If y varies directly with x, and If y = 12 when x = 15, how do you find x when y = 21?

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Answer 1 · 2016-02-19T21:57:21

$x = 26 \frac{1}{4}$ $x=26 1/4$

Use ratios such that $\frac{x_{1}}{y_{1}} = \frac{x_{2}}{y_{2}}$ $" " (x_1)/(y_1) = (x_2)/(y_2)$

Also note that the condition of 'TRUE' also holds for

$\frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}}$ $" "(y_1)/(x_1)=(y_2)/(x_2)$

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Need to find $x_{2}$ $x_2$ so it works much easier if you use $x$ $x$ as the numerator, giving:

$\frac{15}{12} = \frac{x}{21}$ $" " 15/12 = x/21$

Multiply both sides by 21 and we have:

$21 \times \frac{15}{12} = x \times \frac{21}{21}$ $" "21xx15/12= x xx21/21$

But $\frac{21}{21} = 1$ $" " 21/21 = 1$ giving:

$x = \frac{21 \times 15}{12} = 26 \frac{1}{4}$ $" "x=(21xx15)/12 = 26 1/4 " "$ which in decimal form is 26.25