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Let's go through each question step by step:

### 15. Find the value of \( k \) for which the following pair of linear equations have infinitely many solutions:
\[
2x + 3y = 7
\]
\[
(k+1)x + (2k - 1)y = 4k + 1
\]
For two linear equations to have infinitely many solutions, their corresponding coefficients must be proportional. 

That means:
\[
\frac{2}{k+1} = \frac{3}{2k-1} = \frac{7}{4k+1}
\]

From the first two ratios:
\[
\frac{2}{k+1} = \frac{3}{2k-1}
\]
Cross-multiplying:
\[
2(2k - 1) = 3(k + 1)
\]
Expanding:
\[
4k - 2 = 3k + 3
\]
Solving for \( k \):
\[
4k - 3k = 3 + 2
\]
\[
k = 5
\]
Thus, the value of \( k \) is 5.

### 16. If \( 51x + 23y = 116 \) and \( 23x + 51y = 106 \), then find the value of \( (x - y) \).

We can solve this system of equations by elimination or substitution.

Start with the two equations:
1. \( 51x + 23y = 116 \)
2. \( 23x + 51y = 106 \)

We can solve these by adding the equations to eliminate one of the variables. Adding the two equations:

\[
(51x + 23y) + (23x + 51y) = 116 + 106
\]
\[
74x + 74y = 222
\]
\[
x + y = \frac{222}{74} = 3
\]

Now subtract the second equation from the first:
\[
(51x + 23y) - (23x + 51y) = 116 - 106
\]
\[
28x - 28y = 10
\]
\[
x - y = \frac{10}{28} = \frac{5}{14}
\]

Thus, \( x - y = \frac{5}{14} \).

### 17. Solve \( 2x + 3y = 11 \) and \( x - 2y = -12 \) algebraically and hence find the value of \( m \) for which \( y = mx + 3 \).

We first solve the system of equations:
1. \( 2x + 3y = 11 \)
2. \( x - 2y = -12 \)

From the second equation, solve for \( x \):
\[
x = 2y - 12
\]
Substitute this into the first equation:
\[
2(2y - 12) + 3y = 11
\]
Expanding:
\[
4y - 24 + 3y = 11
\]
\[
7y = 35
\]
\[
y = 5
\]

Substitute \( y = 5 \) into \( x = 2y - 12 \):
\[
x = 2(5) - 12 = 10 - 12 = -2
\]

So, \( x = -2 \) and \( y = 5 \).

Now, we need to find \( m \) such that \( y = mx + 3 \). Substituting \( x = -2 \) and \( y = 5 \) into the equation \( y = mx + 3 \):
\[
5 = m(-2) + 3
\]
\[
5 = -2m + 3
\]
\[
2m = -2
\]
\[
m = -1
\]

Thus, \( m = -1 \).

### 18. Represent the system of linear equations \( 3x + y = 5 \) and \( 2x + y = 5 \) graphically. From the graph, find the points where the lines intersect the y-axis.

We can solve for \( y \)-intercepts by setting \( x = 0 \) in both equations.

1. For \( 3x + y = 5 \), when \( x = 0 \):
\[
y = 5
\]

2. For \( 2x + y = 5 \), when \( x = 0 \):
\[
y = 5
\]

Both lines intersect the y-axis at \( (0, 5) \).

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Do you want more details on any of these solutions or further clarification?

Here are five follow-up questions based on this information:
1. How do you determine if two lines are parallel, intersecting, or coincident?
2. What are the necessary conditions for two linear equations to have no solution?
3. Can you derive the general formula for solving a system of equations using substitution?
4. What happens geometrically when two equations represent the same line?
5. How can you verify if a point is a solution to both equations in a system?

**Tip:** When solving systems of equations, graphing is a useful tool to visualize whether the system has one solution, no solutions, or infinitely many solutions.

15) Find the value of k for which the following pair of linear equations have infinitely many solutions. 2x + 3y = 7, (k+1)x + (2k-1)y = 4k + 1.
16) If 51x + 23y = 116 and 23x + 51y = 106, then find the value of (x - y).
17) Solve 2x + 3y = 11 and x - 2y = -12 algebraically and hence find the value of m for which y = mx + 3.
18) Represent the system of linear equations 3x + y = 5 and 2x + y = 5 graphically. From the graph, find the points where the lines intersect the y-axis.

Learn how to solve systems of linear equations using algebraic methods, including finding values for infinitely many solutions and graphing intersections. The problems involve proportionality, substitution, and graphical methods, suitable for students in Grades 9-11.

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