Question

Subtract the following:
a.4/3-4/3x²-5 from 2/3x³+6/7x²+2x-3​

Answer 1

Answer:

Here is your answer with step by step explanation.

Answer 2

Answer:

Subtracting $a\frac{4}{3}-\frac{4}{3x^{2} } -5$ from $\frac{2}{3x^{3} }+\frac{6}{7x^{2} } +2x-3$ we get $\frac{2}{3x^{3} } +\frac{46}{21x^{2} } +2x+\frac{2}{3}$.

Step-by-step explanation:

In the given question we have two polynomials, we are asked to subtract one from another. For this we need to operate the terms together which have same power of x, for example if we have a term with power of x as 2 in both the equations, then we need to subtract one from the other.

We can write the operation as below:

$\frac{2}{3x^{3} }+\frac{6}{7x^{2} } +2x-3-(\frac{4}{3} -\frac{4}{3x^{2} } -5)\\=\frac{2}{3x^{3} }+\frac{6}{7x^{2} } +2x-3-\frac{4}{3}+\frac{4}{3x^{2} } +5\\=\frac{2}{3x^{3} }+(\frac{6}{7x^{2} }+\frac{4}{3x^{2} } )+2x-(3+\frac{4}{3} -5)\\=\frac{2}{3x^{3} } +(\frac{18+28}{21x^{2} } )+2x-(\frac{9+4-15}{3})\\=\frac{2}{3x^{3} } +\frac{46}{21x^{2} } +2x+\frac{2}{3}$

Therefore the final result that we get is $\frac{2}{3x^{3} }+\frac{46}{21x^{2} } +2x+\frac{2}{3}$.