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, $b=1.29$.

Updated on: 10-Oct-2022

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