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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

Find the point of intersection of the two straight lines in each of the following: a. y = 2x - 1 and y = 3 - x; b. x + y - 16 = 0 and x + 2y = 10; c. 2y = x - 5 and y = -2x + 5; d. 3y = 6 - 4x and 6y = x - 12.

Learn how to find the point of intersection between two straight lines with step-by-step examples. This problem involves solving systems of linear equations and is suitable for students in Grades 8-10.

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