Given:
It is given that,
PQ ⊥ PS and
∠QPR = 7x-9
∠RPS = 4x+22
To find the value of ∠QPR.
Formula
As per the given problem PR lies between PQ and PS,
So,
∠QPR+∠RPS = 90°
Now,
Putting the values of ∠RPS and ∠QPR we get,

or, 
or, 
or, 
or, 
Substituting the value of
in ∠QPR we get,
∠QPR = 
or, ∠QPR = 
Hence,
The value of ∠QPR is 40°.
Answer:
x = 8 - 3y
Step-by-step explanation:

Hope this helps.
Answer:
x = 16
m<Y = 34°
Step-by-step explanation:
∆XYZ is an isosceles ∆. An isosceles ∆ has two equal sides, as well as the bases of the isosceles triangle are congruent. In this case, therefore:
<X = <Z
(6x - 23)° = (4x + 9)
Solve for x
6x - 23 = 4x + 9
Collect like terms
6x - 4x = 23 + 9
2x = 32
Divide both sides by 2
x = 16
m<Y = 180° - (m<X + m<Z) (sum of ∆)
m<Y = 180 - ((6x - 23) + (4x + 9))
Plug in the value of x
m<Y = 180 - ((6(16) - 23) + (4(16) + 9))
m<Y = 180 - (73 + 73)
m<Y = 34°