Answer:
in short, Yes they are always rational.
here's why...
E.g. suppose 
and 
are fractions, that means that a,b,c,d are all integers, and b and d are not zero. finding the sum the numerator, and denominator would also have to be integers. the denominator of the sum can't be zero since the denominators of the fractions were not zero, and would give 
and since they are bound by addition (sum means addition) they must also be rational since eit would equal a bigger integer than initially had
Step-by-step explanation:
Ok , with that information we can write the equations
x + y = 6
5x + 4y = 28
Where x = how many people ordered chicken
and y = how many people ordered egg salad
Through elimination , we can set one of the variables in both equations equal so we can eliminate it :
(4)x + (4)y = (4)6
5x + 4y = 28
4x + 4y = 24. equation 1
5x + 4y = 28. equation 2
Now we can subtract the second equation by the first equation and isolate one variable:
equation 2 - equation 1
5x - 4x + 4y - 4y = 28 - 24
x = 4
Now that we discovered our x value ( How many people ordered chicken salad ) , we can apply it to one of the equations and discover y ( how many people ordered egg salad)
x + y = 6
x= 4
4 + y = 6
We can shift 4 to the other side of the equation by subtracting 4 from both sides of the equation:
4 - 4 + y = 6 - 4
y = 2
x=4 and y=2
So the awnser is :
4 people ordered chicken salad and 2 people ordered egg salad!
I hope you understood my brief explanation!!
p.s if you want to know how to use another method to solve these problems ( Substition) , just let me know in a comentary down here
Answer:
A math test
Step-by-step explanation:
Answer:
Let x = the charge in 1st city before taxes
Let y = the charge in 2nd city before taxes
Set up equation before taxes.
y = x - 1500 eq1
Set up equation for total tax paid.
0.065x + 0.06y = 378.75 eq2
Substitute eq1 into eq2.
0.065x + 0.06(x - 1500) = 378.75
0.065x + 0.06x - 90 = 378.75
0.125x - 90 = 378.75
0.125x = 468.75
x = 3750
Substitute this value of x into eq1.
y = 3750 - 1500
y = 2250
The hotel charge in city one is $3750 and the hotel charge in city two is $2250
