Couple math problems.

N

NuiHime

New member
Joined
Jul 6, 2009
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6
Alright the first one I gotta simplfy.

8 - 12 y^3 all over 4

What I don't know what to do with this one is the y. Someone told me to divide it all by 4. But I'm not sure how.


The second one I gotta solve. Its

x^2 - 3x + 5 = x(x - 8) +20

I'm not sure how to get the x's on one side. After that I'll have it.
 
\(\displaystyle \frac{8-12y^3}{4}=\frac{\rlap{/}4(2-3y^3)}{\rlap{/}4}=?\)
 
\(\displaystyle x^2 - 3x + 5 = x(x - 8) +20\)

Step 1. Eliminate the grouping symbols.
Step 2. To get rid of the x[sup:14lc522i]2[/sup:14lc522i] on the right side of the equation add or subtract (whichever does the job) it from both sides of the equation.
Step 3. Get all the "x" terms on the left side of the equation by adding or subtracting the term containing x on the right side to/from both sides of the equation.
Step 4. Get rid of the 5 on the left side of the equation by adding or subtracting 5 (whichever does the job) to/from both sides of the equation.
Step 6. Divide both sides of the equation by the numerical coefficient of the x term.
Step 7. Check your answer by substituting your answer back into the original equation and doing the arithmetic to see if the equation tells the truth. For instance, suppose you get an answer of x=2. Check it this way.

x^2 - 3x + 5 = x(x - 8) +20
(2)^2 - 3(2) + 5 = 2(2 - 8) +20???
4 - 6 + 5 = 2(-6) + 20???
-2 + 5 = -12 + 20???
3 = 8??? <<< Nope. x=2 is NOT the correct solution. When you find the correct solution you will end up with something like 7 = 7, or 5=5, or 0=0, or -3=-3, or etc.
 
This second part is kinda confusing me.

I thought you were suppose to get the x's on one side. Do you need to substitute with a number?


Edit: Actually I worked it out to this.

x^2-3x+5=x(x-8)+20
x^2-3x+5=x^2+8x+20
-x^2 -x^2
------------------------
-3x+5=8x+20

and I"m stuck there.
 
You have completed step 2. Step 3 is next. Then continue with step 4, etc.
 
-3x+5=8x+20
+3x +3x
---------------
5=11x+20
-20= -20
-------------------
-15=11x
and i get -1.36
I don't think thats right though.
 
NuiHime said:
This second part is kinda confusing me.

I thought you were suppose to get the x's on one side. Do you need to substitute with a number?


Edit: Actually I worked it out to this.

x^2-3x+5=x(x-8)+20
x^2-3x+5=x^2+8x+20<<<error. Correct it and continue.
-x^2 -x^2
------------------------
-3x+5=8x+20

and I"m stuck there.
Click to expand...
 
NuiHime said:
Alright the first one I gotta simplfy.

8 - 12 y^3 all over 4

\(\displaystyle \frac{8 \, - \, 12y^3}{4} \, = \, \frac{8}{4} \, - \, \frac {12y^3}{4} \, = \, 2 \, - \, \frac {12}{4}y^3 \,= \, 2\, - \, 3y^3\)

What I don't know what to do with this one is the y. Someone told me to divide it all by 4. But I'm not sure how.


The second one I gotta solve. Its

x^2 - 3x + 5 = x(x - 8) +20

First do the multiplication on the right hand side by distributing 'x'. If you do not understand that - you need face-to-face help.

I am afraid you are getting into higher level of algebra without understanding the basics.


I'm not sure how to get the x's on one side. After that I'll have it.
Click to expand...
 
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