Hi

How would I be able to find the value of $\displaystyle a$ that would give the equations an infinite number of solutions

$\displaystyle x-4y+z=14$

$\displaystyle -4x-3y+3z=8$

$\displaystyle 17x-30y+az=110$

Thanks

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- Sep 17th 2010, 06:48 AMbobredSimultaeous equation with infinite solutions
Hi

How would I be able to find the value of $\displaystyle a$ that would give the equations an infinite number of solutions

$\displaystyle x-4y+z=14$

$\displaystyle -4x-3y+3z=8$

$\displaystyle 17x-30y+az=110$

Thanks - Sep 17th 2010, 07:06 AMPlato
I see that you have over fifty postings.

By now you should understand that this is not a homework service

So you need to either post some of your work on this problem or you need to explain what you do not understand about the question. - Sep 17th 2010, 07:11 AMmathaddict
1 x (-4) - 2

19 y -7z =-64 5

1 x 17 - 3

-38 y +(17-a)z=128 6

Divide equation 6 by -2 : 19y - (17-a)/2 = -64

By comparing equation 6 to equation 5, we see that (17-a)/2=7 so a=3.

In a case where you are given 3 equations and only two of them are useful, the system is said to have infinite solutions. - Sep 17th 2010, 08:08 AMSoroban
Hello, bobred!

Quote:

How would I be able to find the value of $\displaystyle \,a$

that would give the equations an infinite number of solutions?

. . $\displaystyle \begin{array}{ccc}x-4y+z &=& 14 \\

\text{-}4x-3y+3z&=&8 \\

17x-30y+az &=& 110 \end{array}$

A system of equations has no solution or an infinite number of solutions

. . if the determinant of its coefficients is zero.

$\displaystyle D \;=\; \begin{vmatrix}1 & \text{-}4 & 1 \\ \text{-}4 & \text{-}3 & 3 \\ 17 & \text{-}30 & a \end{vmatrix} \;=\;1\begin{vmatrix}\text{-}3&3\\\text{-}30 &a \end{vmatrix} + 4\begin{vmatrix}\text{-}4 & 3\\17&a\end{vmatrix} + 1\begin{vmatrix}\text{-}4&\text{-}3\\17&\text{-}30\end{vmatrix} $

. . .$\displaystyle =\;(\text{-}3a + 90) + 4(\text{-}4a-51) + (120 + 51) \;=\;\text{-}19a + 57$

If $\displaystyle D = 0$, then: .$\displaystyle \text{-}19a + 57 \:=\:0 \quad\Rightarrow\quad \boxed{a \:=\:3}$

We have: .$\displaystyle \begin{Bmatrix} x - 4y + z &=& 14 & [1] \\

\text{-}4x - 3y + 3z &=& 8 & [2] \\

17x - 30y + 3z &=& 110 & [3] \end{Bmatrix}$

We see that: .$\displaystyle 9\1\cdot\1\text{[equation 1]} - 2\!\cdot\1\text{[equation 2]} \;=\;\text{[equation 3]}$

The system of equations is.*dependent*

There is an infinite number of solutions.

- Sep 17th 2010, 08:42 AMbobred
Hi

Thanks all, I had been staring at it for ages and wasn't sure where to start. I started doing somethig like mathaddict

Thanks - Sep 18th 2010, 03:38 AMHallsofIvy
Let me add to Plato's request that you start showing some attempt of your own on these problems!

Note that after Soroban found "a" that made the determinant 0, he then checked that the equations, with that a, were dependent. If the determinant of coefficients of a system is 0, then**either**there are an infinite number of solutions or there is no solution.