How to solve the system of quadratic equations x + 8x = - 15 and X - 5x = - 16

How to solve the system of quadratic equations x + 8x = - 15 and X - 5x = - 16

Is it a quadratic equation with one variable?
x^2+8x=-15
If the square (8 / 2) ^ 2 of half of the coefficient of the first-order term of two sides is equal to 16, then,
x^2+8x+16=-15+16
（x+4）^2=1
X + 4 = soil 1
X = - 4 soil 1
∴x1=-3,x2=-5
x^2-5x=-16
x^2-5x+16=0
△=b^2-4ac=5^2-4×1×16＜0
The original equation has no real solution

3 (1 + x) = 3x + 7 3 (1 + x) = x (3x + 7) PX + x-4 = x (px-1) which are univariate quadratic equations?

3 (1 + x) = 3x + 7 is a univariate quadratic equation. 3 (1 + x) = x (3x + 7) PX + x-4 = x (px-1) is a univariate quadratic equation after simplification

On the univariate quadratic equation x ^ 2 + px-p ^ 2 = 0 of X, if one root is - 2, then p=

X = - 2 is substituted into the equation
4-2p-p²=0
p²+2p-4=0
The derivation of Weida theorem
p=(-2±√(4+4*4))/2=-1±√5

(equations and arithmetic) hurry!
Use the same rope to measure the same tree. The first time you fold the rope in half, you can make it around the tree for another 1 meter. The second time you fold the rope in three, you can make it around the tree for another 1.5 meters. Find the circumference of the tree and the length of the rope

y=x/2-1
y=x/3+1.5
x=15 y=6.5

How to use parallelogram rule to calculate resultant acceleration

Take velocity V1 and V2 as the side to make parallelogram, and make the starting point of vector the same. The diagonal line which has the same starting point with velocity is the resultant acceleration, and the direction is to point from this starting point to another starting point

One math problem
1 judgment
The length of line segments between parallel lines is equal
2 is a number that is a multiple of 2, which is both an even number and a composite number

1. True, if the length of the segments between parallel lines is not equal, then they intersect
2. Incorrect. 2 is a multiple of 2, but 2 is not a composite

If the function y = f (x) is an increasing function on R, it is proved that KF (x) is also an increasing function on R when k > 0
Do you want to use the definition? He didn't say to use the definition. I don't know whether to use other methods

It is generally proved by definition
Let X1 > X2, X1 and X2 belong to R
Because y = f (x) is an increasing function on R, then f (x1) - f (x2) > 0
kf(x1) - kf(x2) = k[f(x1) - f(x2)]
Because, k > 0, f (x1) - f (x2) > 0
So, K [f (x1) - f (x2)]
Then KF (x) is also an increasing function on R

The value of function f (x) = x-lnx + a (a belongs to R) is always greater than zero in the domain of definition
If the value of F (x) is always greater than zero in the domain of definition, what is the relationship between F '(x)?
The derivative is f '(x) = 1 - (1 / x). What should we do next

When the derivative is f '(x) = 1 - (1 / x) = 0, x = 1, and
When 0 < x < 1, f '(x) < 0, the function decreases;
When x > 1, f '(x) > 0, the function increases
So f (1) = 1-0 + a = 1 + A is the minimum of the function
Because the value of function f (x) = x-lnx + a (a belongs to R) is always greater than zero in the domain of definition
f(1) = 1+a > 0,
Then a > - 1

Simplify the absolute value of C minus the absolute value of C-B plus the absolute value of a + C

|a+b|-|a-c|-|c-b|
Because a

The ratio of x2 + Y2 + 1 to 2 (x + Y-1) (inequality)

x^2+y^2+1-2x-2y+2
=x^2-2x+1+y^2-2y+1+1=(x-1)^2+(y-1)^2+1>0