3y(2 - y) + 8y
6y - 3y^2 + 8y
-3y^2 + 14 y
        
             
        
        
        
Do you will need to know that for what.
        
             
        
        
        
<span>-3x + 2y = 2  eq (1)
 
7x - 3y = 7    eq (2) 
Mutiply eq (1) by 3 and eq (2) by 2:
-9x + 6y = 6
14x - 6y = 14
Add th two new equations:
-9x + 14x = 6 + 14
=> 5x = 20
=> x = 20/5
=> x = 4
Now from eq (1) 
-3(4) + 2y = 2
=> 2y = 2 + 12
=> 2y = 14
=> y = 14 / 2
=> y = 7
So the ordered pair is (4,7)
Answer: option B. (4,7)
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Answer: 14x^2-93xy+60y^2 Hope that helps!
Step-by-step explanation: 
1. Expand by distributing terms 
(20x-12y)(x-4y)-(3x-4y)(2x+3y)
2. Use the Foil method:(a+b)(c+d)= ac+ad+bc+bd
20x^2-80xy-12yx+48y^2-(3x-4y)(2x+3y)
3. Use the Foil method : (a+b)(c+d)= ac+ad+bc+bd
20x^2-80xy-12yx+48y^2-(6x^2+9xy-8yx-12y^2)
4. Remove parentheses 20x^2-80xy-12yx+48y^2-6x^2-9xy+ 8yx+12y^2
5. Collect like terms (20x^2-6x^2)+(-80xy-12xy-9xy+8xy)+(48y^2+12y^2)
6. Simplify.
And your answer would be 14x^2-93xy+60y^2
 
        
             
        
        
        
Let's define variables:
 s = original speed
 s + 12 = faster speed
 The time for the half of the route is:
 60 / s
 The time for the second half of the route is:
 60 / (s + 12)
 The equation for the time of the trip is:
 60 / s + 60 / (s + 12) + 1/6 = 120 / s
 Where,
 1/6: held up for 10 minutes (in hours).
 Rewriting the equation we have:
 6s (60) + s (s + 12) = 60 * 6 (s + 12)
 360s + s ^ 2 + 12s = 360s + 4320
 s ^ 2 + 12s = 4320
 s ^ 2 + 12s - 4320 = 0
 We factor the equation:
 (s + 72) (s-60) = 0
 We take the positive root so that the problem makes physical sense.
 s = 60 Km / h
 Answer:
 The original speed of the train before it was held up is:
 s = 60 Km / h