Question

Solve each given equation and show your work. Tell whether it has one solution, an infinite number of solutions, or no solutions, and identify each equation as an identity , a contradiction, or niether.

(a) 6x + 4x - 6 + 24 + 9x
(b) 25 - 4x + 15 - 3x + 10 - x
(c) 4x + 8 = 2x + 7 +2x -20

answer:

Answer 1

Let's solve each equation step by step:

### (a) \(6x + 4x - 6 + 24 + 9x\)

1. Combine like terms:
\[
6x + 4x + 9x - 6 + 24 = (6x + 4x + 9x) + (-6 + 24)
\]
\[
= 19x + 18
\]

Since this expression is not set equal to anything, we interpret \(6x + 4x - 6 + 24 + 9x = 0\), which simplifies to:
\[
19x + 18 = 0
\]

2. Solve for \(x\):
\[
19x = -18
\]
\[
x = -\frac{18}{19}
\]

Since we found a single solution, the answer has **one solution** and is classified as **neither an identity nor a contradiction**.

---

### (b) \(25 - 4x + 15 - 3x + 10 - x\)

1. Combine like terms:
\[
25 + 15 + 10 - 4x - 3x - x = (25 + 15 + 10) + (-4x - 3x - x)
\]
\[
= 50 - 8x
\]

Again, interpret expression as equaling zero (\(50 - 8x = 0\)):
\[
50 - 8x = 0
\]

2. Solve for \(x\):
\[
-8x = -50
\]
\[
x = \frac{50}{8} = \frac{25}{4}
\]

This gives us **one solution** and is classified as **neither an identity nor a contradiction**.

---

### (c) \(4x + 8 = 2x + 7 + 2x - 20\)

1. Simplify the right-hand side:
\[
2x + 7 + 2x - 20 = (2x + 2x) + (7 - 20)
\]
\[
= 4x - 13
\]

So the equation becomes:
\[
4x + 8 = 4x - 13
\]

2. Subtract \(4x\) from both sides:
\[
8 = -13
\]

This is a contradiction because \(8\) cannot equal \(-13\). Thus, the equation has **no solutions** and is classified as a **contradiction**.

---

### Summary:
- (a) One solution: \(x = -\frac{18}{19}\) (Neither)
- (b) One solution: \(x = \frac{25}{4}\) (Neither)
- (c) No solution (Contradiction)