3x=5y 2y-2=x

3x=5y 2y-2=x

3X=5Y…… one
2Y-2=X…… two
From 2, 6y-6 = 3x three
Substituting 3 into 1, the solution is y = 6
Take y = 6 into 2 to get x = 10

(3x-5y+1)^2-(x+2y)(x-2y)

(3x-5y+1)^2-(x+2y)(x-2y)
=9x²＋25y²＋1-30xy＋6x-10y-x²＋4y²
=8x²＋29y²-30xy＋6x-10y＋1

Can the four numbers of 6666 be 24 o'clock

6+6+6+6=24 6X6-6-6=24
Remember to adopt it

It is known that {x = - 1, y = 2} is the solution of the equations {3x + 2Y = a, 5x + 2Y = B}, and the value of (a + 2b) is obtained

It is known that {x = - 1, y = 2} is the solution of {3x + 2Y = a, 5x + 2Y = B} three thousand eight hundred and one

A car drove 195 kilometers at 3:00. At this speed, the car drove from a to B for 9:00. How many kilometers is the distance between the two places?

195 △ 3 × 9 = 585km

If the lengths of two sides of a triangle are 2 and 7 respectively, and the third side is even, what is the length of the third side

7-2 ＜ third side ＜ 7 + 2
5 ＜ third side ＜ 9
Because it's even
So it can be: 6 and 8

A couple of cars drove from city a to city B for 3 / 10 of the whole journey in 4 / 5 hours?

Given that AB is a rational number and ab is different sign, the size relation of / A + B /, / A-B /, / A / ten / B / is

/a+b/

The US aircraft carrier "Kennedy" is equipped with an ejection system to help the aircraft take off (the ejection system can make the aircraft have an initial speed). It is known that the maximum acceleration of the "f-a15" fighter when accelerating on the runway is 5.0 M / S2, and the take-off speed (the speed at which the aircraft leaves the aircraft carrier) is 50 m / S. (1) if the aircraft taxis for 100 m and then takes off, the ejection system must be used What initial speed must the aircraft have? (the root sign can be retained) (2) assuming that the aircraft carrier is not equipped with ejection system, but the aircraft is required to take off normally on it, what is the minimum length of the aircraft carrier?

(1) From the formula of velocity and displacement, we can get: v2-v02 = 2aX, we can get: V0 = 1015m / S; (2) from the law of uniform linear motion, we can get that the runway length of the aircraft needs at least: S = v22a = 25002 × 5 = 250m; (1) the initial velocity is 1015m / S; (2) the runway length of the aircraft needs at least 250m

6 (2x minus 7) is equal to 5 (x plus 8) plus 2, use the equation solution, thank you

6 (2x minus 7) equals 5 (x plus 8) plus 2
12x-42=5x+40+2
7x=84
x=12