The original equation has no solution in the real number domain because there's no value of
x that would make both x−5 and 2x+7 positive at the same time and satisfy the equation.
Frequently asked questions (FAQs)
Question: Find the area of a right-angled triangle if one leg measures 8 units and the hypotenuse is 10 units.
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What is the vertex form of the quadratic function f(x) = x^2 and what are the coordinates of the vertex?
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Question: Graph the inequality y ≤ 2x - 4, where x and y are real numbers.