Question

algebra 1
usd4 types of solutions
solve each equation.

1. 4 + 8x = 8x - 2 \t\t\t 2) 13x - 13 = 2(9x - 5) - (3 + 5x)

1. 4 = 10x - 6 \t\t\t 4) 28 = -2x - 4

1. 3h + 5 = 3h - 1 \t\t\t 6) -7 + 4a = -19

Explanation:

Problem 1: \( 4 + 8x = 8x - 2 \)

Step 1: Subtract \( 8x \) from both sides

\( 4 + 8x - 8x = 8x - 2 - 8x \)
Simplifying both sides, we get \( 4 = -2 \).

Step 2: Analyze the result

The equation \( 4 = -2 \) is a contradiction (it is never true). So, this equation has no solution.

Step 1: Expand the right - hand side

First, expand \( 2(9x - 5) \) to get \( 18x-10 \), and expand \( -(3 + 5x) \) to get \( -3-5x \).
Then the right - hand side becomes \( 18x - 10-3 - 5x \).
Combine like terms: \( 18x-5x=13x \) and \( - 10-3=-13 \). So the equation is \( 13x - 13=13x - 13 \).

Step 2: Subtract \( 13x \) from both sides

\( 13x - 13-13x=13x - 13-13x \)
Simplifying both sides, we get \( - 13=-13 \).

This is an identity (it is true for all values of \( x \)). So, the solution is all real numbers.

Step 1: Add 6 to both sides

\( 4 + 6=10x-6 + 6 \)
Simplifying, we have \( 10 = 10x \).

Step 2: Divide both sides by 10

\( \frac{10}{10}=\frac{10x}{10} \)
Simplifying, we get \( x = 1 \).

Answer:

No solution

Problem 2: \( 13x - 13 = 2(9x - 5) - (3 + 5x) \)