substitution method

[jshaziza]

I was given the following problem to solve:

4x-y=9
2x+3y=-27

This is how I worked the problem out:

3y=-2x-27
y=-2/3x-9

4x-(-2/3x-9)=9
4x+2/3x+9=9
4/2/3x+9=9
4/2/3x=0

I already tried to make x=-4/2/3 go back into the equation and then figure out y, but it didn't work out. So my question is did I work it out wrong or is there no solution for this problem?

 

[pka]

It is much easier to use the first equation for y:
\(\displaystyle \begin{array}{l}
y = 4x - 9 \\
2x + 3\left( {4x - 9} \right) = - 27 \\
\end{array}\)

 

[soroban]

Hello, jshaziza!

As pka pointed out, there is an easier way.
What you did was correct (despite the strange fractions) up to the last step.

\(\displaystyle \begin{array}{ccc}4x\,-\,y &\,=\, & 9 \\2x\,+\,3y &\,=\, &-27\end{array}\)

This is how I worked the problem out:

\(\displaystyle 3y\:=\:-2x\,-\,27\)
\(\displaystyle y\:=\:-\frac{2}{3}x\,-\,9\)

\(\displaystyle 4x\,-\,\left(-\frac{2}{3}x\,-\,9)\:=\:9\)

\(\displaystyle 4x\,+\,\frac{2}{3}\,+\,9\:=\:9\)

\(\displaystyle 4\frac{2}{3}x\,+\,9\:=\:9\)

\(\displaystyle 4\frac{2}{3}x\:=\:0\)

We don't use "mixed numbers" in algebra. .**

It should be written: \(\displaystyle \:\frac{14}{3}x\:=\:0\)


Here's your error . . .

Multiply both sides by \(\displaystyle \frac{3}{14}:\;\;\frac{3}{14}\left(\frac{14}{3}x\right)\:=\:\frac{3}{14}(0)\)

Therefore: \(\displaystyle \:x\,=\,0\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

Do you know why "mixed numbers" aren't used in Algebra?
. . Because they're dangerous!


Multiply: \(\displaystyle \L\:\left(\frac{7}{3}x\right)\left(\frac{15}{14}x^2\right)\)

. . We have: \(\displaystyle \L\:\left(\frac{7}{3}\cdot\frac{15}{14}\right)(x\cdot x^2) \:=\:\frac{5}{2}x^3\)


Suppose it was written: \(\displaystyle \L\:\left(2\frac{1}{3}x\right)\left(1\frac{1}{14}x^2\right)\)


Would you do this?\(\displaystyle \L\:\left(2\frac{1}{3}\right)\left(1\frac{1}{14}\right) \:=\:2\frac{1}{42}\) . . . WRONG!


Why can't we do that?

Because \(\displaystyle 2\frac{1}{3}\) means: \(\displaystyle \,2\,+\,\frac{1}{3}\)
. . .and \(\displaystyle 1\frac{1}{14}\) means: \(\displaystyle \,1\,+\,\frac{1}{14}\)

You see, there are invisible plus-signs in there.
We have enough trouble with "invisibles", don't we?
. . \(\displaystyle x\) could be written: \(\displaystyle \L\,\frac{1x^^{1}}{1}\;\) which has an invisible times-sign, too.

We are actually multiplying two "binomials": \(\displaystyle \:\left(2\,+\,\frac{1}{3}\right)\left(1\,+\,\frac{1}{14}\right)\)

If you're adventurous, you can do the "FOIL" on it.
. . You should come up with: \(\displaystyle \,2\frac{1}{2}\,=\,\frac{5}{2}\)