How do you solve $3 x + 6 y = 9$ $3x+6y=9$ and $2 x + 8 y = 10$ $2x+8y=10$ using substitution?

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How do you solve $3 x + 6 y = 9$ $3x+6y=9$ and $2 x + 8 y = 10$ $2x+8y=10$ using substitution?

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Answer 1 · 2017-03-30T05:41:05

$x = 1; y = 1$ $x=1; y=1$

Explanation:

$[Eqn 1] 3 x + 6 y = 9; [Eqn 2] 2 x + 8 y = 10$ $["Eqn 1"] 3x+6y=9; ["Eqn 2"] 2x+8y=10$

Multiply [Eqn 1] by 2; [Eqn 2] by 3, so $x$ $x$ can be in the same multiple:

$[Eqn 3] 6 x + 12 y = 18$ $["Eqn 3"] 6x+12y=18$

$[Eqn 4] 6 x + 24 y = 30$ $["Eqn 4"] 6x+24y=30$

Hence, we can find $y$ $y$ by $[Eqn 4] - [Eqn 3]$ $"[Eqn 4]"-"[Eqn 3]"$ :
$6 x - 6 x + 24 y - 12 y = 30 - 18$ $6x-6x+24y-12y=30-18$
$12 y = 12$ $12y=12$
$y = 1$ $y=1$

Now, just substitute $y$ $y$ into either [Eqn 1] or [2]:
For example, I will subst. into [Eqn 1].

$3 x + 6 (1) = 9$ $3x+6(1)=9$
$3 x = 9 - 6 = 3$ $3x=9-6=3$
$x = 1$ $x=1$

Hence, the solution:
$x = 1; y = 1$ $x=1; y=1$

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How do you solve $3 x + 6 y = 9$ $3x+6y=9$ and $2 x + 8 y = 10$ $2x+8y=10$ using substitution?

1 Answer

Explanation:

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How do you solve $3 x + 6 y = 9$ $3x+6y=9$ and $2 x + 8 y = 10$ $2x+8y=10$ using substitution?