Question

1. Solve the system.

{j+k=3j−k=7{j+k=3j−k=7

Use the substitution method.

The solution is (5, −2)(5, −2).

There is no solution.

The solution is (8, 1)(8, 1)


There are an infinite number of solutions.








2.Solve the system of equations.

{y=2x−64x−2y=14{y=2x−64x−2y=14

Use the substitution method.

The solution is (4, 14)(4, 14).

The solution is (14, 4)(14, 4).

There is no solution.

There are an infinite number of solutions.








3. Solve the system of equations.

{y=x−82x+3y=1{y=x−82x+3y=1

Use the substitution method.

(2, −6)(2, −6)

(4, −4)(4, −4)

(0, −8)(0, −8)

(5, −3)








4. What is the y

Answer 1

Here... I only have one answer.

Answer 2

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 :)