Multiply the numerator and denominator of the first question by 2 in order for both of the denominators to be the same.
1*2=2
3*2=6
Now that the denominators are the same, we can simply add the numerators and simplify if possible.
(2/6)+(1/6)= (3/6)
(3/6)= (1/2)
Final answer: 1/2
Step-by-step explanation:
Area of Rectangle at the top =
Length x Width

Area of Quarter- Circle = 1/4 x Area of Circle

Area of Rectangle at the right = Length x Width

Total Area = Area of 2 Rectangles + Area of Quarter Circle

I will leave the answer in terms of pi as I'm not sure if you need to round off your answer.
I'd suggest using "elimination by addition and subtraction" here, altho' there are other approaches (such as matrices, substitution, etc.).
Note that if you add the 3rd equation to the second, the x terms cancel out, and you are left with the system
- y + 3z = -2
y + z = -2
-----------------
4z = -4, so z = -1.
Next, multiply the 3rd equation by 2: You'll get -2x + 2y + 2z = -2.
Add this result to the first equation. The 2x terms will cancel, leaving you with the system
2y + 2z = -2
y + z = 4
This would be a good time to subst. -1 for z. We then get:
-2y - 2 = -2. Then y must be 0. y = 0.
Now subst. -1 for z and 0 for y in any of the original equations.
For example, x - (-1) + 3(0) = -2, so x + 1 = -2, or x = -3.
Then a tentative solution is (-3, -1, 0).
It's very important that you ensure that this satisfies all 3 of the originale quations.
90-42=48
48/6 = 8
The 6 student each handed out 8 invitation