obtained as follows:

$\begin{array}{c}5y-21=3y+7\\ 5y-3y=7+21\\ 2y=28\end{array}$

Step 3- Step description.

Divide both sides of the equation by 2.

$\begin{array}{l}2y=28\\ y=\frac{28}{2}\\ y=14\end{array}$

Therefore, the solution is $y=14$.

Step 4- Verify the solution.

Substitute $y=14$ in the equation and simplify as follows:

$\begin{array}{c}\frac{5}{7}y-3=\frac{3}{7}y+1\\ \frac{5}{7}\left(14\right)-3=\frac{3}{7}\left(14\right)+1\\ \frac{70}{7}-3=\frac{42}{7}+1\\ \frac{70-21}{7}=\frac{42+7}{7}\\ \frac{49}{7}=\frac{49}{7}\\ 7=7\end{array}$

Since the left-hand side of the equation is equal to the right-hand side of the equation, therefore, the solution is $y=14$.