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Consider the system \begin{align*} x + y + 2z &= 2 \\ 2x + 3y - z &= 5 \\ 3x + 4y + z &= b \end{align*} (a) For what values of $b$ does the system have a solution? Using this value of $b$ find the solution.

(b) For what values of $h$ and $k$ does the system \begin{align*} x + hy &= 2 \\ 4x + 8y &= k \end{align*} have (i) No solutions and (ii) Infinitely many solutions?

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What are you having difficulty with here? –  Mark Bennet Feb 2 '13 at 19:46
    
I am having difficult with finding b first in question part (a). And all question part (b) –  Mike Feb 2 '13 at 19:49
    
the first line is correct as from the question : x + y + 2z = 2 –  Mike Feb 2 '13 at 19:51
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How would you normally go about solving equations like this, if they had numbers rather than $b, h, k$? And can you check the signs of the terms in $z$ in the first question, which looks as though the second (or the first and third) might be negative in the original question ... –  Mark Bennet Feb 2 '13 at 19:54
    
You are right Mark apology the negative sign should have been on the second equation i.e ( 2x + 3y - z = 5 ) –  Mike Feb 2 '13 at 19:58
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4 Answers

Hint for b: A system containing of equations of two lines:

$$ax+by=c\\a'x+b'y=c'$$ has always no soultion iff $$\frac{a}{a'}=\frac{b}{b'}\neq\frac{c}{c'}$$ and has infinite many solutions iff $$\frac{a}{a'}=\frac{b}{b'}=\frac{c}{c'}$$

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I like this approach! +1 –  amWhy Feb 3 '13 at 0:21
    
@amWhy: Many times, I think simple while the others here do thru formal ways. Maybe, my method of considering and constructing a proof for a mathematical questions is wrong. ;) –  B. S. Feb 3 '13 at 3:41
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Hint: for the first question, what happens if you add the first two equations? For the second, you can have no solutions if the lines are parallel and infinite solutions if the lines are the same. What values of $h$ and $k$ make one equation a multiple of the other?

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I hope, I could show your words in the following fractions and relations. + –  B. S. Feb 3 '13 at 3:43
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Another way of doing it :

Using Augmented Matrix Method \begin{align} [A|RHS]&= \begin{pmatrix} 1 &1 &2 &2\\ 2 &3 &-1 &5\\ 3 &4 &1 &b\\ \end{pmatrix}\\ \end{align} To Find : Reduced Form of the Augmented Matrix \begin{align} \text{Step 1:}\\ &\begin{pmatrix} 1 &1 &2 &2\\ 0 &1 &-5 &1\\ 0 &1 &-5 &b-6\\ \end{pmatrix}\\ \text{Step 2:}\\ &\begin{pmatrix} 1 &1 &2 &2\\ 0 &1 &-5 &1\\ 0 &0 &0 &b-7\\ \end{pmatrix}\\ \end{align} Rank of Original Matrix is 2. Rank of Augmented Matrix must also be 2 in order for solutions to exist. So, what must be the value of b?

Similarly, you can do the second one.

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In the 2nd question it has infinite soln.for $h=2$ and $k=8$. You can think it of as two lines in the x-y plane. So they will have infinite no. Of soln. if they represent the same line and no solution if they are parallel lines.

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And no solution for h=2and k$\neq$ 8 –  Abhra Abir Kundu Feb 2 '13 at 20:13
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