# Thread: Simultaneous Equations with way too many unknowns..

1. ## Simultaneous Equations with way too many unknowns..

Yea, so equations aren't my bestfriends, and I'm now stuck with two simultaneous equations, and I just don't know where to start, so if someone could help me and show me what to do, I'd appreciate it.

'Find the values of m such that these equations have no solutions':

3x - my = 4
x + y = 12

'Find the values of m and a such that these equations have infinite solution sets':

4x - my = a
2x + y = 4

So if someone could patiently guide me through step by step, that'd be awesome.. I hate not knowing what I'm supposed to do.
Thanks a lot

2. Originally Posted by Fnus
Yea, so equations aren't my bestfriends, and I'm now stuck with two simultaneous equations, and I just don't know where to start, so if someone could help me and show me what to do, I'd appreciate it.

'Find the values of m such that these equations have no solutions':

3x - my = 4
x + y = 12

'Find the values of m and a such that these equations have infinite solution sets':

4x - my = a
2x + y = 4

So if someone could patiently guide me through step by step, that'd be awesome.. I hate not knowing what I'm supposed to do.
Thanks a lot
If the equations have an infinite solution set then basically the two equations say the same thing: that is to say, one equation is a multiple of the other.

So since the coefficient of x in the first equation is twice the coefficient of x in the second equation, then "-m" must be twice the coefficient of y in the second equation, so
$-m = 2 \cdot 1$
or m = -2.

Similarly a = 8.

-Dan

3. there are no solutions if both equations are lines with the same slope (parallel) and different y-intercepts. thus they will never cross.

Question 1:
'Find the values of m such that these equations have no solutions':

3x - my = 4
x + y = 12

Question 2:
'Find the values of m and a such that these equations have infinite solution sets':

4x - my = a
2x + y = 4

Does that change anything? <.<

5. $x+y=12\Rightarrow y=-x+12$ so the slope is -1.

$3x-my=4\Rightarrow y=\frac{3}{m}x-\frac{4}{m}$

now solve $\frac{3}{m}=-1$ you should get $m=-3$

6. for the second one we have

$4x-my-a=0$
and
$2(2x+y)=2(4)\Rightarrow 4x+2y-8=0$

from here it should be obvious what values to make a and m.

7. Hello, Fnus!

Find the values of $m$ such that these equations have no solutions:
. . $\begin{array}{cc}(1)\\(2)\end{array}
\begin{array}{cc}3x - my \\ x + y \end{array}
\begin{array}{cc} = \\ = \end{array}
\begin{array}{cc}4 \\ 12\end{array}$

If you are familiar with determinants, there is a simple solution.

A system has no solution if its determinant equal zero.

The determinant is: . $\begin{vmatrix} 3 & \text{-}m \\ 1 & 1\end{vmatrix} \:=\:(3)(1) - (\text{-}m)(1) \:=\:3 + m$

Therefore: . $3 + m \:=\;0\quad\Rightarrow\quad\boxed{m = -3}$

Otherwise, we can try to solve the system.

We have equation (1): . $3x - my \:=\:4$
. . .Multiply (2) by $m:\;\;mx + my \:=\:12m$

. . . . . . . . . . . . Add: . $mx + 3x \:=\:12m + 4$

. . . . . . . . . . .Factor: . $(m + 3)x \:=\:12m + 4$

. . . . . . . . . . . Then: . . . . . . $x\:=\:\frac{12m + 4}{m + 3}$

We see that $x$ is undefined if $m = -3.$

Therefore, the system has no solutions for $m = -3.$

8. Originally Posted by topsquark
If the equations have an infinite solution set then basically the two equations say the same thing: that is to say, one equation is a multiple of the other.

So since the coefficient of x in the first equation is twice the coefficient of x in the second equation, then "-m" must be twice the coefficient of y in the second equation, so
$-m = 2 \cdot 1$
or m = -2.

Similarly a = 8.

-Dan
Originally Posted by putnam120
there are no solutions if both equations are lines with the same slope (parallel) and different y-intercepts. thus they will never cross.
But if one equation is merely a multiple of the other, they represent the same line, so any point (x,y) on the line is a solution. So we have an infinite number of solutions.

-Dan