Q32. - Algebra 1

Question

Solve each equation. State any extraneous solutions.

$\frac{\mathbf{h}\mathbf{-}\mathbf{2}}{\mathbf{h}}\mathbf{=}\frac{\mathbf{h}\mathbf{-}\mathbf{2}}{\mathbf{h}\mathbf{-}\mathbf{5}}$

Step-by-Step Solution

Answer

The solution of the equation is 2.

Step 1. State the concept for observational studies.

The linear equations in one variable is an equation which is expressed in the form of $ax+b=0$, where a and b are two integers, and x is a variable and has only one solution.

Step 2. Solve the equation.

The equation can be solved as follows:

$\begin{array}{c}\frac{h-2}{h}=\frac{h-2}{h-5}\\ \left(h-2\right)\left(h-5\right)=\left(h-2\right)h\\ h\left(h-5\right)-2\left(h-5\right)=\left(h-2\right)h\\ h^2-5h-2h+10=h^2-2h\\ -7h+10=-2h\\ -7h+2h+10=-2h+2h\\ -5h+10=0\\ -5h=-10\\ h=2\end{array}$

Step 3. Check the solution.

To check the solution, substitute $h=2$ for in the original equation,

$\begin{array}{c}\frac{h-2}{h}=\frac{h-2}{h-5}\\ \frac{2-2}{2}=\frac{2-2}{2-5}\\ \frac{0}{2}=\frac{0}{-3}\\ 0=0\\ LHS=RHS\end{array}$

Since both sides are equal, the equation is correct.