To solve the equations: 3x − 4Y = 1012x + 6y = 5

To solve the equations: 3x − 4Y = 1012x + 6y = 5

3x − 4Y = 10, ① 12x + 6y = 5, ② substitute ① × 3 + ② × 2 to get: 10x = 40, x = 4. Substitute ① to get: 3 × 4-4y = 10, y = 12. So the solution of the original equations is x = 4Y = 12

If the equation of X, 3x + a = ax + 3, has a solution, find the value of A

3X+a=ax+3
If a is not equal to 3, there is a solution

Y = C squared D + D squared-c divided by d-c + 3. C and D are reciprocal

y=c^2d+d^2-d/c-c+3
c=1/d
y=1/d+d^2-d^2-1/d+3
=3

The distance between a and B is 300 km. A passenger car and a freight car drive from a to B respectively. The passenger car arrives in 6 hours and the freight car arrives in 5 hours. The speed of the freight car is faster than that of the passenger car
What percentage faster?

Bus speed = 300 △ 6 = 50 km / h
Truck speed = 300 △ 5 = 60 km / h
How much faster is a freight car than a passenger car?
（60-50）÷50×100%=20%

Given (x ^ 2 + 1) (y ^ 2 + 1) = 4xy, find the value of the algebraic formula x ^ 2-3y + 2

(X^2+1)(Y^2+1)=4XY
x^2y^2+(x^2+y^2)+1-4xy=0
(x^2y^2-2xy+1)+(x^2-2xy+y^2)=0
(xy-1)^2+(x-y)^2=0
Square nonnegative
The only way to make the equation true is to
xy-1=0
x-y=0
X = 1, y = 1 or x = - 1, y = - 1
X^2-3Y+2
=1-3+2
=0
perhaps
X^2-3Y+2
=1+3+2
=6

A. When car a is 40 kilometers away from place B, car B is 50 kilometers away from place B. when car a reaches place B, how many kilometers can car B reach place B?

If the speed doesn't change
When car a is 40 km away from ground B, that is 300-40 = 260 km
At this time, B car line: 300-50 = 250 km
So the speed of B is 250 / 260 of a = 25 / 26
When car a arrives, car B runs 25 / 26 of the whole journey, leaving 1-25 / 26 = 1 / 26
That is 300 * 1 / 26 = 150 / 13 (km)

Who can solve the quadratic equation of one variable: (29 + 2x) (22 + 2x) = 29 × 22 × (4 / 3)
The result is kept to one decimal place

（29+2X)(22+2X)=29×22×（4/3)
29*22+(29+22)2X+4X^2=29*22*4/3
4X^2+102X=29*22/3
X^2+25.5X=29^22/3
(X+12.75)^2=212.67+162.5625=375.23
X+12.75=±√375.23
x1=√375.23-12.75=6.6
x2=-√375.23-12.75=-32.1

The two cars are 450 kilometers apart. They are facing each other in five hours. The speed ratio of car a and car B is 4:5. How many kilometers do the two cars travel per hour?

The speed ratio of car a and car B is 4:5
A speed of 4 / (4 + 5) = 4 / 9
B speed 5 / (4 + 5) = 5 / 9
So the speed of a is 4 / 9 * 450 / 5 = 4 / 9 * 90 = 40 km / h
B. the speed is 5 / 9 * 450 / 5 = 5 / 9 * 90 = 50 km / h

If you move a decimal point one place to the left, it will be 2.34 less than the original number. What is the original number?

2.6

The passenger car and the freight car leave from a and B at the same time. The passenger car runs 50km per hour, and the speed of the freight car is 80% of that of the passenger car. After meeting, the passenger car continues to travel for 3.2 hours to reach B. how many kilometers are there between a and B?

(50 × 3.2) △ (50 × 80%) = 160 △ 40 = 4 (hours) (50 + 50 × 80%) × 4 = (50 + 40) × 4 = 90 × 4 = 360 (kilometers) answer: A and B are 360 kilometers apart