Level of solution complexity - Solutions - Algebra Equations

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EMTeachline mathematics software offers 11 levels of math problems complexity, from basic to advanced. These examples enable you to view solutions of varied complexity in arithmetic, algebra, trigonometry and hyperbolic trigonometry. All EMTeachline programs offer the full range of solution complexity.

Topic	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	Algebra Equations
Complexity	1	2	3	4	5	6	7	8	9	10	11					Level of solution complexity 1

Help

Solve the equation

\frac{- 4 s - 3}{1} = \frac{- 7 s - 2}{8} + \frac{s - 4}{3} + \frac{- 4 s - 5}{- 5}

1. The considered expression contains no forbidden operations (division by zero, extraction of the even roots from negative numbers, etc.). Therefore, the domain of definition is:

s \in R

2. Using the addition principle of equivalence of equations, let us move the first addend from the right-hand side of the equation to its left-hand side

\frac{- 4 s - 3}{1} - \frac{- 7 s - 2}{8} = \frac{s - 4}{3} + \frac{- 4 s - 5}{- 5}

3. Let's add up fractions by applying the rule of algebraic addition (with allowance made for the sign) of fractions

\frac{(- 4 s - 3) \cdot (8) - (- 7 s - 2) \cdot (1)}{(1) \cdot (8)} = \frac{s - 4}{3} + \frac{- 4 s - 5}{- 5}

4. Let's multiply polynomials by each other, using the distributive law

\frac{(- 32 s - 24) - (- 7 s - 2)}{8} = \frac{s - 4}{3} + \frac{- 4 s - 5}{- 5}

5. Let's combine polynomials, using the definition of operation of addition

\frac{- 32 s - 24 + 7 s + 2}{8} = \frac{s - 4}{3} + \frac{- 4 s - 5}{- 5}

6. Let's collect similar terms of polynomial

\frac{- 32 s + 7 s - 24 + 2}{8} = \frac{s - 4}{3} + \frac{- 4 s - 5}{- 5}

7. Let's add up coefficients at the like terms

\frac{- 25 s - 22}{8} = \frac{s - 4}{3} + \frac{- 4 s - 5}{- 5}

8. Using the addition principle of equivalence of equations, let us move the first addend from the right-hand side of the equation to its left-hand side

\frac{- 25 s - 22}{8} - \frac{s - 4}{3} = \frac{- 4 s - 5}{- 5}

9. Let's add up fractions by applying the rule of algebraic addition (with allowance made for the sign) of fractions

\frac{(- 25 s - 22) \cdot (3) - (s - 4) \cdot (8)}{(8) \cdot (3)} = \frac{- 4 s - 5}{- 5}

10. Let's multiply polynomials by each other, using the distributive law

\frac{(- 75 s - 66) - (8 s - 32)}{24} = \frac{- 4 s - 5}{- 5}

11. Let's combine polynomials, using the definition of operation of addition

\frac{- 75 s - 66 - 8 s + 32}{24} = \frac{- 4 s - 5}{- 5}

12. Let's collect similar terms of polynomial

\frac{- 75 s - 8 s - 66 + 32}{24} = \frac{- 4 s - 5}{- 5}

13. Let's add up coefficients at the like terms

\frac{- 83 s - 34}{24} = \frac{- 4 s - 5}{- 5}

14. Let's move expressions from one side of equation to the other

\frac{- 83 s - 34}{24} - \frac{- 4 s - 5}{- 5} = 0

15. Let's add up fractions by applying the rule of algebraic addition (with allowance made for the sign) of fractions

\frac{(- 83 s - 34) \cdot (5) - (4 s + 5) \cdot (24)}{(24) \cdot (5)} = 0

16. Let's multiply polynomials by each other, using the distributive law

\frac{(- 415 s - 170) - (96 s + 120)}{120} = 0

17. Let's combine polynomials, using the definition of operation of addition

\frac{- 415 s - 170 - 96 s - 120}{120} = 0

18. Let's clear the fraction, using the multiplication principle of equivalence of equations (inequalities)

- 415 s - 170 - 96 s - 120 = 0

19. Let's collect similar terms of polynomial

- 415 s - 96 s - 170 - 120 = 0

20. Let's add up coefficients at the like terms

- 511 s - 290 = 0

21. Let's multiply both sides of equation by "-1"

511 s + 290 = 0

22. Using the addition principle of equivalence of equations, let us move the numeric addend from the left-hand side of the equation to its right-hand side

511 s = - 290

23. Using the multiplication principle of equivalence of equations, let us divide both sides of the equation by numerical coefficient at the argument

s = - \frac{290}{511}

24. Answer

\forall s \in {- \frac{290}{511}}

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