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On considère les équations et inéquations suivantes :
$\begin{array}{l}
\left(E_1\right):-2(x+3)=x+6 \\
\left(E_2\right): 2 x-\frac{3}{2}=\frac{7}{9} \\
\left(I_1\right): 5 x+3<10 \\
\left(I_2\right):-3 x+4>-5 x+6
\end{array}$
1) Résous dans $Q$ les équations $\left(E_1\right)$ et $\left(E_2\right)$
2) Résous dans $Q$ les inéquations $\left(I_1\right)$ et $\left(I_2\right)$
Answer (2)
Les solutions aux équations et inéquations données sont : Pour l'équation (E_1), x = − 4 et pour l'équation (E_2), x = 36 41 . Pour l'inégalité (I_1), x < 5 7 et pour l'inégalité (I_2), 1"> x > 1 .
;
Solve equation ( E 1 ) : − 2 ( x + 3 ) = x + 6 , which simplifies to x = − 4 .
Solve equation ( E 2 ) : 2 x − 2 3 = 9 7 , which simplifies to x = 36 41 .
Solve inequality ( I 1 ) : 5 x + 3 < 10 , which simplifies to x < 5 7 .
Solve inequality ( I 2 ) : -5x + 6"> − 3 x + 4 > − 5 x + 6 , which simplifies to 1"> x > 1 .
The solutions are x = − 4 , x = 36 41 , x < 5 7 , and 1}"> x > 1 .
Explanation
Problem Analysis We are given two equations and two inequalities. Our goal is to solve for x in each of these, finding the solution within the set of rational numbers, denoted as Q .
Solving Equation (E1) Let's solve the first equation, ( E 1 ) : − 2 ( x + 3 ) = x + 6 . First, we distribute the − 2 on the left side: − 2 x − 6 = x + 6 Next, we want to isolate the x terms on one side and the constants on the other. We can add 2 x to both sides and subtract 6 from both sides: − 6 − 6 = x + 2 x This simplifies to: − 12 = 3 x Finally, we divide both sides by 3 to solve for x : x = 3 − 12 = − 4 So the solution to equation ( E 1 ) is x = − 4 .
Solving Equation (E2) Now, let's solve the second equation, ( E 2 ) : 2 x − 2 3 = 9 7 . We want to isolate x , so we first add 2 3 to both sides: 2 x = 9 7 + 2 3 To add the fractions, we need a common denominator, which is 18 . So we rewrite the fractions: 2 x = 18 14 + 18 27 Adding the fractions gives: 2 x = 18 41 Now, we divide both sides by 2 to solve for x : x = 18 41 ÷ 2 = 18 41 × 2 1 = 36 41 So the solution to equation ( E 2 ) is x = 36 41 .
Solving Inequality (I1) Next, we solve the first inequality, ( I 1 ) : 5 x + 3 < 10 . We subtract 3 from both sides: 5 x < 10 − 3 5 x < 7 Then, we divide both sides by 5 : x < 5 7 So the solution to inequality ( I 1 ) is x < 5 7 .
Solving Inequality (I2) Finally, we solve the second inequality, -5x + 6"> ( I 2 ) : − 3 x + 4 > − 5 x + 6 . We add 5 x to both sides: 6"> − 3 x + 5 x + 4 > 6 6"> 2 x + 4 > 6 Then, we subtract 4 from both sides: 6 - 4"> 2 x > 6 − 4 2"> 2 x > 2 Finally, we divide both sides by 2 : \frac{2}{2}"> x > 2 2 1"> x > 1 So the solution to inequality ( I 2 ) is 1"> x > 1 .
Final Answer In summary, we have solved the two equations and two inequalities. The solutions are:
Equation ( E 1 ) : x = − 4 Equation ( E 2 ) : x = 36 41 Inequality ( I 1 ) : x < 5 7 Inequality ( I 2 ) : 1"> x > 1
Examples
Understanding how to solve equations and inequalities is crucial in many real-world scenarios. For instance, imagine you're managing a budget. Equations help you balance income and expenses to ensure they are equal, while inequalities help you determine how much you can spend on certain items without exceeding your budget. Similarly, in science, you might use equations to calculate the amount of reactants needed for a chemical reaction and inequalities to determine the range of temperatures for a successful experiment. These mathematical tools provide a framework for making informed decisions and solving problems in various aspects of life.
