On equation of a line...?

Question

Find the equation of a straight line passing through origin and the point of intersection of the lines x+2y=7 and x-y=4.Please show the full process also.

Answer 1

x+2y=7----(1)
x-y=4-----(2)

(1)-(2)

2y-(-y)=7-4
2y+y=3
3y=3
y=1

substitute y=1 into (1)
x+2y=7
x+2(1)=7
x=7-2
x=5

another point:(5,1)

now we have two point

(5,1) and (0,0)

gradient of line
=(y2-y1)/(x2-x1)
=(1-0)/(5-0)
=1/5

y=mx+c
0=1/5(0)+c
c=0

line equation:
y=x/5
5y=x

Answer 2

First find the intersection. To do this, we can subtract the two equations to get y=1, and from back-substitution, x= 5. The line wanted passes through (0,0) and (5,1). You should be able to work that out.

Answer 3

x + 2y = 7
x - y = 4

x - y = 4
x = y + 4

y + 4 + 2y = 7
3y = 3
y = 1

x = 1 + 4 = 5

(5,1) and (0,0)
m = (0 - 1)/(0 - 5)
m = (1/5)

0 = (1/5)(0) + b
0 = 0 + b
b = 0

ANS : y = (1/5)x

Proof, go to http://www.calculator.com/calcs/GCalc.html and type in
y = x - 4
y = (-1/2)x + (7/2)
y = (1/5)x

just leave out the y = part because it isn't need, and if you do then the graph will not show up.