Can’t solve system of linear equations (that need simplification first)

algebra-precalculusself-learning

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?

Best Answer

The first mistake I see is here:

$$9x + 6 - 6x - 12 y = x - 3$$ $$9x -6x -x - 12y = -3 +6$$

The second line should read

$$9x-6x-x-12y=-3-6$$

you forgot to invert the sign when moving $6$ to the other side of the $=$ sign.


You were equally sloppy with the second equation, when you did this:

$$5x - 6x + 8y - 15y = 2 - 4$$

We get our second simplified linear equation: $$-x -7y = 2$$

In fact, $2-4$ is not equal to $2$.

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