How do you solve $8 y - 1 = x, 3 x = 2 y$ $8y-1=x, 3x=2y$ ?

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How do you solve $8 y - 1 = x, 3 x = 2 y$ $8y-1=x, 3x=2y$ ?

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Answer 1 · 2016-11-26T13:37:46

I found: $x = \frac{1}{11} and y = \frac{3}{22}$ $x=1/11 and y=3/22$

Explanation:

We can trysubstituting the first equation, for $x$ $x$ , intothe second to get:
$3 \cdot (8 y - 1) = 2 y$ $3*(color(red)(8y-1))=2y$
Solve for $y$ $y$ :
$24 y - 3 = 2 y$ $24y-3=2y$
$22 y = 3$ $22y=3$
So: $y = \frac{3}{22}$ $y=3/22$
Using this value back into the first equation:
$8 \cdot (\frac{3}{22}) - 1 = x$ $8*(3/22)-1=x$
$x = \frac{24 - 22}{22} = \frac{2}{22} = \frac{1}{11}$ $x=(24-22)/22=2/22=1/11$

Answer 2 · 2016-11-26T13:41:12

$(x, y) = (\frac{1}{11}, \frac{3}{22})$ $(x,y)=(1/11,3/22)$

Explanation:

We have two equations:

$8 y - 1 = x$ $8y-1=x$
$3 x = 2 y$ $3x=2y$

We can substitute in the value of x (in terms of y) from the first equation into the second, and then solve for y. Like this:

$3 (8 y - 1) = 2 y$ $3(8y-1)=2y$

$24 y - 3 = 2 y$ $24y-3=2y$

$22 y = 3$ $22y=3$

$y = \frac{3}{22}$ $y=3/22$

And then we substitute into either of the initial equations (I'll do both to show we'll get the same answer for x):

$8 y - 1 = x$ $8y-1=x$

$8 (\frac{3}{22}) - 1 (1) = x$ $8(3/22)-1(1)=x$

$\frac{24}{22} - 1 (\frac{22}{22}) = x$ $24/22-1(22/22)=x$

$\frac{24}{22} - \frac{22}{22} = \frac{2}{22} = \frac{1}{11} = x$ $24/22-22/22=2/22=1/11=x$

or:

$3 x = 2 y$ $3x=2y$

$3 x = 2 (\frac{3}{22})$ $3x=2(3/22)$

$3 x = \frac{3}{11}$ $3x=3/11$

$x = \frac{1}{11}$ $x=1/11$

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How do you solve $8 y - 1 = x, 3 x = 2 y$ $8y-1=x, 3x=2y$ ?

2 Answers

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How do you solve 8y-1=x, 3x=2y8y−1=x,3x=2y8y-1=x, 3x=2y?

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How do you solve $8 y - 1 = x, 3 x = 2 y$ $8y-1=x, 3x=2y$ ?