Systems of Equations and Inequalties?

**Melcarthus** · November 12th 2012, 08:56 AM

I think I got the first steps right.

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**Melcarthus** · November 12th 2012, 08:57 AM

I think in the first equation i could of solved for y by dividing by x.

**Soroban** · November 12th 2012, 09:21 AM

Hello, Melcarthus!

Your canceling is incorrect.

$35.\;\begin{Bmatrix}xy^2 &=& 10^{11} & [1] \\ \\[-3mm] \dfrac{x^3}{y} &=& 10^{12} & [2]\end{Bmatrix}$

Divide [2] by [1]: . $\dfrac{\frac{x^3}{y}}{xy^2} \:=\:\frac{10^{12}}{10^{11}} \quad\Rightarrow\quad \frac{x^3}{xy^3} \:=\:10 \quad\Rightarrow\quad x^2 \:=\:10y^3 \;\;[3]$

Square [1]: . $(xy^2)^2 \:=\:(10^{11})^2 \quad\Rightarrow\quad x^2y^4 \:=\:10^{22}\;\;[4]$

Substitute [3] into [4]: . $(10y^3)(y^4) \:=\:10^{22} \quad\Rightarrow\quad y^7 \:=\:10^{21} \quad\Rightarrow\quad \boxed{y \:=\:10^3}$

Substitute into [1]: . $x(10^3)^2 \:=\:10^{11} \quad\Rightarrow\quad x(10^6) \:=\:10^{11} \quad\Rightarrow\quad \boxed{x \:=\:10^5}$

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