How do u solve these equations?

4.
Let t be the tens digit of the actual score, and u be the units digit.

(10t + u) - (10u + t) = 36 ..... [1]
t + u = 14 ...... [2]

From [1] :
9t - 9u = 36
t - u = 4 ...... [3]

[2] + [3] :
2t = 18
t = 9

[2] - [3] :
2u = 10
u = 5

Actual test score = 95


5.
Let y L of 90% of hydrochloric acid is used to make 20 L of 60% hydrochloric acid.
Then, volume of 15% hydrochloric acid used = (20 - y) L

y × 90 + (20 - y) × 15% = 20 × 60%
90y + 300 - 15y = 1200
75y = 900
y = 12

12 L of 90% of hydrochloric acid is used to make the 60% hydrochloric acid.
Volume of 90% of hydrochloric acid left in the bottle = (30 - 12) L = 18 L

Strength of the final acid
= (18 × 90% + 5 × 60%) / (18 + 5)
= 83.48%

4.
When entering your test score into the computer,
Mrs McCombes accidentally reversed the two digits.
This error reduced your score by 36 points.
Mrs. McCombes told you that the sum of the digits of your actual score was 14,
corrected the error, and agreed to give you extra credit
if you could determine the actual score without looking back at the test.
What was your actual test score?
(10y + x) - (10x + y ) = 36
9(y - x) = 36
y - x = 4
x + y = 14
y = 9
x = 5
The actual test score was 95.
5.
A chemist has one 30-L bottle of 15% hydrochloric acid and one 30-L bottle of 90% hydrochloric acid.
She mixes 20 L of 60% hydrochloric acid (made by a combination of the two 30-L bottles)
and then pours 5 L of that solution back into the bottle containing the 90% hydrochloric acid.
How strong is the acid in that bottle now?
10(0.15) + 10(0.90) = 17.5(0.60)
20(0.90) + 5(0.60) = 25x
x = 0.84
The hydrochloric acid in that bottle is now 84% strong.