1、 Solve the following equations 1, {x + y = 3, y + Z = 5, x + Z = 4} 2、｛x/2=y/3=z/5,2x+y+3z=88｝ Please, please.

1、 Solve the following equations 1, {x + y = 3, y + Z = 5, x + Z = 4} 2、｛x/2=y/3=z/5,2x+y+3z=88｝ Please, please.

(1)x+y=3,y+z=5,x+z=4
Add three formulas: x + y + y + Z + X + Z = 2 (x + y + Z) = 12
x+y+z=6
x+y=3,z=3
y+z=5,y=2,x+1
(2) Let x = 2K, y = 3k, z = 5K
2x+y+3z=88
4l+3k+15k=88
22k=88,k=4
x=8,y=12,z=20

The roots of some higher one equations and the zeros of functions!
1. Find the zero point of the function
①y=-x^2+x+6
②y=(x^2-2)(x^2-3x+2)
2. The interval of the root of square root lgx + x = 0 is ()
A.(-∞,0)
B.(0,1)
C.(1,2)
D.(2,4)
3. It is known that the image of the function y = f (x) is continuous, if there is a corresponding value table as follows
X 1 2 3 4 5 6
y 123.56 21.45 -7.82 11.45 -53.76 -128.88
Then the function y = f (x) has at least ()
A. 2 B.3 C.4 D 5
A function with zeros in the interval [3,5] is ()
A.f(x)=2xln(x-2)-3;
B.f(x)=-x^3-3x+5
C.f(x)=2^x-4
D.f(x)=-1/x+2
4. If the two zeros of the function f (x) = x ^ 2 - (T-2) x + 5-T are greater than 2, then the value range of T is_____
5. The quadratic equation x ^ 2 + (m-1) x + 1 = 0 has unique solution in the interval [0,2],
Then the value range of the real number M_________
6. Let f (x) = ax + 2A + 1 (a ≠ 0)

Question 1: y = - x ^ 2 + X + 6 = - (x + 2) (x-3) so its two points are x = - 2 and x = 3Y = (x ^ 2-2) (x ^ 2-3x + 2) = (x + √ 2) (x - √ 2) (x-1) (X-2) so its zero points are x = √ 2 and x = - √ 2 and x = 1 and x = 2

The root of the equation of the first grade of higher education and the problem of the zero point of the function
The equation MX2 + 2 "m + 3" x + 2m + 14 = 0 of X has two real roots, one is greater than 4 and the other is less than 4, so we can find the value range of M,

M can't be 0, the equation is as follows:
f(x)=x^2+2(1+3/m)x+2+14/m=0 ,
Because the opening of F (x) is upward, only f (4) is needed