Andrew M. answered • 07/14/15

Tutor

New to Wyzant
Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors

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.