# Label the following line equations as parallel or coinciding: 3/5x-7/5y=1x-7/3y=5/3

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

(3/5)x - (7/5) y = 1

We will write the equation in the slope fom.

Let us multiply by 5:

==> 3x - 7y = 5

==> 7y = 3x - 5

==> y= (3/7)x - 5/7  ==> m= 3/7

x  - (7/3) y = 5/3

Multiply by 3:

==> 3x - 7y = 5

==> 7y = 3x - 5

==> y= (3/7)x - 5/7 ==>m= 3/7

Both the slopes and y-intercepts are equal, so both lines are coinciding.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll put the given equations into the general form, that is:

ax + by + c = 0

We'll work on the first equation to put it in the general form:

3/5x-7/5y=1

We'll subtract 1 both sides:

(3/5)x - (7/5)y - 1 = 0

We'll multiply -1 by 5:

3x - 7y - 5 = 0 (3) - the general form of the 1st equation

We'll work on the second equation:

x-7/3y=5/3

We'll subtract 5/3 both sides:

x - (7/3)y - (5/3) = 0

We'll multiply x by 3:

3x - 7y - 5 = 0 (4) - the general form of the 2nd equation

We can remark that (3) and (4) are identically so the first line coincides with the second line.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

3/5x -7/5y = 1......(1)

x-7/3y = 5/3..........(2)

From the 2nd equation we get x = 5/3 -7/3y = (5y-7)/3y.

We substitute  x =  (5y-7)/3y in (1).

3*3y/5(5y-7) -7/5y = 1. Multiply by 5y(5y-7).

9y*y - 7(5y-7)  = 5y(5y-7)

9y^2 -35y+49 = 25y^2 -35y

49 = 25y^2. So y^2 =49/25.

y = 7/5 or y = -7/5.

Substituting in (2)   y values , we get: x = 5/3+7/3y.

So x= 5/3 +7/3(7/5) = 5/3 +5/3 = 10/3.

x = 5/3 +7/3(-7/5) = 5/3 -5/3 = 0.

So (x ,y) = (0 , -7/5) , Or (x,y) = (10/3 , 7/5)