How do you solve the following linear system: $3 x - 6 y = 9, y = \frac{1}{2} x - \frac{3}{2}$ $3x - 6y = 9 , y = 1/2x - 3/2$ ?

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How do you solve the following linear system: $3 x - 6 y = 9, y = \frac{1}{2} x - \frac{3}{2}$ $3x - 6y = 9 , y = 1/2x - 3/2$ ?

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Answer 1 · 2016-02-07T08:35:09

Solve by substitution:

$3 x - 6 y = 9, y = \frac{1}{2} x - \frac{3}{2}$ $3x - 6y = 9 , y = 1/2x - 3/2$

In the second equation the value of $y$ $y$ can be substituted to the first equation:

$\to 3 x - 6 (\frac{1}{2} x - \frac{3}{2}) = 9$ $rarr3x-6(1/2x-3/2)=9$

Use distributive property ( $a (b + c) = a b + a c$ $a(b+c)=ab+ac$ ):

$\to 3 x - (3 x - 9) = 9$ $rarr3x-(3x-9)=9$

$\to 3 x - 3 x + 9 = 9$ $rarr3x-3x+9=9$

$\to 9 = 9$ $rarr9=9$

So,if you see a equation ending like this,they mean that the value is possible for all real numbers.The value of $x$ $x$ is all the real numbers in mathematics.

Now we know that $x$ $x$ corresponds to all real numbers.So, $y$ $y$ is as same as the second equation:

$\Rightarrow y = \frac{1}{2} x - \frac{3}{2}$ $=>y=1/2x-3/2$

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How do you solve the following linear system: $3 x - 6 y = 9, y = \frac{1}{2} x - \frac{3}{2}$ $3x - 6y = 9 , y = 1/2x - 3/2$ ?

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How do you solve the following linear system: $3 x - 6 y = 9, y = \frac{1}{2} x - \frac{3}{2}$ $3x - 6y = 9 , y = 1/2x - 3/2$ ?