Question

Solve, using the substitution method.

j + k = 3
j – k = 7

A.The solution is (8, 1)
B.The solution is (5, –2)
C.There is no solution.
D.There are an infinite number of solutions.

Answer 1

if j-k=7, what is j?

Answer 2

If it was substituted it would be A.8 - 1 = 7
And the other one on top would be 5 - 2 = 3

Answer 3

but x and y are same throut

Answer 4

so if j-k=7 j=7+k, and j+k=3 so 7+2k=3, now what is k

Answer 5

I messed up
j + k = 3
j - k = 7

cancel out the ks you end up with
2j = 10
j = 5

then plug back the 5 into one of the equations and solve for k

Answer 6

Wait, Okay. I don't understand.

Answer 7

first equation
(5) + k = 3
k = -5 +3
= -2

or
second equation
(5) - k = 7
-k = 7-5
= 2
BUT k is negative and it can't be negative so you bring the negative sign onto the other side and you end up with
k = -2

Answer 8

The answer has to be C

Answer 9

no it is b

Answer 10

No its B according to my calculations.
Then what's the answer and please show me the work.

Answer 11

How can the solution be (5, –2) ?

Answer 12

I just showed you all the math up there.....

Answer 13

@DanJS what do you think?

Answer 14

been remembering that the the product of the x coefficeients + product of the y coefficients, will be zero, perpendicular lines

but that is not what you do here..

Answer 15

if you need any help learning the topic more use the link here https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-systems-topic/cc-8th-systems-with-substitution/v/the-substitution-method it will help you learn more on it

Answer 16

j + k = 3
j - k = 7

there will be some solution where they intersect at point (j,k) from looking at it,

( k + 7) + k = 3
2k = -4
k = -2

and then j will be 5

Answer 17

that is the unique solution, the two lines cross at the point (j,k) = (5, -2)

Answer 18

try it in the problem

5 - 2 = 3
and
5+2 = 7

Answer 19

See I was right! thank you @DanJS