Question

Solve the system of equations:
1/x + 1/y = 8
3/x - 5/y = 0
The textbook gave me a hint that I'm supposed to use substitution. I tried that, but I couldn't get anywhere.

Answer 1

Let u = 1/x and v = 1/y
Make the substitutions. Solve for u and v
then put back u = 1/x; v = 1/y.
Take the reciprocal to get x and y.

Answer 2

Yeah, that's what the books says, but I don't really know how that works. Should I be getting u = 8 - v and v = 8-u?

Answer 3

After the substitution, you get 2 equations:
u + v = 8
3u - 5v = 0

Solve for u and v. Use elimination method or substitution method -- whichever is easy for you.

Answer 4

Ok! When I solved for u and v I got u=20 and v=-12. So from there I plug it back into u=1/x and v=1/y?
20=1/x => x=1/20
12=1/y => y=-1/12
I know I'm doing something wrong because the back says it's (1/5, 1/3)

Answer 5

You got the wrong values for u and v.

u + v = 8
3u - 5v = 0
To eliminate v, multiply first equation by 5 and add it to the second:
5u + 5v = 40
3u - 5v = 0
add
8u = 40
divide both sides by 8
u = 5
put this back in first equation:
5 + v = 8
subtract 5 from both sides
v = 8 - 5 = 3

u = 5; v = 3
put back u = 1/x and v = 1/y
1/x = 5
take reciprocal on both sides:
x = 1/5

1/y = 3
y = 1/3

Answer 6

write it as (x,y): (1/5, 1/3)

You may want to get some practice solving 2 equations and 2 unknowns because you will come across it many times in algebra.

Answer 7

Oh! Thank you! I didn't notice that I wrote down 3u + 5v instead of 3u - 5v.
This was really the only problem that confused me; I just need to keep watching out for careless mistakes. Thanks again!

Answer 8

you are welcome.