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

Re: Help using two equations to solve each other

Hello, Scryer!

Quote:

$\displaystyle \begin{array}{ccccc}\frac{a}{2}+b&=&1 & [1] \\ 3a-\frac{b}{3} &=&\frac{31}{2} & [2] \end{array}$

I see that you are using the Substitution Method.

First, I would get rid of the fractions.

$\displaystyle \begin{array}{ccccccc}\text{Multiply [1] by 2:} & a + 2b &=& 2 & [3] \\ \text{Multiply [2] by 6:} & 18a - 2b &=& 93 & [4] \end{array}$

Solve [1] for $\displaystyle a\!:\;a + 2b \:=\:2 \quad\Rightarrow\quad a \:=\:2 - 2b$

Substitute into [2]: .$\displaystyle 18(2-2b)-2b \:=\:93 \quad\Rightarrow\quad 36 - 36b - 2b \:=\:93$

. . . . . . . . . . . . . . . $\displaystyle \text{-}38b \:=\:57 \quad\Rightarrow\quad b \:=\:\tfrac{57}{\text{-}38}\quad\Rightarrow\quad \boxed{b \:=\:\text{-}\tfrac{3}{2}}$

Substitute into [1]: .$\displaystyle a + 2\left(\text{-}\tfrac{3}{2}\right)-2 \:=\:2 \quad\Rightarrow\quad a - 3 \:=\:2 \quad\Rightarrow\quad \boxed{a \:=\:5}$

Re: Help using two equations to solve each other

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

Re: Help using two equations to solve each other

Quote:

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

Re: Help using two equations to solve each other

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

Re: Help using two equations to solve each other

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.

Re: Help using two equations to solve each other

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