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Let $b$ be a positive number such that the system $ax+3y=15; x+ay=b$ has an infinite number of solutions. Find the value of $b$ to the nearest hundredth.
Given: Let $b$ be a positive number such that the system $ax+3y=15;\ x+ay=b$ has an infinite number of solutions.
To do: To find the value of $b$ to the nearest hundredth.
Solution:
As given $ax+3y=15$
$x+ay=b$
The system of equations have infinitely many solutions.
$\Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\Rightarrow \frac{a}{5}=\frac{3}{a}=\frac{1}{b}$
$a^2=15$
$\Rightarrow a=\sqrt{15}=3.872$
$3b=a$
$\Rightarrow 3b=3.872$
$\Rightarrow b=1.29$
Thus, $b=1.29$.
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