Systems of equations: solve x - y = -5, x + 3y = 3 in 3 ways

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mizkej
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Systems of equations: solve x - y = -5, x + 3y = 3 in 3 ways

Postby mizkej » Thu May 07, 2009 9:36 am

Hi everyone! Thanks for all of your help on the last question. I ended up with an A on my last test. Talking to you really helped and when I saw the questions on the test I knew what to do. :thumb:

I am stuck again. The question says to use three methods to solve for the following system of equations.

x-y=-5
x+3y=3

I need to solve by graphing, the addition method and the substitution method.

I think I have solved by graphing, please let me know if I have done this part correctly

the points I came up with are (0,-5) (-5,0) and for the second one (0,1) (1,0)

the addition method is where I beging to really struggle... this is what I have written

x-y=-5
x+3y=3
_______
2x+2y=-2

:shock:


as for the substitution method... which was easy in class.... I now.. dont have a clue :confused:

thanks in advance

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stapel_eliz
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Joined: Mon Dec 08, 2008 4:22 pm
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Postby stapel_eliz » Thu May 07, 2009 2:25 pm

mizkej wrote:...use three methods to solve for the following system of equations.

x-y=-5
x+3y=3

I need to solve by graphing, the addition method and the substitution method.

To solve by graphing, draw the two lines and look for where they cross.

To learn how to solve by addition and elimination, try here and here. :wink:

For instance:

. . . . .x + 3y = 3

. . . . .x = -3y + 3

Substituting:

. . . . .(-3y + 3) - y = -5

. . . . .-3y - y + 3 = -5

. . . . .-4y = -8

...and so forth. On the other hand:

1x - 1y = -5
1x + 3y = 3

-1x + 1y = 5
1x + 3y = 3
------------
4y = 8

...and so forth, by "elimination" of the x-variable. :D


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