There are two ways to do this but the way I prefer is to make one of the equations in terms of one variable and then 'plug this in' to the second equation. I will demonstrate
Look at equation 1,

this can quite easily be manipulated to show

.
Then because there is a y in the second equation (and both equations are simultaneous) we can 'plug in' our new equation where y is in the second one

which can then be solved for x since there is only one variable

and then with our x solution we can work out our y solution by using the equation we manipulated

.
So the solution to these equations is x=-2 when y=6
Answer:
$2100
Step-by-step explanation:
1/3 of 4500 is 1500
to find 1/3 of a number is to multiply by 1/3
1/3 * 4500 = 1500
$1500 is for savings
subtract 1500 from 4500
4500 - 1500 = 3000
the remaining money is $3000
to find 50% of something, divide by 2
3000/2 = 1500
$1500 is spent on food.
20% of 3000 is 600
to find 20% of something, multiply the decimal value (0.2)
0.2 * 3000 = 600
$600 are spent on rent.
$450 leftover
1500 + 600 = 2100
Food and rent cost $2100
hope this helps:)
(7-3+4^3/2) / (9-5)
7-3+64/2) / 4
(7-3 + 32) / 4
(4 + 32) / 4
36/4 = 9
48 + 72
24(2 + 3)