Thread: Help using two equations to solve each other

Help using two equations to solve each other
(1) a/2+b=1
(2) 3a-b/3=31/2

Im lost on what to do here. I know I need to pick the easier equation to produce a equaling so I choose equation 1
what im trying to do is first clearing the fractions in eq1 I listed below what my thinking looks like but I know its wrong Please help

a/2+b=1 multiply both a/2 by 2 and the right side by 2 and i get
a+b=2 then I want to find a so I -b
a=2-b

in the second eq i substitute for a now
3a-b/3=31/2
3(2-b)-b/3=31/2 now i clear the brackets
6-3b+b/3=31/2 now i clear the fractions bymultiplying both sides by 3. 3x b/3 and the right side 31/2=15.5 x3
6-3b+b=15.5x3 simplify to
6-3b+b=46.5 now i get the bs alone
-3b+b=40.5 merge the bs
-2b=40.5
-b=40.5/2

Hello, Scryer!

I see that you are using the Substitution Method.

First, I would get rid of the fractions.



Solve [1] for

Substitute into [2]: .

. . . . . . . . . . . . . . .

Substitute into [1]: .

I see you are multiplying through i out i forgot to do that thanks and I see how you would multiply (1) by 2 But how did you figure to multiply (2) by 6 I seem to be having trouble with that

Originally Posted by Scryer

I see you are multiplying through i out i forgot to do that thanks and I see how you would multiply (1) by 2 But how did you figure to multiply (2) by 6 I seem to be having trouble with that

That was done to get 2b in the second equation and have b eliminated in summing equation 1 and 2

but how did you decide to multiply (1) by 2 and (2) by 6

Those are the lowest common denominators. (1) has one denominator of 2 , so multiplying by 2 will clear it, while (2) has a denominator of 2 and a denominator of 3 and the LCM of 2 and 3 is 6, so 6 is the smallest number that will clear both denominators.

Ah I see I forgot completly about the LCM thanks to all of you