Question

Here are two equations:

Equation 1: 6x + 4y = 34
Equation 2: 5x – 2y = 15
Decide if each point is a solution to one equation, both
equations, or neither equation.
Show your work by sketching your work for each point.

Point (2, 5)
A. Solution to Equation 1
B. Solution to Equation 2
C. Solution to both equations
D. Neither equation

Point (7, 4)
A. Solution to Equation 1
B. Solution to Equation 2
C. Solution to both equations
D. Neither equation

Point (6, 7)
A. Solution to Equation 1
B. Solution to Equation 2
C. Solution to both equations
D. Neither equation

Point (3, 6)
A. Solution to Equation 1
B. Solution to Equation 2
C. Solution to both equations
D. Neither equation

Point (1, 2)
A. Solution to Equation 1
B. Solution to Equation 2
C. Solution to both equations
D. Neither equation

Answer 1

Points are tested for the two linear equations by substituting their coordinates into the equations. (2, 5) is not a solution to either equation, (7, 4) and (3, 6) are solutions to Equation 1, (1, 2) is a solution to Equation 2, and (6, 7) is not a solution to either equation.

Step-by-step explanation:

The subject deals with determining if a given point is a solution for one or both of the two provided linear equations. A point is a solution to a linear equation if, when substituted into the equation, it satisfies the equation (both sides of the equation are equal).

To test the point (2, 5) for Equation 1 (6x + 4y = 34), we substitute x = 2 and y = 5:

• 6(2) + 4(5) = 12 + 20 = 32, which is not equal to 34.

So point (2, 5) is not a solution to Equation 1. Now testing the same point for Equation 2 (5x – 2y = 15):

• 5(2) - 2(5) = 10 - 10 = 0, which is not equal to 15.

Therefore, point (2, 5) is not a solution to Equation 2 either. Similar steps are followed for each point with the respective equations.

• For the point (7, 4), the calculations reveal that it is a solution to Equation 1 only.
• The point (6, 7) does not satisfy either equation.
• The point (3, 6) is a solution to Equation 1 only.
• The point (1, 2) satisfies Equation 2 only.