Question

write a system of linear equations in which (3,-5) is a solution of equation 1 but not a solution of equation 2 and (-1,7) is a solution of the system

Answer 1

To create a system of linear equations with specific solutions, we can use the given points to check if they satisfy the equations. For the given points (3,-5) and (-1,7), (3,-5) does not satisfy any of the equations but (-1,7) does. The system of linear equations is 7y = 6x + 8 and 4y = 8.

Step-by-step explanation:

To create a system of linear equations with (3,-5) as a solution for equation 1 but not a solution for equation 2, and (-1,7) as a solution for the system, we can start by writing the equations in slope-intercept form: y = mx + b.

Equation 1: 7y = 6x + 8 (or y = (6/7)x + 8/7)

Equation 2: 4y = 8 (or y = 2)

We want (3,-5) to satisfy equation 1 but not equation 2. Plugging in the values of x=3 and y=-5 into equation 1:

-5 = (6/7)*3 + 8/7

-5 = 18/7 + 8/7

-5 = 26/7

This equation is not true, so (3,-5) is not a solution for equation 1. Now, let's check if (3,-5) satisfies equation 2:

4*(-5) = 8

-20 = 8

This equation is also not true, so (3,-5) is not a solution for equation 2. Now, let's check if (-1,7) satisfies the system:

7*7 = 6*(-1) + 8

49 = -6 + 8

49 = 2

This equation is true, so (-1,7) is a solution for the system of equations. Therefore, the system of linear equations is:

7y = 6x + 8

4y = 8

Answer 2

Examine this system of linear equations.

y – 3x = –2,

y = 4

Which is a solution of the system of equations?

(0, 4)

(2, 2)

(2, 4)=this is the correct answer

(4, 2)

Step-by-step explanation:

thats the answer

(2,4)