Q13. - Algebra 1

Question

Solve. Assume that $\mathit{y}$ varies inversely as $\mathit{x}$.

If $\mathit{y}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}$ when $\mathit{x}\mathbf{=}\mathbf{6}$, find $\mathit{y}$ when $\mathit{x}\mathbf{=}\mathbf{-}\mathbf{16}$.

Step-by-Step Solution

Answer

When $x=-16, \mathit{y}\mathbf{=}\mathbf{-}\frac{\mathbf{9}}{\mathbf{16}}$. 

Step 1. Product rule for inverse variation.

If $(a_1,b_1)$ and $(a_2,b_2)$ are solutions of an inverse variation, then the products $a_1{b}_1$ and $a_2{b}_2$ are equal.

Step 2. Apply product rule.

Apply product rule for inverse variation.

$\begin{array}{c}{x}_1{y}_1=x_2{y}_2\\ \left(6\right)\left(1.5\right)=(-16)\left(y\right)\end{array}$

Step 3. Simplify.

Simplify the above obtained equation.

$\begin{array}{l}\frac{9}{-16}=\frac{-16y}{-16}\\ y=-\frac{9}{16}\end{array}$

Therefore, when $x=-16, y=-\frac{9}{16}$.