What is solvefor t,x^2-(1+t)x+t>0 ?

The solution to solvefor t,x^2-(1+t)x+t>0 is t>x;\quad x<1

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solvefor t,x^2-(1+t)x+t>0

solve for

t, x^{2} - (1 + t) x + t > 0

t > x; x < 1

Solution steps

x^{2} - (1 + t) x + t > 0

Move

x^{2}

to the right side

- (1 + t) x + t > - x^{2}

Expand

- (1 + t) x + t : - x - t x + t

- x - t x + t > - x^{2}

Move

x

to the right side

- t x + t > - x^{2} + x

Factor

- t x + t : t (- x + 1)

t (- x + 1) > - x^{2} + x

Divide both sides by

- x + 1; x < 1

t > x; x < 1

\frac{( x + 1 )}{4} \geq 8 - 2 x

lo g_{10} (7) (6 x) > 7 (8 - 4 x)

x, x + 9 < - y + 9

- 6 \leq \frac{- 9 + ( - n )}{4}

4 (- 3 i + 4) > - 1 (36 u + 16)

What is solvefor t,x^2-(1+t)x+t>0 ?
The solution to solvefor t,x^2-(1+t)x+t>0 is t>x;\quad x<1