Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (hellokitty17):

If the quantity 4 times x times y cubed plus 8 times x squared times y to the fifth power all over 2 times x times y squared is completely simplified to 2xayb + 4xcyd, where a, b, c, and d represent integer exponents, what is the value of a? _______

OpenStudy (anonymous):

Do you have any idea?

OpenStudy (mathstudent55):

Can you use the equation editor or the draw tool to show what you mean? Is it this? \(\Large \dfrac{4xy^3 + 8x^2y^5}{2xy^2} = 2x^ay^b + x^cy^d\)

OpenStudy (hellokitty17):

sort of but the question is asking what is the value of a

oregonduck (oregonduck):

Value of a is zero 2xy2(2y+4xy3)2xy2 cancels the 2xy^2 on top and bottom =2y+4xy3=2x0y1+4x1y3 so x^a= x^0 a=0

OpenStudy (mathstudent55):

I understand what the question is asking. My question is if the above expression is correct.

OpenStudy (mathstudent55):

You can separate the left side into two fractions. \(\Large \dfrac{4xy^3 + 8x^2y^5}{2xy^2} = 2x^ay^b + x^cy^d\) \(\Large \dfrac{4xy^3}{2xy^2} + \dfrac{8x^2y^5}{2xy^2} = 2x^ay^b + x^cy^d\) Now reduce each fraction. When you divide powers with the same base, subtract the exponents.

OpenStudy (mathstudent55):

\(\Large \dfrac{4}{2} \times \dfrac{x}{x} \times \dfrac{y^3}{y^2} + \dfrac{8}{2} \times \dfrac{x^2}{x} \times \dfrac{y^5}{y^2} = 2x^ay^b + x^cy^d\)

OpenStudy (mathstudent55):

\(\Large 2x^{1-1}y^{3-2} + 4x^{2-1}y^{5-2} = 2x^ay^b + 4x^cy^d\)

OpenStudy (mathstudent55):

\(\Large 2x^{0}y^{1} + 4x^{1}y^{3} = 2x^ay^b + 4x^cy^d\)

OpenStudy (mathstudent55):

Now you can compare each exponent on the left side with each variable, a, b, c, and d, on the right side to see what value each of those variables has.

OpenStudy (mathstudent55):

\(\Large 2x^{\color{red}{0}}y^{\color{green}{1}} + 4x^{\color{blue}{1}}y^{\color{brown}{3}} = 2x^\color{red}{a}y^\color{green}{b} + 4x^\color{blue}{c}y^\color{brown}{d}\)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!