Answer:
(6, - 3 )
Step-by-step explanation:
Given the 2 equations
2x + 5y = - 3 → (1)
2x + 2y = 6 → (2)
Subtracting (1) from (2) term by term will eliminate the x- term
(2x - 2x) + (2y - 5y) = 6 - (- 3), that is
- 3y = 9 ( divide both sides by - 3 )
y = - 3
Substitute y = - 3 in either of the 2 equations and solve for x
Substituting y = - 3 in (1)
2x + 5(- 3) = - 3
2x - 15 = - 3 ( add 15 to both sides )
2x = 12 ( divide both sides by 2 )
x = 6
Solution is (6, - 3 )
730/27 or improper fractions 27 1/27
Answer:
The width of the piece is 11 inches
Step-by-step explanation:
Let
x ----> the length of the rectangular piece of aluminum in inches
y ----> the width of the rectangular piece of aluminum in inches
we know that
The perimeter of the rectangular piece of aluminum is equal to

we have

so

simplify
----> equation A
----> equation B
Solve the system by substitution
substitute equation B in equation A
solve for y
therefore
The width of the piece is 11 inches
You have 3 unknowns: a, b, and c. That means you have to have 3 equations to solve for the values of them. 3 unknowns needs 3 different equations. We will use the first 3 points in the table and thank God that one of them has an x value of 0. We will replace the x and y in the general form of the quadratic with the x and y from the table, 3 times, to find each variable. Watch how it works. We will start with (0, 15).

. That gives us right away that c = 15. We will do the same thing again with the second value in the table along with the fact that c = 15 to get an equation in a and b.

which simplifies to
4a+2b=.5. Now do the same for the third set of coordinates from the table.

which simplifies down to
16a+4b=2. Solve those simultaneously. Multiply the first bolded equation by -4 and then add that one to the second bolded one.

gives us
-16a-8b=-2. Add that to the second bolded equation and the a terms cancel out giving us -4b=0 so b = 0. Subbing that back in we solve for a: 16a+4(0)=2 and 16a = 2. Therefore, a = 1/8. The quadratic then is