$\frac{7x}{{x}^{2}+2xy+y}\ufeff+\frac{3x}{{x}^{2}+xy}=0$

Baladdaa9
2022-08-01
Answered

$\frac{7x}{{x}^{2}+2xy+y}\ufeff+\frac{3x}{{x}^{2}+xy}=0$

You can still ask an expert for help

Dominique Ferrell

Answered 2022-08-02
Author has **18** answers

$\frac{7x}{{x}^{2}+2xy+y}+\frac{3x}{{x}^{2}+xy}=0$

$\frac{7x}{{x}^{2}+2xy+y}=-\frac{3x}{{x}^{2}+xy}$

doing cross multiplication:

$7x({x}^{2}+xy)=({x}^{2}+2xy+y)(-3x)$

$7{x}^{3}+7{x}^{2}y=-3{x}^{3}-6{x}^{2}y-3xy$

isolate anything with y on one side and rest on other

$7{x}^{2}y+6{x}^{2}y+3xy=-3{x}^{3}-7{x}^{3}$

$13{x}^{2}y+3xy=-10{x}^{3}$

$y(13{x}^{2}+3x)=-10{x}^{3}$

$y=\frac{-10{x}^{3}}{13{x}^{2}+3x}$

$\frac{7x}{{x}^{2}+2xy+y}=-\frac{3x}{{x}^{2}+xy}$

doing cross multiplication:

$7x({x}^{2}+xy)=({x}^{2}+2xy+y)(-3x)$

$7{x}^{3}+7{x}^{2}y=-3{x}^{3}-6{x}^{2}y-3xy$

isolate anything with y on one side and rest on other

$7{x}^{2}y+6{x}^{2}y+3xy=-3{x}^{3}-7{x}^{3}$

$13{x}^{2}y+3xy=-10{x}^{3}$

$y(13{x}^{2}+3x)=-10{x}^{3}$

$y=\frac{-10{x}^{3}}{13{x}^{2}+3x}$

Bruno Thompson

Answered 2022-08-03
Author has **9** answers

$(7x)/({x}^{2}+2xy+y)+(3x)/({x}^{2}+xy)=0$

$(7x)/({x}^{2}+2xy+y)+(3x)/(x(x)+x(y))=0$

$(7x)/({x}^{2}+2xy+y)+(3x)/(x(x+y))=0$

Least common denominator: $x({x}^{2}+2xy+y)(x+y)$

$(7x)/({x}^{2}+2xy+y)\ast x({x}^{2}+2xy+y)(x+y)+(3x)/(x(x+y))\ast x({x}^{2}+2xy+y)(x+y)=0\ast x({x}^{2}+2xy+y)(x+y)$

$10{x}^{3}+13{x}^{2}y+3xy=0\ast x({x}^{2}+2xy+y)(x+y)$

$10{x}^{3}+13{x}^{2}y+3xy=0$

$13{x}^{2}y+3xy=-10{x}^{3}$

$13{x}^{2}y+3xy+10{x}^{3}=0$

$x(10{x}^{2})+x(13xy)+x(3y)=0$

$x(10{x}^{2}+13xy+3y)=0$

x=0

$10{x}^{2}+13xy+3y=0$

$13xy+3y=-10{x}^{2}$

$y(13x)+y(3)=-10{x}^{2}$

$y(13x+3)=-10{x}^{2}$

$(y(13x+3))/(13x+3)=-(10{x}^{2})/(13x+3)$

$y=-(10{x}^{2})/(13x+3)$

ANSWER $y=0,-(10{x}^{2})/(13x+3)$

$(7x)/({x}^{2}+2xy+y)+(3x)/(x(x)+x(y))=0$

$(7x)/({x}^{2}+2xy+y)+(3x)/(x(x+y))=0$

Least common denominator: $x({x}^{2}+2xy+y)(x+y)$

$(7x)/({x}^{2}+2xy+y)\ast x({x}^{2}+2xy+y)(x+y)+(3x)/(x(x+y))\ast x({x}^{2}+2xy+y)(x+y)=0\ast x({x}^{2}+2xy+y)(x+y)$

$10{x}^{3}+13{x}^{2}y+3xy=0\ast x({x}^{2}+2xy+y)(x+y)$

$10{x}^{3}+13{x}^{2}y+3xy=0$

$13{x}^{2}y+3xy=-10{x}^{3}$

$13{x}^{2}y+3xy+10{x}^{3}=0$

$x(10{x}^{2})+x(13xy)+x(3y)=0$

$x(10{x}^{2}+13xy+3y)=0$

x=0

$10{x}^{2}+13xy+3y=0$

$13xy+3y=-10{x}^{2}$

$y(13x)+y(3)=-10{x}^{2}$

$y(13x+3)=-10{x}^{2}$

$(y(13x+3))/(13x+3)=-(10{x}^{2})/(13x+3)$

$y=-(10{x}^{2})/(13x+3)$

ANSWER $y=0,-(10{x}^{2})/(13x+3)$

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A room can have up to 800 people. There are two types of tables: one, rectangular, seating 20 people; another, circular, that seats 9. If the organizers want to have approximately the same amount of rectangular and circular tables to seat the maximum of people possible, what combination of tables can they use?

Initially, I started off with a simple let statement, stating that $x$ will represent rectangular tables, and $y$ will represent circular tables. So my inequality so far would be something like $20x+9y\le 800$. I also know that $x\ge 0$ and $y\ge 0$. However, with this equation, I am not sure if I can actually find out an equal amount of rectangular and circular tables. I have also considered that this is a maximum/minimum problem that I have learned before, but the inequality I came up with is not a quadratic equation.

The question is multiple choice, and, substituting the correct values in, I can see that it works, but I have no idea how they arrived at the numbers. It seems to me that the question can be solved without this guess-and-check, so I'm interested in what you have to say about this problem.

Initially, I started off with a simple let statement, stating that $x$ will represent rectangular tables, and $y$ will represent circular tables. So my inequality so far would be something like $20x+9y\le 800$. I also know that $x\ge 0$ and $y\ge 0$. However, with this equation, I am not sure if I can actually find out an equal amount of rectangular and circular tables. I have also considered that this is a maximum/minimum problem that I have learned before, but the inequality I came up with is not a quadratic equation.

The question is multiple choice, and, substituting the correct values in, I can see that it works, but I have no idea how they arrived at the numbers. It seems to me that the question can be solved without this guess-and-check, so I'm interested in what you have to say about this problem.