So assuming that the total =27
r+b=27
r=-5+3b
r=3b-5
subsitute 3b-5 for r in first equation
3b-5+b=27
4b-5=27
add 5
4b=32
divide by 4
b=8
subsitute
b+r=27
8+r=27
subtracct 8
r=19
red=19
blue=8
Answer:
(x, y) = (2, 5)
Step-by-step explanation:
I find it easier to solve equations like this by solving for x' = 1/x and y' = 1/y. The equations then become ...
3x' -y' = 13/10
x' +2y' = 9/10
Adding twice the first equation to the second, we get ...
2(3x' -y') +(x' +2y') = 2(13/10) +(9/10)
7x' = 35/10 . . . . . . simplify
x' = 5/10 = 1/2 . . . . divide by 7
Using the first equation to find y', we have ...
y' = 3x' -13/10 = 3(5/10) -13/10 = 2/10 = 1/5
So, the solution is ...
x = 1/x' = 1/(1/2) = 2
y = 1/y' = 1/(1/5) = 5
(x, y) = (2, 5)
_____
The attached graph shows the original equations. There are two points of intersection of the curves, one at (0, 0). Of course, both equations are undefined at that point, so each graph will have a "hole" there.
We find the area of the total square which is 8in • 8in=64in^2 . Now we find the area of the small quadrilateral which is 4in•4in=16 in^2.
Now we subtract the area of the small quadrilateral from the big quadrilateral’s area 64in^2 -16in^2 leaving us with the are of 48in^2 of the figure
Answer:
1 parallelogram and 1 rectangle
Step-by-step explanation:
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