Learning techniques online - Training problem solving skills - algebraic inequalities

This page was created by EMMentor_Algebra_Inequalities

These pages allow you to try 15 learning techniques out of 57 techniques available in the program EMMentor. Each learning technique applies to the solution of math problem. Select a technique and do what is required at each solution step. If doubt, click an arrow and get this step done. An icon of magnifier allows seeing all steps done. All examples include an option of translation to German (a globe icon). The program EMMentor offers all shown learning techniques. The first two learning techniques are available in the programs EMSolution and EMMentor. At the bottom of this page we offer you to see examples of performance/error analysis available in the program EMMentor.

Learning techniques

Selecting a formula for the next step

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Description

For the current step of solution, the program provides complete info and asks to select a formula that will underlie the next solution step. This algorithm trains an understanding of cause-and-effect relationships and active reasoning skills.

Rational inequalities

Weak linear inequalities

Weak inequalities reduced to linear inequalities. Method of substitution. Properties of powers

Solve the inequality

{(\frac{- 35 u - 5}{6 u - 8} + 3)}^{2} + {(\frac{- 21 u - 3}{6 u - 8} + 7)}^{2} \geq \frac{(16 \cdot {(\frac{- 7 u - 1}{6 u - 8})}^{2} + \frac{- 168 u - 24}{6 u - 8} + 9) \cdot (- \frac{- 7 u - 1}{6 u - 8} - 6)}{\frac{28 u + 4}{6 u - 8} - 3} + 2 \cdot (\frac{- 35 u - 5}{6 u - 8} + 3) \cdot (\frac{- 21 u - 3}{6 u - 8} + 7)

1. Let's find the domain of definition D(f) for the considered expression, taking into account that the division by zero is not defined

{\begin{matrix} 6 u - 8 \neq 0 \\ \frac{28 u + 4}{6 u - 8} - 3 \neq 0 \end{matrix}

Methodical task: Scheme "To select for transformation 2"

To select the formula allowing to accomplish the next step of transformation

Select the continuation

	$a \cdot (b + c) = a \cdot b + a \cdot c$
	$\frac{a}{b} + \frac{c}{d} = \frac{a d + b c}{b d}$
	$\begin{array}{l} {(a + b)}^{n} = \sum_{k = 0}^{n} C_{n}^{k} \cdot a^{(n - k)} \cdot b^{k}, \\ {\begin{cases} C_{n}^{k} = \frac{n!}{(n - k)! \cdot k!} \\ 0! = 1 n! = (n - 1)! \cdot n \end{cases} \\ E x a m p l e 1 : {(a \pm b)}^{5} = C_{5}^{0} \cdot a^{5} \pm C_{5}^{1} \cdot a^{4} \cdot b + C_{5}^{2} \cdot a^{3} \cdot b^{2} \pm C_{5}^{3} \cdot a^{2} \cdot b^{3} + C_{5}^{4} \cdot a \cdot b^{4} \pm C_{5}^{5} \cdot b^{5} \\ {(a \pm b)}^{5} = a^{5} \pm 5 \cdot a^{4} \cdot b + 10 \cdot a^{3} \cdot b^{2} \pm 10 \cdot a^{2} \cdot b^{3} + 5 \cdot a \cdot b^{4} \pm b^{5} \\ E x a m p l e 2 : C_{7}^{3} = \frac{7!}{(7 - 3)! \cdot 3!} = \frac{4! \cdot 5 \cdot 6 \cdot 7}{4! \cdot 3!} = \frac{5 \cdot 6 \cdot 7}{1 \cdot 2 \cdot 3} = 35 \end{array}$

Hint

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