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Last updated date: 01st Dec 2023
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# If the product of $\left( \dfrac{7}{9}a{{b}^{2}} \right)\times \left( \dfrac{15}{7}a{{c}^{2}}b \right)\times \left( -\dfrac{3}{5}{{a}^{2}}c \right)$ is $\dfrac{-1}{x}{{a}^{4}}{{b}^{3}}{{c}^{3}}$ , then what is the value of $x$?

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Hint: To solve this question we need to know the concept of multiplication of algebraic numbers. We will be multiplying all the fractions and will be finding the numerator and denominator of the fraction. All the constant terms which are $a,b,c$ having certain powers which are given in the question will be added as the terms are being multiplied.

Complete step by step solution:
The question ask us to find the value of $x$, if the product of the three terms $\left( \dfrac{7}{9}a{{b}^{2}} \right)$,$\left( \dfrac{15}{7}a{{c}^{2}}b \right)$ and $\left( -\dfrac{3}{5}{{a}^{2}}c \right)$ which is given to us is $\dfrac{-1}{x}{{a}^{4}}{{b}^{3}}{{c}^{3}}$. We are asked to find the value of $x$ by equating the product to the fraction $\dfrac{-1}{x}{{a}^{4}}{{b}^{3}}{{c}^{3}}$. The first step to solve it this problem will be to calculate the product of the terms $\left( \dfrac{7}{9}a{{b}^{2}} \right)$,$\left( \dfrac{15}{7}a{{c}^{2}}b \right)$ and $\left( -\dfrac{3}{5}{{a}^{2}}c \right)$. On calculating it we get:
$\Rightarrow \left( \dfrac{7}{9}a{{b}^{2}} \right)\times \left( \dfrac{15}{7}a{{c}^{2}}b \right)\times \left( -\dfrac{3}{5}{{a}^{2}}c \right)$
We will have fractional terms having numbers, and since each constant term which is being multiplied has a certain power, the powers on the constant number will be added. On solving this we get:
$\Rightarrow \dfrac{7}{9}\times \dfrac{15}{7}\times \left( -\dfrac{3}{5} \right)\times {{a}^{1+1+2}}\times {{b}^{2+1+0}}\times {{c}^{0+2+1}}$
$\Rightarrow \dfrac{7\times 15\times \left( -3 \right)}{9\times 7\times 5}\times {{a}^{1+1+2}}\times {{b}^{2+1+0}}\times {{c}^{0+2+1}}$
Now we will be cancelling the common terms in the numerator and the denominator, on doing this we get:
$\Rightarrow \dfrac{-1}{1}\times {{a}^{4}}\times {{b}^{3}}\times {{c}^{3}}$
So the product we get is:
$\Rightarrow \dfrac{-1}{1}{{a}^{4}}{{b}^{3}}{{c}^{3}}$
The second step is to equate the product to the term $\dfrac{-1}{x}{{a}^{4}}{{b}^{3}}{{c}^{3}}$. Now equating the two we get:
$\Rightarrow \dfrac{-1}{1}{{a}^{4}}{{b}^{3}}{{c}^{3}}=\dfrac{-1}{x}{{a}^{4}}{{b}^{3}}{{c}^{3}}$
On analysing the two, the value of $x$ we get is $1$.
$\therefore$ If the product of $\left( \dfrac{7}{9}a{{b}^{2}} \right)\times \left( \dfrac{15}{7}a{{c}^{2}}b \right)\times \left( -\dfrac{3}{5}{{a}^{2}}c \right)$ is $\dfrac{-1}{x}{{a}^{4}}{{b}^{3}}{{c}^{3}}$ , then the value of $x$ is $1$.

Note: When a constant number having different powers are multiplied then the powers are added having the common base. For example if the three terms being multiplied are ${{a}^{m}},{{a}^{n}}$ and ${{a}^{p}}$ then the product result into ${{a}^{m+n+p}}$. On writing it mathematically we get:
$\Rightarrow {{a}^{m}}\times {{a}^{n}}\times {{a}^{n}}={{a}^{m+n+p}}$