Chapter 5.2, Problem 22P

Suppose $y$ varies directly with x. Write and graph a direct variation equation that relates $x$ and $y$. $y=\frac{4}{3}$

when $x=-4$.

To determine

To find : direct variation equation and its graph that relates x and y

Answer to Problem 22P

The direct variation equation that relations x and y is $\boxed{y=\frac{1}{3}x}$

Explanation of Solution

Given information : $y=-\frac{4}{3}$ when $x=-4$

Concept Involved:

Use the fact that if “ y varies directly with x ” then the equation relating x and y is $y=kx$ where $k$ is constant of proportionality to set up an equation.

Substitute the given values of x and y in to the equation.

Solve the equation for k .

Back Substitute the k value in equation $y=kx$ to give the relationship between x and y .

To graph the equation, identify any two points satisfying the equation by picking some x values substituting in equation and finding the corresponding y value.

Calculation: Use the fact that if “ y varies directly with x ” then the equation relating x and y is $y=kx$ where $k$ is constant of proportionality to set up an equation.

$y=kx$

Substitute $y=-\frac{4}{3}$ & $x=-4$ in to the equation

$-\frac{4}{3}=k\left(-4\right)$

Solve for k

Use the symmetric property of identity which states that if a=b, then b=a to rewrite the equation

$-4k=-\frac{4}{3}$

To get k alone divide by -4 on both sides of the equation

$\frac{-4k}{-4}=\frac{-4/3}{-4}$

Simplify fraction in left side and flip the bottom fraction and multiply with top in the right side of the equation

$k=\frac{-4}{3}\cdot \frac{1}{-4}$

Multiply numerator with numerator and denominator with denominator in right side of the equation

$k=\frac{-4}{-12}$

Divide the numerator and denominator by common factor of the fraction

  $k=\frac{-4÷-4}{-12÷-4}$

Simplify fraction in right side of the equation

$k=\frac{1}{3}$

Back Substitute the k value in equation $y=kx$ to give the relationship between x and y .

$y=\frac{1}{3}x$

To graph the equation, identify any two points satisfying the equation by picking some x values substituting in equation and finding the corresponding y value.

$x$	$y=\frac{1}{3}x$	$\left(x,y\right)$
$-3$	$\frac{1}{3}\cdot -3=-1$	$\left(-3,-1\right)$
$0$	$\frac{1}{3}\cdot 0=0$	$\left(0,0\right)$
$3$	$\frac{1}{3}\cdot 3=1$	$\left(3,1\right)$

Graph:

Conclusion:

The direct variation equation that relations x and y is $\boxed{y=\frac{1}{3}x}$