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Let b be a positive number such that the system $ax+3y=1,\ 5x+ay=b$ has an infinite number of solutions. Find the value of b to the nearest hundredth.
Given: $b$ is a positive number such that the system $ax+3y=1,\ 5x+ay=b$ has an infinite number of solutions.
To do: To find the value of b to the nearest hundredth.
Solution:
Condition for pair of linear equations $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$
to have infinite solutions is:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\therefore \frac{a}{5}=\frac{3}{a}=\frac{1}{b}$
$\Rightarrow a^2=15$
$\Rightarrow a=\sqrt{15}$
and $b=\frac{a}{3}=\frac{\sqrt{15}}{3}=\frac{\sqrt{5×3}}{3}$
$=\frac{\sqrt{5}}{\sqrt{3}}$
$\Rightarrow b\approx 1.29$
Therefore, $a^2=15$ and $b\approx 1.29$.
Updated on: 10-Oct-2022
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