How do you solve the system $y - \frac{4}{5} x = 1$ $y-4/5x=1$ and $y - x = 1$ $y-x=1$ using substitution?

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How do you solve the system $y - \frac{4}{5} x = 1$ $y-4/5x=1$ and $y - x = 1$ $y-x=1$ using substitution?

1 Answer

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Answer 1 · 2018-03-13T18:38:57

$x = 0$ $x=0$
$y = 1$ $y=1$

Explanation:

$y - x = 1 ∣ + x$ $y-x=1|+x$
$y = 1 + x$ $y=color(red)(1+x)$

$y - \frac{4}{5} x = 1$ $color(red)(y)-4/5x=1$
$1 + x - \frac{4}{5} x = 1 ∣ - 1$ $color(red)(1+x)-4/5x=1|-1$
$\frac{1}{5} x = 0 ∣ \cdot 5$ $1/5x=0|*5$
$x = 0$ $x=color(blue)(0)$

$y - x = 1 ∣ + x$ $y-color(blue)(x)=1|+x$
$y - 0 = 1$ $y-color(blue)(0)=1$
$y = 1$ $y=1$

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How do you solve the system $y - \frac{4}{5} x = 1$ $y-4/5x=1$ and $y - x = 1$ $y-x=1$ using substitution?

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How do you solve the system $y - \frac{4}{5} x = 1$ $y-4/5x=1$ and $y - x = 1$ $y-x=1$ using substitution?