Solve the equation, 2x + 3Y = 1, x-2y = 3, try to judge the sign of the solution

Solve the equation, 2x + 3Y = 1, x-2y = 3, try to judge the sign of the solution

(1) 2X + 3Y = 1, (2) x-2y = 3 (2) the equation is expanded twice 2x-4y = 6. By synthesizing (1), we can conclude that y is negative and X is positive

It is known that the equation y & # 178; + 2Y = a + 9 (1) about y has no real root. Try to judge the root of the equation x & # 178; + ax-2a + 5 = 0 (2) about X

y²+2y=a+9
y²+2y-(a+9)=0
No real root
Then: △ = 4 + 4 (a + 9)

Given the square of the equation AX + (2a + 1) x + (a + 1) = 0, (a ≠ 0) prove: this equation has two unequal roots!

∵a≠0
This is a quadratic equation of one variable
∴△=(2a+1)²-4a(a+1)
=4a²+4a+1-4a²-4a
=1＞0
The equation has two unequal roots
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a=___ The root of the equation AX + 3 = x + 2a is 1

Substituting x = 1 into the equation AX + 3 = x + 2A
a+3=1+2a
a-2a=1-3
-a=-2
a=2