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# m10 Q02

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Status: And the Prep starts again...
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22 Apr 2012, 22:57
If $$2X + 3Y = 12$$ , what is $$X$$ ?

1) $$X = 6 - (\frac{3}{2})Y$$
2) $$Y = 4 - (\frac{3}{2})X$$

Why is statement 1 not sufficient?

Can I not find the value of Y and substitute in $$2X + 3Y = 12$$. I did this, please tell me what mistake I am making

$$2X = 12-3Y$$
From Statement 1 $$X = 6 - (\frac{3}{2})Y$$
Which can be simplified to $$X = (\frac{9}{2})Y$$
Therefore, $$Y = (\frac{2}{9})X$$

Now if i can substitute this Y in $$2X = 12-3Y$$
$$2X = 12-\frac{6}{9}X$$
$$2X = 12-\frac{2}{3}X$$
$$2X+\frac{2}{3}X = 12$$
$$\frac{8X}{3} = 12$$
$$8X = 36$$
$$X = \frac{36}{8}$$
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22 Apr 2012, 23:35
ENAFEX wrote:
If $$2X + 3Y = 12$$ , what is $$X$$ ?

1) $$X = 6 - (\frac{3}{2})Y$$
2) $$Y = 4 - (\frac{3}{2})X$$

Why is statement 1 not sufficient?

Can I not find the value of Y and substitute in $$2X + 3Y = 12$$. I did this, please tell me what mistake I am making

$$2X = 12-3Y$$
From Statement 1 $$X = 6 - (\frac{3}{2})Y$$
Which can be simplified to $$X = (\frac{9}{2})Y$$
Therefore, $$Y = (\frac{2}{9})X$$

Now if i can substitute this Y in $$2X = 12-3Y$$
$$2X = 12-\frac{6}{9}X$$
$$2X = 12-\frac{2}{3}X$$
$$2X+\frac{2}{3}X = 12$$
$$\frac{8X}{3} = 12$$
$$8X = 36$$
$$X = \frac{36}{8}$$

The red part is not correct: $$x = 6 - \frac{3}{2}y$$ --> $$x=\frac{12-3y}{2}$$, not $$x=\frac{9}{2}y$$

If $$2x + 3y = 12$$, what is $$x$$ ?

(1) $$x = 6 - (\frac{3}{2})y$$ --> multiply both parts by 2: $$2x=12-3y$$ --> rearrange: $$2x + 3y = 12$$, so we have the same equation as in the stem. Not sufficient.

(2) $$y = 4 - (\frac{3}{2})x$$ --> $$2y=8-3x$$ --> $$3x+2y=8$$. We have two distinct linear equations, hence we can get the single numerical value of $$x$$. Sufficient.

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23 Apr 2012, 02:22
That was silly!!

Sorry to waste your time. :D GGRRR
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Re: m10 Q02   [#permalink] 23 Apr 2012, 02:22
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# m10 Q02

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