Part 1)
we have
------> equation A
------> equation B
Multiply by
the equation A
------> equation C
Multiply by
the equation B

-------> equation D
Adds equation C and equation D

therefore
<u>the answer Part 1) is the option A </u>

Part 2)
we have
------> equation A

Simplify Divide by
both sides

------> equation B
the lines A and B are parallel lines, because the slope m is equal
so
The system has no solution
therefore
<u>the answer Part 2) is the option D</u>
There is no x value as there is no solution to the system.
Part 3)
we have
------> equation A

------> equation B
substitute equation B in equation A
![4x+2[x-3]=6](https://tex.z-dn.net/?f=4x%2B2%5Bx-3%5D%3D6)



therefore
<u>the answer part 3) is the option D</u>

Part 4)
Let
x---------> The number of one-step equations
y---------> The number of two-step equations
we know that

-------> equation A
------> equation B
substitute equation A in equation B
![3[1,120-y]-2y=1,300](https://tex.z-dn.net/?f=3%5B1%2C120-y%5D-2y%3D1%2C300)




therefore
<u>the answer part 4) is the option D</u>

In order to find height from where ball is dropped, you have to find height or h(t) when time or t is zero.So plug in t=0 into your quadratic equation:h(0) = -16.1(0^2) + 150h(0) = 0 +150h(0) = 150 ft is the height from where ball is dropped. When ball hits the ground, the height is zero. So plug in h(t) = 0 and solve for t.0 = -16.1t^2 + 15016.1 t^2 = 150t^2 = 150/16.1t = sqrt(150/16.1)t = ± 3.05Since time cannot be negative, your answer is positive solution i.e. t = 3.05