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Sagot :
Sure, let's solve the equation [tex]\(\left|\frac{1}{2} x + 3\right| = x + 3\)[/tex].
An absolute value equation [tex]\(|A| = B\)[/tex] holds true if and only if [tex]\(A = B\)[/tex] or [tex]\(A = -B\)[/tex]. Therefore, we split the given equation into two cases:
1. [tex]\(\frac{1}{2} x + 3 = x + 3\)[/tex]
2. [tex]\(\frac{1}{2} x + 3 = -(x + 3)\)[/tex]
Let's solve each case separately:
### Case 1: [tex]\(\frac{1}{2} x + 3 = x + 3\)[/tex]
Subtract 3 from both sides:
[tex]\[ \frac{1}{2} x = x \][/tex]
Subtract [tex]\(\frac{1}{2} x\)[/tex] from both sides:
[tex]\[ 0 = \frac{1}{2} x \][/tex]
Divide both sides by [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ x = 0 \][/tex]
Now, verify the solution [tex]\(x = 0\)[/tex] by substituting it back into the original equation:
[tex]\[ \left|\frac{1}{2}(0) + 3\right| = 0 + 3 \][/tex]
Simplifying both sides:
[tex]\[ |3| = 3 \][/tex]
This is a true statement, so [tex]\(x = 0\)[/tex] is a valid solution.
### Case 2: [tex]\(\frac{1}{2} x + 3 = -(x + 3)\)[/tex]
Distribute the negative sign on the right-hand side:
[tex]\[ \frac{1}{2} x + 3 = -x - 3 \][/tex]
Add [tex]\(x\)[/tex] to both sides:
[tex]\[ \frac{3}{2} x + 3 = -3 \][/tex]
Subtract 3 from both sides:
[tex]\[ \frac{3}{2} x = -6 \][/tex]
Divide both sides by [tex]\(\frac{3}{2}\)[/tex]:
[tex]\[ x = -6 \div \frac{3}{2} = -6 \times \frac{2}{3} = -4 \][/tex]
Now, verify the solution [tex]\(x = -4\)[/tex] by substituting it back into the original equation:
[tex]\[ \left|\frac{1}{2}(-4) + 3\right| = -4 + 3 \][/tex]
Simplifying both sides:
[tex]\[ |-2 + 3| = -1 \][/tex]
[tex]\[ |1| = -1 \][/tex]
This is not a true statement, hence [tex]\(x = -4\)[/tex] is not a valid solution.
### Conclusion
The only solution to the equation [tex]\(\left|\frac{1}{2} x + 3\right| = x + 3\)[/tex] is:
[tex]\[ \boxed{0} \][/tex]
An absolute value equation [tex]\(|A| = B\)[/tex] holds true if and only if [tex]\(A = B\)[/tex] or [tex]\(A = -B\)[/tex]. Therefore, we split the given equation into two cases:
1. [tex]\(\frac{1}{2} x + 3 = x + 3\)[/tex]
2. [tex]\(\frac{1}{2} x + 3 = -(x + 3)\)[/tex]
Let's solve each case separately:
### Case 1: [tex]\(\frac{1}{2} x + 3 = x + 3\)[/tex]
Subtract 3 from both sides:
[tex]\[ \frac{1}{2} x = x \][/tex]
Subtract [tex]\(\frac{1}{2} x\)[/tex] from both sides:
[tex]\[ 0 = \frac{1}{2} x \][/tex]
Divide both sides by [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ x = 0 \][/tex]
Now, verify the solution [tex]\(x = 0\)[/tex] by substituting it back into the original equation:
[tex]\[ \left|\frac{1}{2}(0) + 3\right| = 0 + 3 \][/tex]
Simplifying both sides:
[tex]\[ |3| = 3 \][/tex]
This is a true statement, so [tex]\(x = 0\)[/tex] is a valid solution.
### Case 2: [tex]\(\frac{1}{2} x + 3 = -(x + 3)\)[/tex]
Distribute the negative sign on the right-hand side:
[tex]\[ \frac{1}{2} x + 3 = -x - 3 \][/tex]
Add [tex]\(x\)[/tex] to both sides:
[tex]\[ \frac{3}{2} x + 3 = -3 \][/tex]
Subtract 3 from both sides:
[tex]\[ \frac{3}{2} x = -6 \][/tex]
Divide both sides by [tex]\(\frac{3}{2}\)[/tex]:
[tex]\[ x = -6 \div \frac{3}{2} = -6 \times \frac{2}{3} = -4 \][/tex]
Now, verify the solution [tex]\(x = -4\)[/tex] by substituting it back into the original equation:
[tex]\[ \left|\frac{1}{2}(-4) + 3\right| = -4 + 3 \][/tex]
Simplifying both sides:
[tex]\[ |-2 + 3| = -1 \][/tex]
[tex]\[ |1| = -1 \][/tex]
This is not a true statement, hence [tex]\(x = -4\)[/tex] is not a valid solution.
### Conclusion
The only solution to the equation [tex]\(\left|\frac{1}{2} x + 3\right| = x + 3\)[/tex] is:
[tex]\[ \boxed{0} \][/tex]
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