VIC's Guide Chapter 1: Basic Equation Solutions

by **KTsincere** Thu Mar 06, 2008 2:39 pm

Hello, I have a pretty stupid question but it is making me so extremely confused and I was wondering if anybody can point out what I'm missing...

Question: Solve for x

4a+7b+9x=17
3a+3b+3x=3
9a+(b/4)+(x/9)=9

I understand how to solve the problem however I am confused on a part of the algebra! I know you want to establish a common denominator in order to eliminate the fractions in the third equation but the book says you should multiply : (81) [9a+(b/4)+(x/9)=9} and then it shows the outcome as :729a+20.25b+9x=9.

My question is don't you have to multiply both sides by 81 which would then give you: 729a+20.25b+9x=729
I am sure MGMAT is correct but I am confused, please help...

by **StaceyKoprince** Wed Mar 12, 2008 5:37 pm

I'm traveling and can't look this up myself right now but will forward to someone who can. Based on what you described above, it sounds like a typo on our part.

In any event, yes: whatever you do to one side of an equation you also have to do the other side. (Also, we do sometimes have typos, so do check whenever you're not sure - no such thing as a "dumb" question!)

by **KTsincere** Wed Mar 12, 2008 6:43 pm

Thanks Stacey, FYI Page #25

by **RonPurewal** Wed Mar 19, 2008 3:43 am

KTsincere Wrote:Thanks Stacey, FYI Page #25

you are correct; the right-hand side of the equation should be 81 times 9.

thanks!

by **kimd6746** Fri Apr 11, 2008 2:34 pm

Hello,

Another stupid question regarding the common denominator. Could I have chosen 36 as a common denominator? What's so special about using 81? I thought that since dividing 81 by 4 doesn't give you an integer, it's best to use common denominators that give you integer numerators. Does it matter? The point of this question is not to solve it but to prove that all three are distinct equations so would using 36 as a common denominator for equation #3 work?

36(9a)+36(b/4)+36+(x/9)=36(9)
324a+9b+4x=324 VALID?

KTsincere Wrote:Hello, I have a pretty stupid question but it is making me so extremely confused and I was wondering if anybody can point out what I'm missing...

Question: Solve for x

4a+7b+9x=17
3a+3b+3x=3
9a+(b/4)+(x/9)=9

I understand how to solve the problem however I am confused on a part of the algebra! I know you want to establish a common denominator in order to eliminate the fractions in the third equation but the book says you should multiply : (81) [9a+(b/4)+(x/9)=9} and then it shows the outcome as :729a+20.25b+9x=9.

My question is don't you have to multiply both sides by 81 which would then give you: 729a+20.25b+9x=729
I am sure MGMAT is correct but I am confused, please help...

by **RonPurewal** Wed Apr 16, 2008 4:28 am

kimd6746 Wrote:Hello,

Another stupid question regarding the common denominator. Could I have chosen 36 as a common denominator? What's so special about using 81? I thought that since dividing 81 by 4 doesn't give you an integer, it's best to use common denominators that give you integer numerators. Does it matter? The point of this question is not to solve it but to prove that all three are distinct equations so would using 36 as a common denominator for equation #3 work?

36(9a)+36(b/4)+36+(x/9)=36(9)
324a+9b+4x=324 VALID?

definitely valid.

you can multiply the two sides of the equation by any number you want - even pi times the square root of 17.7, if that's your preference.
the multiplication by 81 is done with the goal of giving x the same coefficient (namely, 9) in all three of the equations; i agree with you that, in isolation, 81 would be a rather absurd choice.