How do you solve the following system of equations?: $10x – 30y = -1, 2x + y = 8$ ?

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Answer 1 · 2016-03-09T06:35:03

$x=239/70$
$y=41/35$

From the given:

$10x-30y=-1" "$ first equation
$2x+y=8" "$ second equation

Solve by method of substitution:
Start with the second equation

$y=8-2x$
Use this in the first equation

$10x-30y=-1" "$ first equation
#10x-30(8-2x)=-1

$10x-240+60x=-1$

$70x=239$

$x=239/70$

Now use this value to solve y:
Go back to $y=8-2x$

$y=8-2(239/70)$

$y=(560-478)/70$

$y=82/70$

$y=41/35$

checking
$10x-30y=-1" "$ first equation
$10(239/70)-30(41/35)=-1$

$239/7-246/7=-1$

$-7/7=-1$
$-1=-1$

checking:
$2x+y=8" "$ second equation

$2(239/70)+41/35=8$

$239/35+41/35=8$
$280/35=8$

$8=8$

God bless....I hope the explanation is useful

How do you solve the following system of equations?: 10x – 30y = -1, 2x + y = 8 10x – 30y = -1, 2x + y = 8 ?