$\frac{1}{2}x+\frac{1}{4}y=6 \ \ \ |\times 2 \\
\frac{1}{5}x+\frac{1}{3}y=1 \ \ \ |\times (-5) \\ \\
x+\frac{1}{2}y=12 \\
\underline{-x-\frac{5}{3}y=-5} \\
\frac{1}{2}y-\frac{5}{3}y=12-5 \\
\frac{3}{6}y-\frac{10}{6}y=7 \\
-\frac{7}{6}y=7 \\
y=7 \times (-\frac{6}{7}) \\
y=-6 \\ \\
x+\frac{1}{2}y=12 \\
x+\frac{1}{2} \times (-6)=12 \\
x-3=12 \\
x=12+3 \\
x=15 \\ \\
\boxed{(x,y)=(15,-6)}$

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3 years ago

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A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$