Is 2x-5 = 2x + 6 a linear equation with one variable

Is 2x-5 = 2x + 6 a linear equation with one variable

yes.
2X - 5 = 2x + 6 is a linear equation with one variable

x/2+x/6+x/12+x/20+x/30=5
X / 2 / is a fractional line

50x/60=5
x=6

1-1 / 2 + 1 / 6 + 1 / 12 + 1 / 20 + 1 / 30 + 1 / 42 must use simple method

Hello, 1-1 / 2 + 1 / 6 + 1 / 12 + 1 / 20 + 1 / 30 + 1 / 42 = 1-1 / 2 + (1 / 2-1 / 3) + (1 / 4-1 / 5) + (1 / 5-1 / 6) + (1 / 6-1 / 7) = 1-1 / 2 + 1 / 2-1 / 3 + 1 / 4-1 / 5 + 1 / 5-1 / 6 + 1 / 6-1 / 7 = 1-1 / 6 = 6 / 7

The square of (into minus 3) minus 16 equals the solution of the zero equation?

The square of (x-3) - 16 = 0
Square of (x-3) = 16
x-3=±4
X = 7 or x = - 1

The solution of the equation 18x minus 39 equals 56

18x-39＝56
18x＝39＋56
18x＝95
x＝95/18

If x = 1, y = 2 is the solution of the system of equations 2x-y = 4P, KX + 2Y = 1, find the value of P and KDE
If x = 1, y = 2 is the solution of the system of equations 2x-y = 4P, KX + 2Y = 1, find the value of P and KDE
2. There are 27 people in an engineering team, each of whom can dig 4 tons of sand or transport 5 tons of sand every day. In order to transport the excavated sand in time, how many people should be allocated to dig and transport sand?

X = 1, y = 2, because 2 * 1-2 = 0, then the formula P is valid when it is equal to 0, so the formula 2 * 1-2 = 4 * 0 is correct, so p = 0k * 1 + 2 * 2 = 1, k = - 3, let's set the man-made x for dredging sand and the man-made y for transporting sand. Because in order to transport sand away, the tonnage of dredging sand and transporting sand should be the same, so 4 * x = 5 * y because 27 = x + y

The common solution of three bivariate linear equations 2x + 5y-6 = 0, 3x-2y-9 = 0, y = kx-9 is k = ()
A. 4B. 3C. 2D. 1

From the meaning of the question: 2x + 5Y − 6 = 03x − 2Y − 9 = 0y = KX − 9, ① × 3 - ② × 2 get y = 0, substitute ① get x = 3, substitute x, y into ③, get 3k-9 = 0, get k = 3

The common solution of three bivariate linear equations 2x + 5y-6 = 0, 3x-2y-9 = 0, y = kx-9 is k = ()
A. 4B. 3C. 2D. 1

From the meaning of the question: 2x + 5Y − 6 = 03x − 2Y − 9 = 0y = KX − 9, ① × 3 - ② × 2 get y = 0, substitute ① get x = 3, substitute x, y into ③, get 3k-9 = 0, get k = 3

Three bivariate linear equations, 2x + 5y-6 = 0 3x-2y-9 = 0 y = kx-9, have the common solution k k =? A.4 B.3 C2. D.1

2x+5y-6=0
3x-2y-9=0
The solution of the simultaneous equation is x = 3. Y = 0
Substituting x = 3. Y = 0 into y = kx-9
0=3k-9
The solution is k = 3

Given that the two roots of the equation are the two right sides of a right triangle, find the area of the right angle
The equation is x ^ 2-7x + 12 = o

x^2-7x+12=o
(x-3)(x-4)=0
The solution is: X1 = 3, X2 = 4
That is, the two right angle sides are 3 and 4, so the area is: 1 / 2 * 3 * 4 = 6