How do you solve by substitution $3 x + 2 y = 6$ $3x + 2y = 6$ and $x + 2 y = - 6$ $x + 2y = -6$ ?

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Answer 1 · 2015-06-13T22:03:48

$x = 6$ $x=6$
$y = - 6$ $y=-6$

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Answer 2 · 2015-06-13T22:05:53

$x = 6$ $x=6$
$y = - 6$ $y=-6$

To solve a system by substitution, you need to replace one unknown by a function of the other, using one of the equations.

Example:
$x + 2 y = - 6$ $x+2y=-6$
becomes
$x = - 2 y - 6$ $x=-2y-6$

Then, in the other equation, you replace $x$ $x$ with its $y$ $y$ -depending expression:

$3 x + 2 y = 6$ $3x+2y=6$
becomes
$2 y - 6 y - 18 = 6$ $2y-6y-18=6$
$\to - 4 y = 24$ $rarr -4y=24$
$\to y = - 6$ $rarr y=-6$

Now that you have the value of $y$ $y$ , you can find the value of $x$ $x$ :

$x = - 2 y - 6$ $x=-2y-6$
becomes
$x = - 2 \cdot (- 6) - 6$ $x=-2*(-6)-6$
$\to x = 12 - 6 = 6$ $rarr x=12-6=6$

How do you solve by substitution 3x + 2y = 63x+2y=63x + 2y = 6 and x + 2y = -6x+2y=−6x + 2y = -6?