SOLVING:, 1. y varies directly as z and inversely as x and y = 8 when z = 4 and x = 2158, 2. y varies directly as x and inversely as z and y = 6 when x = 3 and z = 2

Written by Winona Jul 6 · 19 sec read

SOLVING:

1. y varies directly as z and inversely as x and y = 8 when z = 4 and x = ⅘

2. y varies directly as x and inversely as z and y = 6 when x = 3 and z = 2

✏️COMBINED VARIATIONS

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$nderline{\mathbb{PROBLEMS:}}$

#1. y varies directly as z and inversely as x and y = 8 when z = 4 and x = ⅘.
#2. y varies directly as x and inversely as z and y = 6 when x = 3 and z = 2.

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$nderline{\mathbb{ANSWERS:}}$

$\qquad\LARGE\rm»\:\: 1. \:\green{k=\frac58}$

$\qquad\LARGE\rm»\:\: 2. \:\green{k=4}$

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$nderline{\mathbb{SOLUTIONS:}}$

– Write the equation for combined variations and then find the constant k.

#1.

$y = \frac{kz}{x} \\$

$8 = \frac{k(4)}{ \frac{4}{5} } \\$

$8 = \frac{4k}{ \frac{4}{5} } \\$

$8 = 4k \div \frac{4}{5} \\$

$8 = 4k \cdot \frac{5}{4} \\$

$8 = \frac{20k}{4} \\$

$8 = 5k$

$\frac{8}{5} = \frac{ \cancel5k}{ \cancel5} \\$

$\frac{8}{5} = k \\$

$\therefore$ The constant of the variation is 8/5.

$\rm$

#2.

$y = \frac{kx}{z} \\$

$6 = \frac{k(3)}{2} \\$

$6 = \frac{3k}{2} \\$

$6 \cdot2 = \frac{3k}{ \cancel2} \cdot \cancel2 \\$

$12 = 3k$

$\frac{12}{3} = \frac{ \cancel3k}{ \cancel3} \\$

$4 = k$

$\therefore$ The constant of the variation is 4.

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