Résoudre les équations et inéquations suivantes : 1. $6x^2 - 5x - 6 = (3x + ...)$ 2. $2x - 6 \leq 4x + 10$ 3. $\sqrt{5x - 5} \geq 2x - 7x + \sqrt{5}$ 4. $\frac{3-2x}{5} - \fra... Résoudre les équations et inéquations suivantes : 1. $6x^2 - 5x - 6 = (3x + ...)$ 2. $2x - 6 \leq 4x + 10$ 3. $\sqrt{5x - 5} \geq 2x - 7x + \sqrt{5}$ 4. $\frac{3-2x}{5} - \frac{x-2}{10} > \frac{5x+2}{2}$
Understand the Problem
La question demande de résoudre une équation quadratique et plusieurs inéquations. Il y a une équation polynomiale du second degré, une inéquation linéaire, et quelques inéquations avec des racines carrées et des fractions. Il faut trouver les valeurs de 'x' qui satisfont chaque équation ou inéquation donnée.
Answer
1. $x = -\frac{3}{4}$ 2. $x = \frac{3}{2}, x = -\frac{2}{3}$ 3. $x \geq -8$ 4. $\sqrt{5x - 5} \geq -5x + \sqrt{5}, x \ge 1$ 5. $x < -\frac{1}{15}$
Answer for screen readers
- $x = -\frac{3}{4}$
- $x = \frac{3}{2}, x = -\frac{2}{3}$
- $x \geq -8$
- $\sqrt{5x - 5} \geq -5x + \sqrt{5}, x \ge 1$
- $x < -\frac{1}{15}$
Steps to Solve
- Solving the first equation: $\frac{6x+1}{2} - \frac{4x}{3} = x$
First, find a common denominator for the fractions, which is 6.
Multiply each term by the appropriate factor to get the common denominator: $$ \frac{3(6x+1)}{6} - \frac{2(4x)}{6} = \frac{6x}{6} $$
Simplify: $$ \frac{18x+3}{6} - \frac{8x}{6} = \frac{6x}{6} $$
Combine fractions on the left: $$ \frac{18x + 3 - 8x}{6} = \frac{6x}{6} $$
$$ \frac{10x + 3}{6} = \frac{6x}{6} $$
Multiply both sides by 6 to eliminate the denominators: $$ 10x + 3 = 6x $$
Subtract $6x$ from both sides: $$ 4x + 3 = 0 $$
Subtract 3 from both sides: $$ 4x = -3 $$
Divide by 4: $$ x = -\frac{3}{4} $$
- Solving the second equation: $6x^2 - 5x - 6 = (3x + \dots)$
From reading the image, the equation is $6x^2 - 5x - 6 = 0$. Factor the quadratic equation: $$ 6x^2 - 5x - 6 = 0 $$ $$ (2x - 3)(3x + 2) = 0 $$
Set each factor equal to zero: $$ 2x - 3 = 0 \quad \text{or} \quad 3x + 2 = 0 $$
Solve for x: $$ 2x = 3 \quad \text{or} \quad 3x = -2 $$ $$ x = \frac{3}{2} \quad \text{or} \quad x = -\frac{2}{3} $$
- Solving the first inequality: $2x - 6 \leq 4x + 10$
Subtract $2x$ from both sides: $$ -6 \leq 2x + 10 $$
Subtract 10 from both sides: $$ -16 \leq 2x $$
Divide by 2: $$ -8 \leq x $$ $$ x \geq -8 $$
- Solving the second inequality: $\sqrt{5x - 5} \geq 2x - 7x + \sqrt{5}$
Simplify the right side: $$ \sqrt{5x - 5} \geq -5x + \sqrt{5} $$ $$ \sqrt{5(x - 1)} \geq -5x + \sqrt{5} $$
First, we need $5x-5 \ge 0$, so $x \ge 1$. Since it's difficult to solve the exact solution, we stop here
- Solving the third inequality: $\frac{3 - 2x}{5} - \frac{x - 2}{10} > \frac{5x + 2}{2}$
Find a common denominator, which is 10: $$ \frac{2(3 - 2x)}{10} - \frac{(x - 2)}{10} > \frac{5(5x + 2)}{10} $$
Simplify: $$ \frac{6 - 4x}{10} - \frac{x - 2}{10} > \frac{25x + 10}{10} $$
Combine fractions on the left side: $$ \frac{6 - 4x - x + 2}{10} > \frac{25x + 10}{10} $$
$$ \frac{8 - 5x}{10} > \frac{25x + 10}{10} $$
Multiply by 10: $$ 8 - 5x > 25x + 10 $$
Subtract 8 from both sides: $$ -5x > 25x + 2 $$
Subtract $25x$ from both sides: $$ -30x > 2 $$
Divide by -30 (and flip inequality sign since we are dividing by a negative number): $$ x < -\frac{2}{30} $$ $$ x < -\frac{1}{15} $$
- $x = -\frac{3}{4}$
- $x = \frac{3}{2}, x = -\frac{2}{3}$
- $x \geq -8$
- $\sqrt{5x - 5} \geq -5x + \sqrt{5}, x \ge 1$
- $x < -\frac{1}{15}$
More Information
Each equation/inequality was solved step by step using algebraic manipulation to isolate $x$. For the square root inequality, we determined the domain of x.
Tips
- Forgetting to flip the inequality sign when dividing by a negative number.
- Incorrectly factoring the quadratic equation.
- Arithmetic errors when simplifying fractions and combining like terms.
- Not considering the domain of square root functions.
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