Answer:
3) x = 15; 95 and 85 4) x = 12; 98 for both angles
Step-by-step explanation:
2x + 65 + 3x + 40 = 180 Set the equations equal to 180
5x + 105 = 180 Combine like terms
- 105 - 105 Subtract 105 from both sides
5x = 75 Divide both sides by 5
x = 15
Plug 15 into both equations
2(15) + 65 = 95
3(15) + 40 = 85
4) 5x + 38 = 9x - 10 Set the equations equal to each other
- 5x - 5x Subtract 5x from both sides
38 = 4x - 10
+ 10 + 10 Add 10 to both sides
48 = 4x Divide both sides by 4
12 = x
Plug 12 into both equations
5(12) + 38 = 98
9(12) - 10 = 98
If you're trying to solve for y
-2y = 8
Divide -2 to both sides

Cancel out the -2 on the left, Divide the -2 to the 8 on the right so
y = -4
Answer:
Step-by-step explanation:
this is the same equation, the system has infinitely many solutions
Answer:
(x - 5)² = 41
Step-by-step explanation:
* Lets revise the completing square form
- the form x² ± bx + c is a completing square if it can be put in the form
(x ± h)² , where b = 2h and c = h²
# The completing square is x² ± bx + c = (x ± h)²
# Remember c must be positive because it is = h²
* Lets use this form to solve the problem
∵ x² - 10x = 16
- Lets equate 2h by -10
∵ 2h = -10 ⇒ divide both sides by 2
∴ h = -5
∴ h² = (-5)² = 25
∵ c = h²
∴ c = 25
- The completing square is x² - 10x + 25
∵ The equation is x² - 10x = 16
- We will add 25 and subtract 25 to the equation to make the
completing square without change the terms of the equation
∴ x² - 10x + 25 - 25 = 16
∴ (x² - 10x + 25) - 25 = 16 ⇒ add 25 to both sides
∴ (x² - 10x + 25) = 41
* Use the rule of the completing square above
- Let (x² - 10x + 25) = (x - 5)²
∴ (x - 5)² = 41
One way to do it is with calculus. The distance between any point

on the line to the origin is given by

Now, both

and

attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.

Solving for

, you find a critical point of

.
Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.
You have

so indeed, a minimum occurs at

.
The minimum distance is then