Algebra: Solving 2x - 7 = x + y, x + 3y = 2x - y - 1

josh90

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Aug 10, 2006
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I need to solve for y and x in the following system:

. . .2x - 7 = x + y
. . .x + 3y = 2x - y - 1

Thanks to anyone who can help.
 

pka

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We can rewrite these equations into a system.
\(\displaystyle \
\begin{array}{rcl}
x - y & = & 7 \\
x - 4y & = & 1 \\
\end{array}\)

Now solve the system.
 

stapel

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One method would be to get the variable terms in each equation onto the left-hand sides, and the constant terms onto the right-hand sides. Simplify. (For instance, the first equation would become "x - y = 7".) Then solve the resulting (simplified) system by whichever method you have been taught.

If you get stuck, please reply showing your steps, and specifying the method (Cramer's Rule? Gauss-Jordan elimination? substitution? etc) that you are using.

Thank you.

Eliz.
 

josh90

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Aug 10, 2006
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Algebra

Pka, you said the two equations are x-y=7 and x-4y=1. So I would like you to correct me if I am wrong in figuring it out.

y=-x+7,I am solving for x.

x-4(-x+7)=1

x+4x-28=1
5x-28=1
5x=29
x=5.8
y=1.2
Am I correct?
 

tkhunny

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You can check yourself. Substitute these values into the original equations and see if they work. If you get something like 3 = 7.2, then it's back to the drawing board. If you get 0 = 0, twice, you're done. No more need to doubt.
 

pka

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Or you could do this..
\(\displaystyle \
\begin{array}{rclc}
x - y & = & 7 & \[ 1 \]\\
x - 4y & = & 1 & \[ 2 \]\\
\end{array}\)

Substract [2] from [1].
\(\displaystyle \3 y = 6\)
 

tkhunny

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Re: Algebra

josh90 said:
x-y=7

Add 'y'

x = 7 + y

Subtract 7

x - 7 = y

How did you get what you got? One step at a time. Don't try to jump all over.
 
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