solving by substitution

[mpkoellein]

I've semi-conquered this with whole numbers...now the fractions are !!!

x/2 - y/3 = 3/2
2x/3 - y/9 =3

in this problem 4s + 3t = 3 I isolated the t by multiplying the s in each equation in order to get rid of it. I ended up
5t - 3s =34 with t=5...all well and good....but I don't know how to begin with the fractions...could

you give me the first one or two steps....thanks

 

[Aladdin]

Hint : Multiply the first equation by 6 & the second by 9

 

[wjm11]

I've semi-conquered this with whole numbers...now the fractions are !!!

x/2 - y/3 = 3/2
2x/3 - y/9 =3

There is an easy way to approach equations with fractions. Consider the first equation: x/2 - y/3 = 3/2; the denominators are 2 and 3. Therefore, multiply both sides of the equations by both 2 and 3 (that is, multiply both sides by 6). This results in eliminating the denominators:

6(x/2 - y/3) = 6(3/2)
3x – 2y = 9

In the second equation, the denominators are 3 and 9. Since 3 is a factor of 9, it is sufficient to multiply both sides by 9:

9(2x/3 - y/9) = 9(3)
6x – y = 27

Don’t be afraid of fractions. Just use easy methods like this for dealing with them.

Now you can proceed to solve your system of equations:

3x – 2y = 9
6x – y = 27