To solve the equations, one half x + one third y = 1 3x-y = 2

To solve the equations, one half x + one third y = 1 3x-y = 2

3x-y=2
y=3x-2
1/2x+1/3(3x-2)=1
1/2x+x-2/3=1
3/2x=5/3
x=10/9
y=3*10/9-2=10/3-2=4/3
If you don't understand this question, you can ask,

Given (3x-2y) ^ 2-10 (3x-2y) + 25 = 0, find the value of 9x ^ 2-12xy + 4Y ^ 2 + 1

（3x-2y）^2-10（3x-2y）+25=0,
(3x-2y-5)^2=0
therefore
3x-2y-5=0
3x-2y=5
9x^2-12xy+4y^2+1
=(3x-2y)^2+1
=25+1
=26

Let the proposition be "if M > 0, then the equation x2 + x-m = 0 about X has a real root", try to write its no proposition, inverse proposition and inverse no proposition, and judge their truth and falseness respectively

The negative proposition is "if M ≤ 0, then the equation x2 + x-m = 0 about X has no real roots"; (3) the inverse proposition is "if the equation x2 + x-m = 0 about X has real roots, then M > 0"; (6) the inverse proposition is "if the equation x2 + x-m = 0 about X has no real roots, then m ≤ 0"; (9 points) from the discriminant of equation △ = 1 + 4m, △ 0 is obtained, that is, m ＞− 14, and the equation has real roots. When m ＞ 0 makes 1 + 4m ＞ 0, the equation x2 + x-m = 0 has real roots, and the original proposition is true, so the inverse proposition is true. (10 points) however, when equation x2 + x-m = 0 has real roots, it must be m ＞− 14, and m ＞ 0 cannot be deduced, so the inverse proposition is false. (11 points)