Question

1. Solve the system.

Answer 1

Question (1):
The first given equation is:
j + k = 3
This can be rewritten as:
j = 3 - k .................> I
The second given equation is:
j - k = 7 ............> II
Substitute with I in II and solve as follows:
j - k = 7
(3-k) - k = 7
3 - 2k = 7
-2k = 7-3 = 4
k = 4/-2
k = -2
Substitute with k in I to get the value of J as follows:
j = 3 - k 
j = 3 - (-2) = 3+2
j = 5
Based on the above, the solution is (5,-2)

Question (2):
The first given equation is:
y = 2x - 6 .........> I
The second given equation is:
4x - 2y = 14............> II
Substitute with I in II and solve as follows:
4x - 2y = 14
4x - 2(2x - 6) = 14
4x - 4x + 12 = 14
12 = 14
Since this is not possible, therefore, the system has no solution

Question (3):
The first given equation is:
y = x - 8 .............> equation I
The second given equation is:
2x + 3y = 1 ..............> equation II
Substitute with equation I in equation II and solve as follows:
2x + 3y = 1 
2x + 3(x - 8) = 1
2x + 3x - 24 = 1
5x = 25
x = 25/5
x = 5
Substitute with x in equation I to get y as follows:
y = x - 8
y = 5 - 8
y = -3
Based on the above, the solution of the system is (5,-3)

Hope this helps :)

Answer 2

Here... I only have one answer.