X+Y=10
XY=-300
Solve for one of the variables
X=10-Y
Plug into the other equation
Y(10-Y)=-300
Now, you have a quadratic...
-y²-10y+300=0
I would multiply everything by -1, so there's no coefficient on y²
y²+10y-300=0
And solve...
-10±√(100-4(-300))
----------------------------
2
Y=-5±5√(13)
Plug back onto original equation...
X+(-5±5√(13))=10
X±5√(13)=15
X=15±5√(13)
Note: There are 2 sets of answers, keep the negatives with the negatives, and the positives with the positives when dealing with the ±.
Answer:
i strongly believe the answer is (c)
Step-by-step explanation:
129 divided by 5 is equal to 25 with a remainder of 4.
This means that she will need a total of 26 boxes, but only 25 of them will have 5 cupcakes in them. The last one will have 4 cupcakes.
Answer:
= 3x - 3y
Step-by-step explanation:
Answer:
Solution → (2, -1)
Step-by-step explanation:
Input-output values table for equation (1),
2x + y = 3
x -1 0 1 2
y 5 3 1 -1
Input-output values table for equation (2),
y = x - 3
x -1 0 1 2
y -4 -3 -2 -1
Now plot these points to get the graphs of both the equations.
Common point (2, -1) of both the lines will be the solution of the system of equations.
