How do you solve the system $3x + 6y = 30$ and $x + 6y = 20$ ?

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Answer 1 · 2017-04-08T20:49:23

Solve by elimination to find:

$x=5" "$ and $" "y=5/2$

Given:

${ (3x+6y=30), (x+6y=20) :}$

To eliminate the term in $y$ , we can subtract the second equation from the first to get:

$2x=10$

and hence:

$x=5$

Substituting this value of $x$ in the second equation, we get:

$5+6y=20$

Subtract $5$ from both sides to get:

$6y=15$

Divide both sides by $6$ to get:

$y = 15/6 = 5/2$

How do you solve the system 3x + 6y = 303x + 6y = 30 and x + 6y = 20x + 6y = 20?