Question

How do you solve $\dfrac{9}{{3x}} = \dfrac{4}{{x + 2}}$?

Answer 1

Hint: In the question above, we have an equation, where there exist two fractions, with variables in denominator, which means we cannot check for multiples in order to make the denominators equal. This means we have only one option to solve and find the value of the variable, and that is cross multiplication.
So, we are going to solve this equation with the help of cross multiplication, where usually the denominators of alternate sides shift and get multiplied with the numerators of the opposite sides.

Complete step-by-step solution:
We have an equation $\dfrac{9}{{3x}} = \dfrac{4}{{x + 2}}$, and we are supposed to solve it and find the value of the variable,
So, starting by placing the equation in front of each other,
$ \Rightarrow \dfrac{9}{{3x}} = \dfrac{4}{{x + 2}}$
Now, we will shift the denominators on the opposite sides and multiply it with the numerators of the opposite sides, since it is dividing the numerator in the prior side.
$ \Rightarrow 9(x + 2) = 4(3x)$
Opening the brackets and multiplying the numbers to find the product,
$ \Rightarrow 9x + 18 = 12x$
Shifting variables on one side and constants on the other side, we get,
$ \Rightarrow 12x - 9x = 18$
Subtracting the equation,
$ \Rightarrow 3x = 18$
Shifting the number multiplied with the variable and dividing the constant with it,
$ \Rightarrow x = \dfrac{{18}}{3}$
Finding the final value,
$ \Rightarrow x = 6$
Therefore, after solving an equation $\dfrac{9}{{3x}} = \dfrac{4}{{x + 2}}$ with the help of cross-multiplication, the value will be $x = 6$.

The required value of x is 6.

Note: In algebraic expressions, and equations, when there exist two fractions with an equal to sign in the middle, one can cross multiply by taking the denominator to the numerator of the other side, in order to simplify the given equation and find the value of the variable. Also, this is shown below,
$\dfrac{a}{b} = \dfrac{c}{d}$ where and $d$ are not zero, one can cross-multiply to get, $ad = cd$