<s>[INST] Can't solve system of linear equations (that need simplification first) I'm self-studying from Stroud & Booth's amazing "Engineering Mathematics", and am stuck on a problem at the end of the "Linear Equations" chapter.
I've a system of two linear equations:
$$\frac{3x+2}{4} - \frac{x+2y}{2} = \frac{x-3}{12}$$
$$\frac{2y+1}{5} + \frac{x-3y}{4} = \frac{3x+1}{10}$$
So, these two first need to be simplified. I assume that the LSM for the first one (for 2, 4 and 12) is 12, so we have:
$$12\frac{3x+2}{4} - 12\frac{x+2y}{2} = 12\frac{x-3}{12}$$
Simplifying further, we have:
$$3(3x+2) - 6(x+2y) = (x-3)$$
$$9x + 6 - 6x - 12 y = x - 3$$
$$9x -6x -x - 12y = -3 +6$$
Finally, we get our first simplified linear equation:
$$2x - 12y = 3$$
Now, onto the second one. The LSM of 5, 4 and 10 is 20, so we have:
$$20\frac{2y+1}{5} + 20\frac{x-3y}{4} = 20\frac{3x+1}{10}$$
Simplifying further, we have:
$$4(2y+1) + 5(x-3y) = 2(3x + 1)$$
$$8y + 4 + 5x - 15y = 6x + 2$$
$$5x - 6x + 8y - 15y = 2 - 4$$
We get our second simplified linear equation:
$$-x -7y = 2$$
Now we can solve our system of linear equations:
$$2x - 12y = 3$$
$$-x -7y = 4$$
Multiplying the second one by 2:
$$2x - 12y = 3$$
$$-2x -14y = 4$$
Now, we add the two equations, and get:
$$-26y = 7$$
Solving for $y$, we get:
$$y = -\frac{7}{26}$$
which I'm fairly certain is not a correct answer.
Can anyone see where I'm going wrong here?
 [/INST] Finally, we get our first simplified linear equation: 
This step is wrong:
$2x-12y=-9$  this should be your first equation. 
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