1. ## simoultaneous equation problem

hi guys, i tryed to solve this two equations using substitution but my x's ended up canceling out when trying to solve for x.

$6x - 2y = 14$ eq 1
$-9x + 3y = 12$ eq 2

$y = \frac {6x - 14}{2}$ eq 3 (re-arranged eq 1)

$-9x + 3(\frac {6x - 14}{2}) = 12$ sub eq 3 into eq 2

$-18x + 3(6x - 14)= 24$

$-18x + 18x - 42 = 24$ that leaves me with no x

any help much appreciated

2. Originally Posted by jvignacio
hi guys, i tryed to solve this two equations using substitution but my x's ended up canceling out when trying to solve for x.

$6x - 2y = 14$ eq 1
$-9x + 3y = 12$ eq 2

$y = \frac {6x - 14}{2}$ eq 3 (re-arranged eq 1)

$-9x + 3(\frac {6x - 14}{2}) = 12$ sub eq 3 into eq 2

$-18x + 3(6x - 14)= 24$

$-18x + 18x - 42 = 24$ that leaves me with no x

any help much appreciated
After a little manipulation, we have the system:

$\left\{\begin{array}{rcrcr}3x&-&y&=&7\\-3x&+&y&=&4\end{array}\right.$

Adding the two equations together yields $0=11$, which is not true. Thus there are no solutions.

In the future, if the variable disappears and you end up with some bizarre statement [like -5=3 or 0=2 , etc.], then you can conclude that there is no solution.

Does this make sense?

--Chris

3. You assumed that the two curves intersected, but then you reached to a contradiction. Therefore, your original assumption is false.

4. Originally Posted by Chris L T521
After a little manipulation, we have the system:

$\left\{\begin{array}{rcrcr}3x&-&y&=&7\\-3x&+&y&=&4\end{array}\right.$

Adding the two equations together yields $0=11$, which is not true. Thus there are no solutions.

In the future, if the variable disappears and you end up with some bizarre statement [like -5=3 or 0=2 , etc.], then you can conclude that there is no solution.

Does this make sense?

--Chris
ahh yes so whenever the variable dissappears on me while trying to solve it, it means there are no solutions. okay thanks mate

5. Originally Posted by Chop Suey
You assumed that the two curves intersected, but then you reached to a contradiction. Therefore, your original assumption is false.
yeah i understand, trying to do some tutorial questions set as practice. thanks mate

6. Hello, jvignacio!

Here's another approach . . .

I tried to solve this two equations using substitution
but my x's ended up canceling out when trying to solve for x.

$\begin{array}{cccc}6x - 2y &=& 14 & [1] \\
\text{-}9x + 3y &=& 12 & [2] \end{array}$

Graphic solution: we want the intersection of the two lines.

$\begin{array}{cc}\text{Solve [1] for }y\!: & y\:=\:3x - 7 \\
\text{Solve [2] for }y\!: & y\:=\:3x + 4 \end{array}$

We have two lines: one has y-intercept -7, the other has y-intercept 4.
Both have slope 3 . . . The lines are parallel.
Code:
| /
|/    /
4*    /
/|   /
/ |  /
----/--+-/------
/   |/
-7*
/|
/ |
|

Obviously, the line do not intersect.

Therefore, the system has no solution.

7. Originally Posted by Soroban
Hello, jvignacio!

Here's another approach . . .

Graphic solution: we want the intersection of the two lines.

$\begin{array}{cc}\text{Solve [1] for }y\!: & y\:=\:3x - 7 \\$ $
\text{Solve [2] for }y\!: & y\:=\:3x + 4 \end{array}" alt="
\text{Solve [2] for }y\!: & y\:=\:3x + 4 \end{array}" />

We have two lines: one has y-intercept -7, the other has y-intercept 4.
Both have slope 3 . . . The lines are parallel.
Code:
| /
|/    /
4*    /
/|   /
/ |  /
----/--+-/------
/   |/
-7*
/|
/ |
|
Obviously, the line do not intersect.

Therefore, the system has no solution.
soroban, thanks for that. really appreciate it!!!