2 years ago

Can anyone help me with this Algebra 2 question? Soon as possible!

Mathematics

1 answer:

prohojiy [21]2 years ago

6 0

Answer:

x=4 y=1

Step-by-step explanation:

3×4=12+7×1=19

4+1=5

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3 years ago

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It x, y, and z are positive intergers and 3x = 4y = 7z, then the least possible value of x + y + z is???

Dmitry [639]

$3x = 4y = 7z \\ \Rightarrow \begin{cases}3x - 4y = 0 \\4y - 7z = 0 \\ 7z - 3x = 0\end{cases} \\ \Leftrightarrow \begin{cases}x = \frac{4}{3} y\\4y - 7z = 0 \\ 7z - 3( \frac{4}{3}y) = 0\end{cases} \\ \Leftrightarrow \begin{cases}x = \frac{4}{3} y\\4y - 7z = 0 \\ 7z - 4y= 0\end{cases} \\ \Leftrightarrow \begin{cases}x = \frac{4}{3} y\\ y \\ z = \frac{4}{7}y \end{cases} \\ x,y,z\: are\: intergers\Rightarrow y=LCM(3,7)=21\Rightarrow x = 28,z = 12 \\ \Rightarrow x + y + z = 61$