How do you solve by substitution $4 x + 6 y = 62$ $4x + 6y = 62$ and $10 x - 2 y = 2$ $10x - 2y = 2$ ?

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How do you solve by substitution $4 x + 6 y = 62$ $4x + 6y = 62$ and $10 x - 2 y = 2$ $10x - 2y = 2$ ?

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Answer 1 · 2015-06-05T06:44:04

$10 x - 2 y = 2$ $10x-2y=2$
$10 x = 2 + 2 y$ $10x=2+2y$ ( we add $2 y$ $2y$ on each side )
$10 x - 2 = 2 y$ $10x-2=2y$ ( we substract $2$ $2$ on each side )
$5 x - 1 = y$ $5x-1=y$ ( we divide by $2$ $2$ on each side )

Now that we have $y$ $y$ , we can use substitution in the first equation :

$4 x + 6 \cdot (5 x - 1) = 62$ $4x+6*(5x-1)=62$
$4 x + 30 x - 6 = 62$ $4x+30x-6=62$
$34 x - 6 = 62$ $34x-6=62$
$34 x = 62 + 6$ $34x=62+6$ ( we add $6$ $6$ on each side )
$34 x = 68$ $34x=68$
$x = \frac{68}{34} = 2$ $x=68/34=2$ ( we divide by $34$ $34$ on each side )

Now that we have $x$ $x$ , we can find $y$ $y$ :

We know that $y = 5 x - 1$ $y=5x-1$ , as we calculated it before :

$y = 5 \cdot 2 - 1 = 10 - 1 = 9$ $y=5*2-1=10-1=9$

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How do you solve by substitution $4 x + 6 y = 62$ $4x + 6y = 62$ and $10 x - 2 y = 2$ $10x - 2y = 2$ ?

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How do you solve by substitution 4x + 6y = 62 4x+6y=624x + 6y = 62 and 10x - 2y = 210x−2y=210x - 2y = 2?

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How do you solve by substitution $4 x + 6 y = 62$ $4x + 6y = 62$ and $10 x - 2 y = 2$ $10x - 2y = 2$ ?