We are asked to solve three linear equations. 10. $\frac{x+3}{2} = \frac{x-4}{5}$ 11. $\frac{2x-1}{3} = \frac{x}{2}$ 12. $\frac{3x+1}{5} = \frac{2x}{3}$

1. Problem Description

We are asked to solve three linear equations.

1

0. $\frac{x+3}{2} = \frac{x-4}{5}$

1

1. $\frac{2x-1}{3} = \frac{x}{2}$

1

2. $\frac{3x+1}{5} = \frac{2x}{3}$

2. Solution Steps

1

0. $\frac{x+3}{2} = \frac{x-4}{5}$

Multiply both sides by 10 (the least common multiple of 2 and 5)

$10 \cdot \frac{x+3}{2} = 10 \cdot \frac{x-4}{5}$

$5(x+3) = 2(x-4)$

$5x + 15 = 2x - 8$

$5x - 2x = -8 - 15$

$3x = -23$

$x = -\frac{23}{3}$

1

1. $\frac{2x-1}{3} = \frac{x}{2}$

Multiply both sides by 6 (the least common multiple of 3 and 2)

$6 \cdot \frac{2x-1}{3} = 6 \cdot \frac{x}{2}$

$2(2x-1) = 3x$

$4x - 2 = 3x$

$4x - 3x = 2$

$x = 2$

1

2. $\frac{3x+1}{5} = \frac{2x}{3}$

Multiply both sides by 15 (the least common multiple of 5 and 3)

$15 \cdot \frac{3x+1}{5} = 15 \cdot \frac{2x}{3}$

$3(3x+1) = 5(2x)$

$9x + 3 = 10x$

$3 = 10x - 9x$

$x = 3$

3. Final Answer

1

0. $x = -\frac{23}{3}$

1

1. $x = 2$

1

2. $x = 3$