How do you solve $15 x - 20 y = - 1$ $15x-20y=-1$ and $10 y = 1 + 5 x$ $10y=1+5x$ using substitution?

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How do you solve $15 x - 20 y = - 1$ $15x-20y=-1$ and $10 y = 1 + 5 x$ $10y=1+5x$ using substitution?

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Answer 1 · 2016-08-03T23:05:12

$x = \frac{1}{5}$ $x=1/5$
$y = \frac{1}{5}$ $y=1/5$

Explanation:

$15 x - 20 y = - 1$ $15x-20y=-1$ or $15 x = 20 y - 1$ $15x=20y-1$
Multipling both sides of the equation $10 y = 1 + 5 x$ $10y=1+5x$ by $3$ $3$
we get
$30 y = 3 + 15 x$ $30y=3+15x$
or
$15 x = 30 y - 3$ $15x=30y-3$ And we have $15 x = 20 y - 1$ $15x=20y-1$
$30 y - 3 = 20 y - 1$ $30y-3=20y-1$
or
$30 y - 20 y = 3 - 1$ $30y-20y=3-1$
or
$10 y = 2$ $10y=2$
or
$y = \frac{2}{10}$ $y=2/10$
or
$y = \frac{1}{5}$ $y=1/5$ ===============Ans $1$ $1$
We have $15 x = 20 y - 1$ $15x=20y-1$
so
$15 x = \frac{20}{5} - 1$ $15x=20/5-1$
or
$15 x = 4 - 1$ $15x=4-1$
or
$15 x = 3$ $15x=3$
or
$x = \frac{3}{15}$ $x=3/15$
or
$x = \frac{1}{5}$ $x=1/5$ ================Ans $2$ $2$

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How do you solve $15 x - 20 y = - 1$ $15x-20y=-1$ and $10 y = 1 + 5 x$ $10y=1+5x$ using substitution?

1 Answer

Explanation:

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How do you solve $15 x - 20 y = - 1$ $15x-20y=-1$ and $10 y = 1 + 5 x$ $10y=1+5x$ using substitution?