Answer
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478.2k+ views
Hint: In this question, first we use the cross multiplication that converts fraction form into simple form then after use some algebraic property we will get the required answer.
Complete step-by-step answer:
Given, \[\dfrac{Z}{{Z + 15}} = \dfrac{4}{9}\]
Now, apply cross multiplication
\[
\dfrac{Z}{{Z + 15}} = \dfrac{4}{9} \\
\Rightarrow 9Z = 4\left( {Z + 15} \right) \\
\]
Now, open the bracket
\[
\Rightarrow 9Z = 4Z + 15 \times 4 \\
\Rightarrow 9Z = 4Z + 60............\left( 1 \right) \\
\]
As we can see from (1) equation, fraction form converts into simple form by cross multiplication. Now to solve the equation we use algebraic property.
Now, subtract 4Z from both sides of (1) equation.
\[
\Rightarrow 9Z - 4Z = 4Z + 60 - 4Z \\
\Rightarrow 5Z = 60 \\
\Rightarrow Z = \dfrac{{60}}{5} \\
\Rightarrow Z = 12 \\
\]
So, the solution of the equation is Z = 12.
Note: Whenever we face such types of problems we use some important points. First we convert the fraction form into simple form by using cross multiplication then subtract some expression from both sides of the equation. So, we will get the required answer.
Complete step-by-step answer:
Given, \[\dfrac{Z}{{Z + 15}} = \dfrac{4}{9}\]
Now, apply cross multiplication
\[
\dfrac{Z}{{Z + 15}} = \dfrac{4}{9} \\
\Rightarrow 9Z = 4\left( {Z + 15} \right) \\
\]
Now, open the bracket
\[
\Rightarrow 9Z = 4Z + 15 \times 4 \\
\Rightarrow 9Z = 4Z + 60............\left( 1 \right) \\
\]
As we can see from (1) equation, fraction form converts into simple form by cross multiplication. Now to solve the equation we use algebraic property.
Now, subtract 4Z from both sides of (1) equation.
\[
\Rightarrow 9Z - 4Z = 4Z + 60 - 4Z \\
\Rightarrow 5Z = 60 \\
\Rightarrow Z = \dfrac{{60}}{5} \\
\Rightarrow Z = 12 \\
\]
So, the solution of the equation is Z = 12.
Note: Whenever we face such types of problems we use some important points. First we convert the fraction form into simple form by using cross multiplication then subtract some expression from both sides of the equation. So, we will get the required answer.
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