# Given that 3x+3y-1=0 and x-4y+4=0 calculate y intercepts and the slopes of the lines .

### 3 Answers | Add Yours

To calculate the intercets of the lines: 3x+3y-1 = 0 and x-4y+4 = 0.

Solution: We rewrite the equations of the lines as:

3x+3y = 1........(1)

x-4y = -4 .........(2).

eq (1) - 3eq(2) eliminates x:

(3x+3y) - 3(x-4y) = 1-3(-4) = 13

3y +12y = 13

15y = 13

y = 13/15. Substitute this value of y in eq(2):

x -4y = -4

x = -4+4y = -4 +4(13/15) = (-60+52) = -8/15.

So x = 13/15 and y = -8/15.

When we express the equation of a line in the form:

y = mx + c

Then slope and y intercepts of the line are given by:

Slope = m

y intercept = c

Therefore we will convert the equation of given lines in the formats y = mx + c to find their slope and y intercept:

**Line 1**

3x + 3y -1 = 0

==> 3y = -3x + 1

==> y = (-3/3)x + 1/3

==> y = -x + 1/3

In this equation:

Slope of line = m = -1

y intercept of line = c = 1/3

**Line 2**

x - 4y + 4 = 0

==> - 4y = - x - 4

==> y = (-1/-4)x -4/-4

==> y = (1/4)x + 1

In this equation:

Slope of line = m = 1/4

y intercept of line = c = 1

**Answer:**

Slope of line 1= -1

y intercept of line 1 = 1/3

Slope of line 2 = 1/4

y intercept of line 2 = 1

We'll put the equation of the given lines in the standard form:

y=mx + n, where the coefficients represent:

- m - the slope

- n - y intercept

3x+3y-1=0

We'll isolate 3y to the left side:

3y = -3x+1

We'll divide by 3 both sides:

y=(-3x/3) + 1/3

The standard form of the equation is:

**y = -x + 1/3**

**The slope of the first line is m1 = -1 and the y intercept is n1 = 1/3.**

We'll put the next given equation in the standard form:

y=mx + n, where the coefficients represent:

- m - the slope

- n - y intercept

x-4y+4=0

We'll isolate -4y to the left side:

-4y = -x-4

We'll divide by -4 both sides:

y=(-x/-4) + (-4/-4)

The standard form of the equation is:

**y = (x/4) + 1**

**The slope of the second line is m2 = 1/4 and the y intercept is n = 1.**