Use substitution to solve:

x=3y-2

2x+4y=16

x-3y=2

(+)2x+4y=16

----------------

3x+1y =18

3x+1y-1y+18-y

3x/3 = 18/3

x=6

6-3y=2

6-3y+3y=2+3y

6/6=6y/6

y=0

Is my answer is right?

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- July 20th 2005, 08:19 AMHayabusaPlease check if this is right?
Use substitution to solve:

x=3y-2

2x+4y=16

x-3y=2

(+)2x+4y=16

----------------

3x+1y =18

3x+1y-1y+18-y

3x/3 = 18/3

x=6

6-3y=2

6-3y+3y=2+3y

6/6=6y/6

y=0

Is my answer is right? - July 20th 2005, 08:27 PMMath HelpQuote:

Originally Posted by**Hayabusa**

**eq1**x=3y-2

**eq2**2x+4y=16

eq2 - 2(eq1)

..2x+4y=16

-2(x=3y-2)

-----------

0+4y=16-3y+2

or

4y=18-3y

7y=18

y=18/7, now plug in y to either equation to find x - July 20th 2005, 11:18 PMrgepQuote:

Originally Posted by**Hayabusa**

Since you're told to use substitution I think you were being asked to use x=3y-2 in the second equation, so that 2(3y-2)+4y=16 giving 10y-4=16, 10y=20, y=2 and then x=4. - July 20th 2005, 11:20 PMrgepQuote:

Originally Posted by**Math Help**