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Solve the system of equations $\sqrt{x+y}+\sqrt{2x+y+2}=7, 3x+2y=23$.

$x=\frac{23}{3}-\frac{2}{3} y$ so when I plug it in I get

$\sqrt{\frac{23}{3}-\frac{2}{3}y+y} +\sqrt{2\times\frac{23}{3}-\frac{2}{3}y+y+2}=7$

$\sqrt{\frac{23}{3}+\frac{y}{3}} + \sqrt{\frac{52}{3}-\frac{y}{3}}=7$

$2\sqrt{\frac{1196}{9}+\frac{29}{9}y-\frac{y^{2}}{9}} = 24$

$\frac{1196}{9}+\frac{29}{9}y-\frac{y^{2}}{9}=144$

$y^{2}-29y+100=0$

$y_1=25, x_1=-9$ or $y_2=4, x_2=5$.

The solution is correct, but is everything written properly? Should I put $\iff$ at the beginning of each row?

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  • $\begingroup$ Looks OK. See slightly different approach in solution below. $\endgroup$ – hypergeometric Nov 2 '16 at 8:20
  • $\begingroup$ Thanks, I edited it. Do I need $\iff$ sign in each row? $\endgroup$ – lmc Nov 2 '16 at 8:22
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Put $a=x+y,\; b=2x+y+2$.

Equations:

$$\begin{cases}\begin{align} &\sqrt{a}+\sqrt{b}=7&\Rightarrow a+b+2\sqrt{ab}&=49\\ &a+b-2=23&\Rightarrow a+b&=25\end{align}\bigg\rbrace\Rightarrow 25+2\sqrt{ab}=49\Rightarrow ab=144\\ \end{cases}\\$$

$$\begin{align} a+\frac {144}a&=25\\ a^2-25a+144&=0\\ (a-16)(a-9)&=0\\ a&=16, 9\\ b&=9, 16\\ x=b-a-2&=-9,5\\ y=a-x&=25,4\\ \Rightarrow (x,y)&=(-9,25), (5,4)\;\blacksquare\end{align}$$

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