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Let's solve the given problems one by one to find the point of intersection of each pair of lines.

### Problem a:
**Equations:**
1. \( y = 2x - 1 \)
2. \( y = 3 - x \)

To find the point of intersection, we set the two equations equal to each other:

\[
2x - 1 = 3 - x
\]

Solve for \(x\):

\[
2x + x = 3 + 1 \implies 3x = 4 \implies x = \frac{4}{3}
\]

Substitute \(x = \frac{4}{3}\) into one of the equations to find \(y\). Let's use \(y = 2x - 1\):

\[
y = 2 \left( \frac{4}{3} \right) - 1 = \frac{8}{3} - 1 = \frac{8}{3} - \frac{3}{3} = \frac{5}{3}
\]

So, the point of intersection is:

\[
\left( \frac{4}{3}, \frac{5}{3} \right)
\]

### Problem b:
**Equations:**
1. \( x + y - 16 = 0 \implies y = 16 - x \)
2. \( x + 2y = 10 \)

Substitute \( y = 16 - x \) into the second equation:

\[
x + 2(16 - x) = 10
\]

Simplify and solve for \(x\):

\[
x + 32 - 2x = 10 \implies -x = -22 \implies x = 22
\]

Substitute \(x = 22\) back into the first equation to find \(y\):

\[
y = 16 - 22 = -6
\]

So, the point of intersection is:

\[
(22, -6)
\]

### Problem c:
**Equations:**
1. \( 2y = x + 5 \implies x = 2y - 5 \)
2. \( y = -2x \)

Substitute \( x = 2y - 5 \) into the second equation:

\[
y = -2(2y - 5)
\]

Solve for \(y\):

\[
y = -4y + 10 \implies 5y = 10 \implies y = 2
\]

Substitute \(y = 2\) back into \( x = 2y - 5 \):

\[
x = 2(2) - 5 = 4 - 5 = -1
\]

So, the point of intersection is:

\[
(-1, 2)
\]

### Problem d:
**Equations:**
1. \( 3y = 6 - 4x \implies y = 2 - \frac{4x}{3} \)
2. \( 6y = x - 12 \implies y = \frac{x}{6} - 2 \)

Set the two expressions for \(y\) equal to each other:

\[
2 - \frac{4x}{3} = \frac{x}{6} - 2
\]

Multiply every term by 6 to clear the fractions:

\[
6 \cdot \left( 2 - \frac{4x}{3} \right) = 6 \cdot \left( \frac{x}{6} - 2 \right)
\]

Simplify:

\[
12 - 8x = x - 12
\]

Combine like terms:

\[
12 + 12 = x + 8x \implies 24 = 9x \implies x = \frac{24}{9} = \frac{8}{3}
\]

Substitute \(x = \frac{8}{3}\) back into \(y = \frac{x}{6} - 2\):

\[
y = \frac{8/3}{6} - 2 = \frac{8}{18} - 2 = \frac{4}{9} - 2 = \frac{4}{9} - \frac{18}{9} = -\frac{14}{9}
\]

So, the point of intersection is:

\[
\left( \frac{8}{3}, -\frac{14}{9} \right)
\]

Would you like further details on any of the steps? Here are some related questions:

1. How do you find the intersection of lines graphically?
2. What are some applications of finding line intersections in real life?
3. What methods can be used to solve linear systems of equations?
4. Can two lines intersect more than once in Euclidean geometry?
5. How do parallel and perpendicular lines affect the intersection points?

**Tip:** When solving equations, always double-check each substitution step to ensure accuracy.

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

Learn how to solve systems of linear equations to find the point of intersection of two lines. This problem includes step-by-step solutions for four cases involving different linear equations, using algebraic substitution methods. Suitable for students in Grades 9-12.

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