How do you work x=2y-2 and 3y=x+6 as substitution?

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How do you work x=2y-2 and 3y=x+6 as substitution?

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Answer 1 · 2018-04-13T01:45:10

$\Rightarrow x = 6$ $=>x = 6$
$\Rightarrow y = 4$ $=>y = 4$

Explanation:

Substitution involves putting a known variable expression into another expression.

In this case, you are given $x = 2 y - 2$ $x = 2y-2$ . You can substitute this into the second equation as follows:

$\Rightarrow 3 y = x + 6$ $=>3y = x+6$

$\Rightarrow 3 y = (2 y - 2) + 6$ $=>3y = (2y-2) + 6$

$\Rightarrow 3 y = 2 y + 4$ $=>3y = 2y + 4$

$\Rightarrow y = 4$ $=>y = 4$

Now, you can use this $y = 4$ $y = 4$ to find the value of $x$ $x$ . Going back to our expression for $x$ $x$ :

$\Rightarrow x = 2 y - 2$ $=>x = 2y - 2$

$\Rightarrow x = 2 (4) - 2$ $=>x = 2(4) -2$

$\Rightarrow x = 8 - 2$ $=>x = 8 - 2$

$\Rightarrow x = 6$ $=>x = 6$

So the solution is $x = 6$ $x = 6$ and $y = 4$ $y = 4$ .

Answer 2 · 2018-04-13T01:48:04

$y = 4$ $y=4$
$x = 6$ $x=6$

Explanation:

$x = 2 y - 2$ $x=2y-2$
$3 y = x + 6$ $3y=x+6$

The first one will be easiest to plug into the second, since it is already defined for $x$ $x$ . So, we will be substituting the $x$ $x$ in the second equation with the definition of $x$ $x$ provided in the first.

$3 y = (2 y - 2) + 6$ $3y=(2y-2)+6$

The parenthesis do not prevent you from combining like terms. In this case, the $- 2$ $-2$ and $6$ $6$ .

$3 y = 2 y + 4$ $3y=2y+4$

$3 y - 2 y = 4$ $3y-2y=4$

$y = 4$ $y=4$

Now plug the $y$ $y$ in to the top problem to solve for $x$ $x$ .

$x = 2 (4) - 2$ $x=2(4)-2$

$x = 6$ $x=6$

Results: $y = 4$ $y=4$ and $x = 6$ $x=6$

Answer 3 · 2018-04-13T01:51:12

$x = 6$ $x=6$
$y = 4$ $y=4$

Explanation:

$x = 2 y - 2$ $x=2y-2$
$3 y = x + 6$ $3y=x+6$

To solve for each variable through substitution we have to make one equation have only one variable in it (either $x$ $x$ or $y$ $y$ ). In order to do this, we have to put one variable in terms of the other. This time we got lucky because the first equation already puts $x$ $x$ in terms of $y$ $y$ . We just have to plug in $2 y - 2$ $2y-2$ for $x$ $x$ in the second equation.

$3 y = x + 6$ $3y=x+6$

$3 y = (2 y - 2) + 6$ $3y=(2y-2)+6$

$3 y = 2 y + 4$ $3y=2y+4$

Subtract $2 y$ $2y$ from each side

$y = 4$ $y=4$

Now plug this value into one of the original equations to find $x$ $x$ .

$x = 2 y - 2$ $x=2y-2$

$x = 2 (4) - 2$ $x=2(4)-2$

$x = 8 - 2$ $x=8-2$

$x = 6$ $x=6$

OR

$3 y = x + 6$ $3y=x+6$

$3 (4) = x + 6$ $3(4)=x+6$

$12 = x + 6$ $12=x+6$

$6 = x$ $6=x$

Answer

$x = 6$ $x=6$
$y = 4$ $y=4$

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How do you work x=2y-2 and 3y=x+6 as substitution?

3 Answers

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