Let's solve the problem from the image. You need to find the points of intersection for the given pairs of lines. Let’s go through each part:

a. $y = 2x - 1$ and $y = 3 - x$

1. Set the equations equal to each other to find $x$: $2x - 1 = 3 - x$ Simplifying: $2x + x = 3 + 1$ $3x = 4$ $x = \frac{4}{3}$

2. Substitute $x = \frac{4}{3}$ into one of the equations, say $y = 2x - 1$: $y = 2\left(\frac{4}{3}\right) - 1 = \frac{8}{3} - 1 = \frac{8}{3} - \frac{3}{3} = \frac{5}{3}$

   So, the point of intersection is: $\left(\frac{4}{3}, \frac{5}{3}\right)$

b. $x + y - 16 = 0$ and $x + 2y = 10$

1. Solve the first equation for $y$: $y = 16 - x$

2. Substitute this into the second equation: $x + 2(16 - x) = 10$ Simplifying: $x + 32 - 2x = 10$ $-x + 32 = 10$ $-x = 10 - 32 = -22$ $x = 22$

3. Substitute $x = 22$ into $y = 16 - x$: $y = 16 - 22 = -6$

   So, the point of intersection is: $(22, -6)$

c. $2y = x - 5$ and $y = -2x + 5$

1. Solve the first equation for $y$: $y = \frac{x - 5}{2}$

2. Set this equal to the second equation: $\frac{x - 5}{2} = -2x + 5$ Multiply both sides by 2 to eliminate the denominator: $x - 5 = -4x + 10$ Simplifying: $x + 4x = 10 + 5$ $5x = 15$ $x = 3$

3. Substitute $x = 3$ into $y = -2x + 5$: $y = -2(3) + 5 = -6 + 5 = -1$

   So, the point of intersection is: [ (3