# Revision history [back]

Sorry, I tried to attach an image of it and I messed it up. Here is the code: a,b=var('a b') solve([7.0==asqrt(16)/(1+b/16.0),22==asqrt(4)/(1+b/4)],a,b) print a.n() The result the solve command gives is the ratios of integers.

Sorry, I tried to attach an image of it and I messed it up. Here is the code: a,b=var('a b') b')

solve([7.0==asqrt(16)/(1+b/16.0),22==asqrt(4)/(1+b/4)],a,b) sqrt(4)/(1+b/4)],a,b)

print a.n() The result the solve command gives is the ratios of integers.

Sorry, I tried to attach an image of it and I messed it up. Here is the code: a,b=var('a b')

solve([7.0==asqrt(16)/(1+b/16.0),22==asqrt(4)/(1+b/4)],a,b)

print a.n() a.n()

The result the solve command gives is the ratios of integers.

Sorry, I tried to attach an image of it and I messed it up. Here is the code: a,b=var('a b')

solve([7.0==asqrt(16)/(1+b/16.0),22==asqrt(4)/(1+b/4)],a,b)

print a.n()

The result the solve command gives is the ratios of integers.

Sorry, I tried to attach an image of it and I messed it up. Here is the code: a,b=var('a b')

solve([7.0==asqrt(16)/(1+b/16.0),22==asqrt(4)/(1+b/4)],a,b)

print a.n()

The result the solve command gives is the ratios of integers.

(I typed an asterisk between the a and the sqrt, but it doesn't show up...)