Question

solve each equation. if there is no solution, write no solution.

1. |r - 8| = 5
2. |c + 4| = 6
3. 2 = |g + 3|
4. 3 = |m + 2|
5. -2|7d| = 14
6. -3|2w| = -12
7. 3|v - 3| = 9
8. 2|d + 4| = 8
9. |4f + 1| - 2 = 5
10. |3t - 2| + 6 = 2
11. 4|2y - 3| - 1 = 11
12. 3|x + 2| + 4 = 13
13. -4|k| = 12
14. |-3n| - 2 = 4
15. -4|k + 1| = 16

Answer:

Let's solve these absolute - value equations one by one. We will use the property of absolute value: if \(|x| = a\) (\(a\geq0\)), then \(x = a\) or \(x=-a\).

Problem 17: \(|r - 8|=5\)

Step 1: Apply the absolute - value property

According to the property of absolute value, if \(|r - 8| = 5\), then we have two cases:
Case 1: \(r-8 = 5\)
Case 2: \(r - 8=-5\)

Step 2: Solve for \(r\) in each case

• For Case 1 (\(r - 8=5\)):

Add 8 to both sides of the equation. \(r=5 + 8=13\)

• For Case 2 (\(r - 8=-5\)):

Add 8 to both sides of the equation. \(r=-5 + 8 = 3\)

So the solutions for \(r\) are \(r = 13\) or \(r = 3\).

Problem 18: \(|c + 4|=6\)

Step 1: Apply the absolute - value property

Using the property \(|x|=a\Rightarrow x = a\) or \(x=-a\), we get two cases:
Case 1: \(c + 4=6\)
Case 2: \(c + 4=-6\)

Step 2: Solve for \(c\) in each case

• For Case 1 (\(c + 4=6\)):

Subtract 4 from both sides. \(c=6 - 4=2\)

• For Case 2 (\(c + 4=-6\)):

Subtract 4 from both sides. \(c=-6 - 4=-10\)

So the solutions for \(c\) are \(c = 2\) or \(c=-10\).

Problem 19: \(2=|g + 3|\)

Step 1: Apply the absolute - value property

We can rewrite it as \(|g + 3|=2\). Then we have two cases:
Case 1: \(g + 3=2\)
Case 2: \(g + 3=-2\)

Step 2: Solve for \(g\) in each case

• For Case 1 (\(g + 3=2\)):

Subtract 3 from both sides. \(g=2-3=-1\)

• For Case 2 (\(g + 3=-2\)):

Subtract 3 from both sides. \(g=-2 - 3=-5\)

So the solutions for \(g\) are \(g=-1\) or \(g=-5\).

Problem 20: \(3=|m + 2|\)

Step 1: Apply the absolute - value property

Rewrite it as \(|m + 2|=3\). We get two cases:
Case 1: \(m + 2=3\)
Case 2: \(m + 2=-3\)

Step 2: Solve for \(m\) in each case

• For Case 1 (\(m + 2=3\)):

Subtract 2 from both sides. \(m=3 - 2=1\)

• For Case 2 (\(m + 2=-3\)):

Subtract 2 from both sides. \(m=-3 - 2=-5\)

So the solutions for \(m\) are \(m = 1\) or \(m=-5\).

Problem 21: \(-2|7d|=14\)

Step 1: Isolate the absolute - value expression

Divide both sides of the equation by \(- 2\):
\(|7d|=\frac{14}{-2}=-7\)

But the absolute value of a number is always non - negative (i.e., \(|x|\geq0\) for all real numbers \(x\)). Since \(-7<0\), there is no solution for this equation.

Problem 22: \(-3|2w|=-12\)

Step 1: Isolate the absolute - value expression

Divide both sides of the equation by \(-3\):
\(|2w|=\frac{-12}{-3}=4\)

Step 2: Apply the absolute - value property

We have two cases:
Case 1: \(2w = 4\)
Case 2: \(2w=-4\)

Step 3: Solve for \(w\) in each case

• For Case 1 (\(2w = 4\)):

Divide both sides by 2. \(w=\frac{4}{2}=2\)

• For Case 2 (\(2w=-4\)):

Divide both sides by 2. \(w=\frac{-4}{2}=-2\)

So the solutions for \(w\) are \(w = 2\) or \(w=-2\).

Problem 23: \(3|v - 3|=9\)

Step 1: Isolate the absolute - value expression

Divide both sides of the equation by 3:
\(|v - 3|=\frac{9}{3}=3\)

Step 2: Apply the absolute - value property

We have two cases:
Case 1: \(v - 3=3\)
Case 2: \(v - 3=-3\)

Step 3: Solve for \(v\) in each case

• For Case 1 (\(v - 3=3\)):

Add 3 to both sides. \(v=3 + 3=6\)

• For Case 2 (\(v - 3=-3\)):

Add 3 to both sides. \(v=-3+3 = 0\)

So the solutions for \(v\) are \(v = 6\) or \(v = 0\).

