Question

$\frac{3}{5}p+\frac{1}{5}(40-p)=0$what value makes $p$ true

Answer 1

The value of $p$ that makes the equation true is $-20$

Step-by-step explanation:

To solve an equation of one variable

• Simplify the equation if necessary
• Use the mathematics operations to put the variable in one side and the numerical term in the other side
• Divide both sides by the coefficient of the variable to find its value

Since The equation is $\frac{3}{5}p+\frac{1}{5}(40-p)=0$

Simplify the left hand side

$\begin{array}{rcl}\frac{1}{5}(40-p)& =& \frac{1}{5}40-\frac{1}{5}p\\ & =& 8-\frac{1}{5}p\end{array}$

Substitute $\frac{1}{5}(40-p)$ in the equation by $8-\frac{1}{5}p$

$\therefore \frac{3}{5}p+8-\frac{1}{5}p=0$

Add the like terms

$$\begin{gathered} \left(\frac{3}{5}p-\frac{1}{5}p\right)+8=0 \\ \frac{2}{5}p+8=0 \end{gathered}$$

Subtract $8$ from both sides

$\therefore \frac{2}{5}p=-8$

Divide both sides by $\frac{2}{5}$

$\therefore p=-20$

To check the answer substitute p by $-20$ in the equation, the left hand side is equal to the right hand side, then the answer is right

∵ left hand side

$$\begin{gathered} =\frac{3}{5}(-20)+\frac{1}{5}(40-(-20\left)\right) \\ =\frac{-60}{5}+\frac{1}{5}(40+20) \\ =\frac{-60}{5}+\frac{60}{5} \\ =0=L.H.S \end{gathered}$$

Since Right hand side $=0$

Therefore Left hand side $=$ Right hand side

The value of $p$ that makes the equation true is $-20$