Question

Simplify completely: the quantity 14 times x to the 5th power times y to the 4th power plus 21 times x to the third power times y to the 2nd power all over 7 times x to the third power times y

2x8y5 + 3x6y3

2x2y4 + 3xy

2x2y3 + 3y

2x2y3 + 3y2

Answer 1

so the question appear to be

\[\frac{14x^5y^4 + 21x^3y^2}{7x^3y} = \frac{14x^5y^4}{7x^3y} + \frac{21x^3y^2}{7x^3y}\]

look for the common factors that can be removed.

Answer 2

i don't get it

Answer 3

ok... I'll simplify it further

\[\frac{14}{7} \times \frac{x^5}{x^3} \times \frac{y^4}{y^1} + \frac{21}{7} \times \frac{x^3}{x^3} \times \frac{y^2}{y^1}\]

the index law for division of the same base is subtract the powers

\[\frac{x^a}{x^b} = x^{a - b}\]

just simplify the numbers.... 14/7 and 21/ 7

and apply the index law.

Answer 4

sorry i cant get it

Answer 5

ok... so here we go.. try and simplify this

\[14\div7 \times x^{5 - 3}\times y^{4 -1} + 21\div7 \times x^{3-3}y^{2-1}\]

Answer 6

2x^{2}y^{3}+3xy^{1}

Answer 7

well almost....

the law for a zero power is

\[x^0 = 1\]

so the 2nd term does not have an x since

\[x^{3 -3} = 1\]

Answer 8

hope this has all helped.

Answer 9

so the answer is wht?

Answer 10

ok... so the 1st part is correct

\[2x^2y^3 + 3 \times 1\times y\]

just simplify for your answer

Answer 11

thank you

Answer 12

whats the answer

Answer 13

It's 2x^y^3 + 3y