Question

1. 5y - x = 10; (15, -4)
2. y = -4x + 3; (1, -2)

Explanation:

Problem 13: \( 5y - x = 10 \); \( (15, -4) \)

Step 1: Substitute \( x = 15 \), \( y = -4 \)

Substitute \( x = 15 \) and \( y = -4 \) into the left - hand side of the equation \( 5y - x \).
We get \( 5\times(-4)-15 \).

Step 2: Calculate the left - hand side

First, calculate \( 5\times(-4)=-20 \). Then, \( -20 - 15=-35 \).
The right - hand side of the equation is \( 10 \). Since \( -35
eq10 \), the point \( (15, - 4) \) is not a solution of the equation \( 5y - x = 10 \).

Problem 14: \( y=-4x + 3 \); \( (1,-2) \)

Step 1: Substitute \( x = 1 \), \( y=-2 \)

Substitute \( x = 1 \) and \( y = - 2 \) into the left - hand side and right - hand side of the equation \( y=-4x + 3 \) respectively.
The left - hand side is \( y=-2 \), and the right - hand side is \( -4\times1 + 3 \).

Step 2: Calculate the right - hand side

Calculate \( -4\times1+3=-4 + 3=-1 \).
Since \( -2
eq-1 \), the point \( (1, - 2) \) is not a solution of the equation \( y=-4x + 3 \).

Answer:

s:

1. The point \( (15, - 4) \) is not a solution of \( 5y - x = 10 \).
2. The point \( (1, - 2) \) is not a solution of \( y=-4x + 3 \).