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

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

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Answer 1 · 2016-03-20T11:22:05

$x,y=-1/5$

Explanation:

$color(blue)(15x-20y=1$

$color(blue)(5y=1+10x$

Use the second equation

$rarr5y=1+10x$

Divide both sides by $5$

$rarr(cancel5y)/cancel5=(1+10x)/5$

$rarry=1/5+2x$

Now substitute this value to the first equation

$rarr15x-20(1/5+2x)=1$

Use the distributive property $color(brown)(a(b+c)=ab+ac$

$rarr15x-20/5-40x=1$

$rarr15x-4-40x=1$

$rarr-25x-4=1$

Add $4$ both sides

$rarr-25xcancel(-4+4)=1+4$

$rarr-25x=5$

$color(green)(rArrx=-5/25=-1/5$

Substitute the value of $x$ to the second equation

$rarr5y=1+10(-1/5)$

$rarr5y=1-10/5$

$rarr5y=1-2$

$rarr5y=-1$

$color(green)(y=-1/5$

$:. color(indigo)(ul bar |x=y=-1/5|$

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

1 Answer

Explanation:

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