 #1
 23
 6
 Summary:

Given ## a,b,c,d,e,f \in \mathbb {R}, ad  bc \neq 0 ##, if ##(x_1,y_1)## and ##(x_2,y_2)## are pairs of real numbers satisfying:
## ax_1 + by_1 = e, cx_1 + dy_1 =f ##
## ax_2 + by_2 = e, cx_2 + dy_2 = f ##
then ## (x_1,y_1) = (x_2,y_2). ##
Given ## a,b,c,d,e,f \in \mathbb {R}, ad  bc \neq 0 ##, if ##(x_1,y_1)## and ##(x_2,y_2)## are pairs of real numbers satisfying:
## ax_1 + by_1 = e, cx_1 + dy_1 =f ##
## ax_2 + by_2 = e, cx_2 + dy_2 = f ##
then ## (x_1,y_1) = (x_2,y_2). ##
Here is my attempt at a proof, I have gotten stuck:
1)
## ax_1 + by_1 = ax_2 + by_2 ##
## cx_1 + dy_1 = cx_2 + dy_2 ##
2)
## a(x_1  x_2) = b(y_2  y_1) ##
## c(x_1  x_2) = d(y_2  y_1) ##
3)
## \frac{a}{b} = \frac{y_2  y_1}{x_1  x_2} ##
## \frac{c}{d} = \frac{y_2  y_1}{x_1  x_2} ##
4)
## \frac{a}{b} = \frac{c}{d} ##
5)
## ad = bc ##
6)
## ad  bc = 0 ## but this is false as the hypothesis states that ## ad  bc \neq 0. ##
I am stuck now and am not sure how to proceed.
## ax_1 + by_1 = e, cx_1 + dy_1 =f ##
## ax_2 + by_2 = e, cx_2 + dy_2 = f ##
then ## (x_1,y_1) = (x_2,y_2). ##
Here is my attempt at a proof, I have gotten stuck:
1)
## ax_1 + by_1 = ax_2 + by_2 ##
## cx_1 + dy_1 = cx_2 + dy_2 ##
2)
## a(x_1  x_2) = b(y_2  y_1) ##
## c(x_1  x_2) = d(y_2  y_1) ##
3)
## \frac{a}{b} = \frac{y_2  y_1}{x_1  x_2} ##
## \frac{c}{d} = \frac{y_2  y_1}{x_1  x_2} ##
4)
## \frac{a}{b} = \frac{c}{d} ##
5)
## ad = bc ##
6)
## ad  bc = 0 ## but this is false as the hypothesis states that ## ad  bc \neq 0. ##
I am stuck now and am not sure how to proceed.