Question

5a+5b+5c=-20
4a+3b+3c=-6
-4a+3b+3c=9

Answer 1

The question involves solving a system of linear equations using methods such as substitution or elimination. The irrelevant information provided regarding quadratic equations and certain values is not applicable in this context.

Step-by-step explanation:

The student is working with a set of linear equations that can be solved using methods such as substitution or elimination. The equations provided are:

• 5a + 5b + 5c = -20
• 4a + 3b + 3c = -6
• -4a + 3b + 3c = 9

By examining the equations, one can notice that the second and third equations can be added together to eliminate the variable a, which leads to another equation in b and c that can be used to solve for these variables. Then we can substitute back to find the value of a. It appears there might be some confusion with other irrelevant information provided, as the mention of quadratic equations and substitution of non-related values does not pertain to solving the given system of linear equations.

Answer 2

For 5a+5b+5c=-20

A= -4 - b - c

B= -4 - a - c

C= -4 - a - b

———————————

4a+3b+3c=-6

A= -3/2 - 3b/4 - 3c/4

B= -2 - 4a/3 - c

C= -2 - 4a/3 - b

———————————

-4a+3b+3c=9

A= -9/4 + 3b/4 + 3c/4

B= 3 + 4a/3 - c

C= 3 + 4a/3 - b