Almost got it!
x + 3 = 3(y + 2)/2 [Multiplied both sides by 3]
so x + 3 = (3y + 6)/2
2(x + 3) = 3y + 6 [Multiplied both sides by 2]
2x + 6 = 3y + 6
2x = 3y [Subtracted 6 from both sides]
x = 3y/2 [Divided both sides by 2]
x/3 = y/2 [Divided both sides by 3]
So you wrote y/3 instead of y/2
Hope this helped!
In terms of j: 55j+20 = k
Answer:
The integers are 4 and 7 or -2 and 1.
Step-by-step explanation:
You can make a system of equations with the description of the two integers.
1. x = y + 3
2. 2x + 2 = y^2
The simplest and the fastest way to solve this system in this case is substitution. You can substitute x for y + 3 in the second equation.
1. x = y + 3
2. 2(y + 3) + 2 = y^2
Now simplify and solve the second one. For convenience, I will just disregard the first equation for now.
2y + 6 + 2 = y^2
y^2 - 2y - 8 = 0
You can factor this equation to solve for y.
(y - 4) (y + 2) = 0
y = 4, y = -2
Now we can substitute the value of y for x in the first equation.
x = 7, x = 1
Answer:
1. Slide over the Y-Axis
2. Reflection over the Y-Axis
3. Reflection over the y axis first as well as a rotation
