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Popular >

a^2(x-5)>b^2(x-5)

Solution

a^{2} (x - 5) > b^{2} (x - 5)

Solution

x > 5; x^{2} - b^{2} > 0

+1

Interval Notation

(5, \infty)

Solution steps

a^{2} (x - 5) > b^{2} (x - 5)

Expand

a^{2} (x - 5) : a^{2} x - 5 a^{2}

Expand

b^{2} (x - 5) : b^{2} x - 5 b^{2}

a^{2} x - 5 a^{2} > b^{2} x - 5 b^{2}

Move

5 a^{2}

to the right side

a^{2} x > b^{2} x - 5 b^{2} + 5 a^{2}

Move

b^{2} x

to the left side

a^{2} x - b^{2} x > - 5 b^{2} + 5 a^{2}

Factor

a^{2} x - b^{2} x : x (a^{2} - b^{2})

x (a^{2} - b^{2}) > - 5 b^{2} + 5 a^{2}

Divide both sides by

a^{2} - b^{2}; x^{2} - b^{2} > 0

x > 5; x^{2} - b^{2} > 0

Popular Examples

((a+3))/5-((2a-1))/2 >((4-a))/2-a/(10)

\frac{( a + 3 )}{5} - \frac{( 2 a - 1 )}{2} > \frac{( 4 - a )}{2} - \frac{a}{10}

- 2 z - 3 > - 7

- 3 x + 4 > - 10

(2x)/(5+1)<((x-1))/(3+3)

\frac{2 x}{5 + 1} < \frac{( x - 1 )}{3 + 3}

- 2 x + 3 \leq - 4 x - 5

Frequently Asked Questions (FAQ)

What is a^2(x-5)>b^2(x-5) ?
The solution to a^2(x-5)>b^2(x-5) is x>5;\quad x^2-b^2>0