Elimination method for solving (1/2)x - y = 1, 2x + 3y = 11

[Stine]

Use the elimination method to solve the following system:

. . .(1/2)x - y = 1
. . .2x + 3y = 11

When I multiply the top equation by "3", I get:

. . .1 1/2x - 3y = 3
. . . .2x + 3y = 11
. . .------------------
. . .3 1/2x. . .= 14

Then x = 4, so I plug that back into the first equation:

. . .(1/2)(4) - 3y = 3

. . .2 - 3y = 3

. . .2 - 3y - 2 = 3 - 2

. . .-3y = 1

. . .(-3y) / (-3) = 1 / (-3)

So y = 1/-3 = -1/3.

Right or wrong? Please Help! Thank you!
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Edited by stapel -- Reason for edit: formatting, clarity, etc

 

[stapel]

I'm surprised you haven't yet learned about improper fractions; they're much easier to work with than mixed number, especially in contexts such as this.

The solution to any "solving" question can be checked by plugging your answer back into the original exercise. When you plug "x = 4, y = -1/3" into the two original equations, does the solution work?

Eliz.

 

[Stine]

Thank you Stapel (Eliz) Well it works for the first part of the equation but when I plug it in the second I am not getting 11. I am getting =7

 

[stapel]

Your original second equation was "0.5x - y = 1". After you'd multiplied, you'd had "1.5x - 3y = 3". When you back-substituted, you conflated the two, substituting into "0.5x - 3y = 3".

You have to pick one or the other; this conflation is an entirely different equation, and will not solve the posted system.

Eliz.