Q54. - Algebra 1

Question

Solve each equation. Check your solution. (Lesson 2-3)

$\frac{1}{2}z+7=16-\frac{3}{5}z$

Step-by-Step Solution

Answer

The solution of the given equation is $z=\frac{90}{11}$.

Step 1. Write the given equation.

The given equation is “$\frac{1}{2}z+7=16-\frac{3}{5}z$”.

Step 2. Solve the given equation.

Solving,

 

$\frac{1}{2}z+7=16-\frac{3}{5}z$  (Given equation)

 

$\frac{1}{2}z+7+\frac{3}{5}z=16-\frac{3}{5}z+\frac{3}{5}z$  (Add $\frac{3}{5}z$ to both sides)

 

$\frac{11}{10}z+7=16$  (Simplify)

 

$\frac{11}{10}z+7-7=16-7$  (Subtract 7 from both sides)

 

$\frac{11}{10}z=9$

 

$\frac{11}{10}z\times \frac{10}{11}=9\times \frac{10}{11}$  (Multiply both sides by $\frac{10}{11}$)

 

$z=\frac{90}{11}$  (Simplify)

 

So, $z=\frac{90}{11}$ is the solution of the given equation.

Step 3. Check the obtained solution.

Put $z=\frac{90}{11}$ in the given equation $\frac{1}{2}z+7=16-\frac{3}{5}z$ to get,

 

$\begin{array}{l}\frac{1}{2}\left(\frac{90}{11}\right)+7=16-\frac{3}{5}\left(\frac{90}{11}\right)\\ \frac{45}{11}+7=16-3\left(\frac{18}{11}\right)\\ \frac{45+77}{11}=16-\frac{54}{11}\\ \frac{122}{11}=\frac{176-54}{11}\\ \frac{122}{11}=\frac{122}{11}\end{array}$

 

Since the value obtained on the left-hand side and right-hand side are equal, the obtained value is indeed a solution of the given equation.