Can someone help me with a math problem i just cant solve.





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Discus: College Confidential Café: 2004 Archive: Can someone help me with a math problem i just cant solve.
By Virginia2004 (Virginia2004) on Saturday, August 28, 2004 - 04:43 pm: Edit

I dont know why but i get all the other problems right except this one.

9-(9z+5)=6(4z-5)+3
it says to solve the linear equation and simplify your answer, i know the right answer i just cant get to it.

By Link12 (Link12) on Saturday, August 28, 2004 - 04:51 pm: Edit

Well, as long as you're practicing:
9-(9z+5)=6(4z-5)+3
Expand: 9-9z-5=24z-30+3
Simplify: 4-5z=24z-27
Bring varables to LHS, numbers to RHS: 19z=-31
Divide both sides by 19: z=-31/19.


So, z= -31/19.

By Equestrian (Equestrian) on Saturday, August 28, 2004 - 05:13 pm: Edit

I got z = 31/33

???

By Lisasimpson (Lisasimpson) on Saturday, August 28, 2004 - 06:03 pm: Edit

yea i think its 31/33 .. somehow link simplified 9-9z-5 to get 4-5z

By Virginia2004 (Virginia2004) on Saturday, August 28, 2004 - 06:07 pm: Edit

so how do you simplify it right?

By Link12 (Link12) on Saturday, August 28, 2004 - 06:18 pm: Edit

My mistake...

By Ubercollegeman (Ubercollegeman) on Saturday, August 28, 2004 - 07:36 pm: Edit

I have a problem with why the book calls this a "linear equation." I mean, it is a linear equation, but if you're just learning this stuff, why bog the student down with terms?

I mean, it's like saying:

Solve the POLYNOMIAL equation x-4=0, x AN ELEMENT OF THE REAL NUMBERS on the DOMAIN, {x:x AN ELEMENT OF (-infty, +infty)}.

Why bother? Okay, there ends my rant.

By Equestrian (Equestrian) on Saturday, August 28, 2004 - 08:05 pm: Edit

Virginia, it's basic algebra.

9-(9z+5) = 6(4z-5)+3

First, you want to distribute.

So you then have:
9-9z-5 = 24z-30+3

Then you add 9z to 24z to put the variable on one side.

9-5 = 33z-30+3

then you just continue solving..

4 = 33z-27
31 = 33z
31/33 = z


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