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 · 2016-08-15T14:49:35

$x = \frac{3}{5}$ $x=3/5$ and $y = \frac{2}{5}$ $y=2/5$

Explanation:

We have two equations $15 x - 20 y = 1$ $15x-20y=1$ and $10 y = 1 + 5 x$ $10y=1+5x$ . Here we can easily substitute the value of $10 y$ $10y$ from the second equation into the first equation and doing so we get

$15 x - 20 y = 1$ $15x-20y=1$

$15 x - 2 \times 10 y = 1$ $15x-2xx10y=1$

$15 x - 2 \times (1 + 5 x) = 1$ $15x-2×(1+5x)=1$ or

$15 x - 2 - 10 x = 1$ $15x-2-10x=1$ or

$15 x - 10 x = 1 + 2$ $15x-10x=1+2$ or

$5 x = 3$ $5x=3$ and $x = \frac{3}{5}$ $x=3/5$ .

Hence $10 y = 1 + 5 x$ $10y=1+5x$

= $1 + 5 \times \frac{3}{5} = 1 + 3 = 4$ $1+5×3/5=1+3=4$ or

$y = \frac{4}{10} = \frac{2}{5}$ $y=4/10=2/5$

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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?