Question

Solve the equation $\dfrac{4}{z-1}-\dfrac{5}{z+2}=\dfrac{3}{z},z\ne 1,0,-2$ for z.

Answer 1

Hint: To solve the equation $\dfrac{4}{z-1}-\dfrac{5}{z+2}=\dfrac{3}{z}$ , firstly, we have to simplify the LHS using algebraic rules. Then, we have to use the algebraic identity $\left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab$ on the resultant denominator of the LHS. We have to cross multiply the result and simplify. We will then get a quadratic equation and solve for z.

Complete step by step answer:
We have to solve the equation $\dfrac{4}{z-1}-\dfrac{5}{z+2}=\dfrac{3}{z}$ . We know that when we subtract two fraction, say $\dfrac{a}{b}-\dfrac{c}{d}$ , we will obtain the result in the form $\dfrac{ad-bc}{bd}$ . Therefore, we can write the LHS of the given equation as
$\Rightarrow \dfrac{4\left( z+2 \right)-5\left( z-1 \right)}{\left( z-1 \right)\left( z+2 \right)}=\dfrac{3}{z}$
Let us apply distributive property on the numerator of the LHS.
$\begin{align}
  & \Rightarrow \dfrac{4z+8-\left( 5z-5 \right)}{\left( z-1 \right)\left( z+2 \right)}=\dfrac{3}{z} \\
 & \Rightarrow \dfrac{4z+8-5z+5}{\left( z-1 \right)\left( z+2 \right)}=\dfrac{3}{z} \\
\end{align}$
Let us add the like terms.
$\Rightarrow \dfrac{-z+13}{\left( z-1 \right)\left( z+2 \right)}=\dfrac{3}{z}$
We know that $\left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab$ . Therefore, we can write the denominator of the LHS as follows.
$\begin{align}
  & \Rightarrow \dfrac{-z+13}{{{z}^{2}}+\left( -1+2 \right)z+\left( -1\times 2 \right)}=\dfrac{3}{z} \\
 & \Rightarrow \dfrac{-z+13}{{{z}^{2}}+z-2}=\dfrac{3}{z} \\
\end{align}$
Let us apply cross multiplication.
$\Rightarrow z\left( -z+13 \right)=3\left( {{z}^{2}}+z-2 \right)$
We have to apply the distributive property.
$\begin{align}
  & \Rightarrow z\left( -z+13 \right)=3\left( {{z}^{2}}+z-2 \right) \\
 & \Rightarrow -{{z}^{2}}+13z=3{{z}^{2}}+3z-6 \\
\end{align}$
Let us take $-{{z}^{2}}$ and 13z to the RHS.
\[\begin{align}
  & \Rightarrow 0=3{{z}^{2}}+3z-6+{{z}^{2}}-13z \\
 & \Rightarrow 3{{z}^{2}}+3z-6+{{z}^{2}}-13z=0 \\
\end{align}\]
We have to add the like terms.
\[\Rightarrow 4{{z}^{2}}-10z-6=0\]
We can take the common factor 2 outside.
\[\Rightarrow 2\left( 2{{z}^{2}}-5z-3 \right)=0\]
We have to take 2 to the RHS.
\[\Rightarrow 2{{z}^{2}}-5z-3=0...\left( i \right)\]
Now, we have to solve the above equation. We can see that the above equation is of the form $a{{x}^{2}}+bx+c=0$ , whose solution is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-ac}}{2a}$ . When we compare equation (i) with the standard form, we will get $a=2,b=-5$ and $c=-3$ . Therefore, we can write the values of z as
$\begin{align}
  & z=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\times 2\times \left( -3 \right)}}{2\times 2} \\
 & \Rightarrow z=\dfrac{5\pm \sqrt{25+24}}{4} \\
 & \Rightarrow z=\dfrac{5\pm \sqrt{49}}{4} \\
 & \Rightarrow z=\dfrac{5\pm 7}{4} \\
\end{align}$
We can write z as
$\begin{align}
  & \Rightarrow z=\dfrac{5+7}{4},\dfrac{5-7}{4} \\
 & \Rightarrow z=\dfrac{12}{4},\dfrac{-2}{4} \\
 & \Rightarrow z=3,\dfrac{-1}{2} \\
\end{align}$
Hence, the solution of the equation $\dfrac{4}{z-1}-\dfrac{5}{z+2}=\dfrac{3}{z}$ is $z=3,\dfrac{-1}{2}$ .

Note: Students must thoroughly understand the concept of algebraic expression and the methods and rules to solve algebraic equations. They must know to solve equations and the methods associated with it.