Problem 24: \(2|d + 4|=8\)

Step 1: Isolate the absolute - value expression

Divide both sides of the equation by 2:
\(|d + 4|=\frac{8}{2}=4\)

Step 2: Apply the absolute - value property

We have two cases:
Case 1: \(d + 4=4\)
Case 2: \(d + 4=-4\)

Step 3: Solve for \(d\) in each case

• For Case 1 (\(d + 4=4\)):

Subtract 4 from both sides. \(d=4 - 4=0\)

• For Case 2 (\(d + 4=-4\)):

Subtract 4 from both sides. \(d=-4 - 4=-8\)

So the solutions for \(d\) are \(d = 0\) or \(d=-8\).

Problem 25: \(|4f + 1|-2=5\)

Step 1: Isolate the absolute - value expression

Add 2 to both sides of the equation:
\(|4f+1|=5 + 2=7\)

Step 2: Apply the absolute - value property

We have two cases:
Case 1: \(4f + 1=7\)
Case 2: \(4f + 1=-7\)

Step 3: Solve for \(f\) in each case

• For Case 1 (\(4f + 1=7\)):

Subtract 1 from both sides: \(4f=7 - 1=6\)
Divide both sides by 4: \(f=\frac{6}{4}=\frac{3}{2}\)

• For Case 2 (\(4f + 1=-7\)):

Subtract 1 from both sides: \(4f=-7 - 1=-8\)
Divide both sides by 4: \(f=\frac{-8}{4}=-2\)

So the solutions for \(f\) are \(f=\frac{3}{2}\) or \(f=-2\).

Problem 26: \(|3t - 2|+6=2\)

Step 1: Isolate the absolute - value expression

Subtract 6 from both sides of the equation:
\(|3t - 2|=2 - 6=-4\)

Since the absolute value of a number is non - negative (\(|x|\geq0\) for all real \(x\)) and \(-4<0\), there is no solution for this equation.

Problem 27: \(4|2y - 3|-1=11\)

Step 1: Isolate the absolute - value expression

Add 1 to both sides:
\(4|2y - 3|=11 + 1=12\)
Divide both sides by 4:
\(|2y - 3|=\frac{12}{4}=3\)

Step 2: Apply the absolute - value property

We have two cases:
Case 1: \(2y-3 = 3\)
Case 2: \(2y - 3=-3\)

Step 3: Solve for \(y\) in each case

• For Case 1 (\(2y-3 = 3\)):

Add 3 to both sides: \(2y=3 + 3=6\)
Divide both sides by 2: \(y = 3\)

• For Case 2 (\(2y - 3=-3\)):

Add 3 to both sides: \(2y=-3 + 3=0\)
Divide both sides by 2: \(y = 0\)

So the solutions for \(y\) are \(y = 3\) or \(y = 0\).

Problem 28: \(3|x + 2|+4=13\)

Step 1: Isolate the absolute - value expression

Subtract 4 from both sides:
\(3|x + 2|=13 - 4=9\)
Divide both sides by 3:
\(|x + 2|=\frac{9}{3}=3\)

Step 2: Apply the absolute - value property

We have two cases:
Case 1: \(x + 2=3\)
Case 2: \(x + 2=-3\)

Step 3: Solve for \(x\) in each case

• For Case 1 (\(x + 2=3\)):

Subtract 2 from both sides: \(x=3 - 2=1\)

• For Case 2 (\(x + 2=-3\)):

Subtract 2 from both sides: \(x=-3 - 2=-5\)

So the solutions for \(x\) are \(x = 1\) or \(x=-5\).

Problem 29: \(-4|k|=12\)

Step 1: Isolate the absolute - value expression

Divide both sides by \(-4\):
\(|k|=\frac{12}{-4}=-3\)

Since the absolute value of a number is non - negative (\(|k|\geq0\)) and \(-3<0\), there is no solution for this equation.

Problem 30: \(|-3n|-2=4\)

Step 1: Isolate the absolute - value expression

Add 2 to both sides:
\(|-3n|=4 + 2=6\)

Note that \(|-3n|=| - 3|\times|n| = 3|n|\), so \(3|n|=6\) (or we can directly use the property of absolute value for \(|-3n|\)).
If we consider \(|-3n| = 6\), we have two cases:
Case 1: \(-3n=6\)
Case 2: \(-3n=-6\)

Step 2: Solve for \(n\) in each case

• For Case 1 (\(-3n=6\)):

Divide both sides by \(-3\): \(n=\frac{6}{-3}=-2\)

• For Case 2 (\(-3n=-6\)):

Divide both sides by \(-3\): \(n=\frac{-6}{-3}=2\)

So the solutions for \(n\) are \(n = 2\) or \(n=-2\).

Problem 31: \(-4|k + 1|=16\)

Step 1: Isolate the absolute - value expression

Divide both sides by \(-4\):
\(|k + 1|=\frac{16}{-4}=-4\)

Since the absolute value of a number is non - negative (\(|x|\geq0\) for all real \(x\)) and \(-4<0\), there is no solution for this equation.