y=x+14 line 1
y=3x+2 line 2
These are both the equation of lines written in slope intercept form
y=mx+b where m is the slope and the point (0,b) is the y intercept.
The first line has a slope of m=1. The 2nd line has a slope of m=3
Since these lines have different slopes, they are not parallel, thus they will cross at some point. What you have to determine is where the lines cross, which will be a point (x,y) that is on both lines.
We already have y solved in terms of x from either equation so we can use substitution to solve the system.
Since y=x+14 from line 1, put x+14 in place of y in the equation of line 2.
x+14=3x+2
solve for x.
Subtract x from both sides...
14= 3x-x+2
14=2x+2
subtract 2 from both sides
14-2=2x
12=2x
divide both sides by 2
6=x
We now have the x value of the common point. Plug the value 6 in for x in one of the original equations and solve for y.
y=6+14
y=20
These two lines cross at the point (6,20) which is a point the two lines have in common.
Hope I helped (SharkieOwO)
