Question

Find the value of y, from the equation$\dfrac{{9y + 5}}{{9y - 3}} = \left( {\dfrac{{ - 2}}{3}} \right)$.

Answer 1

Hint: Equation is a mathematical statement which has an equal symbol between the two algebraic expressions, representing that they have the same value. For example,$2x + 3 = 13$ have two algebraic expressions one is $2x + 3$ and the other is $13$and there is an equal symbol between them. In this equation, 2 is the coefficient, x is the variable,$3$ and $13$ are constants.

Complete step-by-step solution:
In this problem, we have to find the value of y from the equation, $ \Rightarrow \dfrac{{9y + 5}}{{9y - 3}} = \left( {\dfrac{{ - 2}}{3}} \right)$. We can solve for y by two ways. The one way is to solve it with a cross multiplication method and the other one is like this.
Now, we will multiply both sides of the equation with$(9y - 3)$,
$ \Rightarrow \dfrac{{9y + 5}}{{9y - 3}} \times (9y - 3) = \left( {\dfrac{{ - 2}}{3}} \right) \times (9y - 3)$
On further solving, we get,
$
   \Rightarrow 9y + 5 = \left( {\dfrac{{ - 2}}{3}} \right) \times (9y - 3) \\
   \Rightarrow 9y + 5 = \left( {\dfrac{{ - 2}}{3}} \right) \times 9y - \left( {\dfrac{{ - 2}}{3}} \right) \times 3 \\
   \Rightarrow 9y + 5 = - 6y - ( - 2) \\
   \Rightarrow 9y + 5 = - 6y + 2 \\
 $
Now, we subtract $5$from both sides,
$ \Rightarrow 9y + 5 - 5 = - 6y + 2 - 5$
After subtracting we get,
$ \Rightarrow 9y = - 6y - 3$
Now, add$6y$to both sides,
$ \Rightarrow 9y + 6y = - 6y - 3 + 6y$
After adding we get,
$ \Rightarrow 15y = - 3$
Now, divide by $15$on both sides,
$ \Rightarrow \dfrac{{15y}}{{15}} = \dfrac{{ - 3}}{{15}}$
On further solving, we get,
$ \Rightarrow y = \dfrac{{ - 1}}{5}$

Hence, the value of y is $\dfrac{{ - 1}}{5}$.

Note: We can also solve it by cross multiplication method $\dfrac{a}{b}$ $ = $ $\dfrac{c}{d}$, in this method a should be multiplied with d , in between there is an equal sign and then c is multiplied with b i.e.$(ad = bc)$ and on further solving we get our answer. Now, we will solve the given equation with this cross multiplication method. Let us consider a as $(9y + 5)$, b as$(9y - 3)$, c as $( - 2)$and d as$(3)$. Now, let us substitute these values into this method and then solve it.
$
   \Rightarrow \dfrac{{9y + 5}}{{9y - 3}} = \left( {\dfrac{{ - 2}}{3}} \right) \\
   \Rightarrow (9y + 5) \times 3 = (9y - 3) \times ( - 2) \\
 $
(Here, cross multiplication is applied.)
On further solving we get,
$
   \Rightarrow 9y \times 3 + 5 \times 3 = 9y \times ( - 2) - 3 \times ( - 2) \\
   \Rightarrow 27y + 15 = - 18y + 6 \\
   \Rightarrow 27y + 18y = 6 - 15 \\
   \Rightarrow 45y = - 9 \\
   \Rightarrow y = \dfrac{{ - 9}}{{45}} \\
   \therefore y = - \dfrac{1}{5} \\
 $