You transformed the function as follows:

All trasformations of the fashion
translate the function horizontally, k units right if k is positive, k units left if k is negative.
In your case, k=6, so you translated the original function y=|x| 6 units to the right.
Answer:
a= 40-3b-c/5
a=-10+3b-2c
a=50+2b-3c/14
Step-by-step explanation:
5a+3b+c-(3b+c)=40-(3b+c)
5a=40-3b-c
5a/5=40/5-3b/5-c/5
a= 40-3b-c/5
a-3b+2c-(3b+2c)=-10-(-3b+2c)
a=-10+3b-2c
14a-2b+3c-(-2b+3c)=50-(-2b+3c)
14a=50+2b-3c
14a/14=50/14+2b/14-3c/14
a=50+2b-3c/14
Write everything on one side
2/3x-y+1=0
multiply by 3
2x-3y+3=0
I hope this is what you're looking for
Question is Incomplete, Complete question is given below.
Prove that a triangle with the sides (a − 1) cm, 2√a cm and (a + 1) cm is a right angled triangle.
Answer:
∆ABC is right angled triangle with right angle at B.
Step-by-step explanation:
Given : Triangle having sides (a - 1) cm, 2√a and (a + 1) cm.
We need to prove that triangle is the right angled triangle.
Let the triangle be denoted by Δ ABC with side as;
AB = (a - 1) cm
BC = (2√ a) cm
CA = (a + 1) cm
Hence,
Now We know that

So;


Now;

Also;

Now We know that




[By Pythagoras theorem]

Hence, 
Now In right angled triangle the sum of square of two sides of triangle is equal to square of the third side.
This proves that ∆ABC is right angled triangle with right angle at B.