- #1

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- TL;DR 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.