CONVERTING SLOPE INTERCEPT FORM TO STANDARD FORM

Convert from the given slope intercept form of a linear equation to the standard form of a linear equation.

Problem 1 :

y = (-11x/10) - 1

Solution :

Given equation is in slope intercept form :

y = (-11x/10) - 1

y = (-11x/10) - (1/1)

LCM(10, 1) is 10.

y = (-11x/10)  - (10/10)

y = (-11x - 10)/10

Multiply by 10 on both sides.

10y = -11x - 10

Add 11x and 10 on both sides.

11x + 10y + 10 = 0

So, the standard form is 11x + 10y + 10 = 0.

Problem 2 :

y = -2x - 9

Solution :

Slope intercept form :

y = -2x - 9

Add 2x on both sides.

2x + y = -9

Add 9 on both sides.

2x + y + 9 = 0

So, the standard form is 2x + y + 9 = 0.

Problem 3 :

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

Solution :

The given equation is in slope intercept form :

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

Both fractions are having the same denominators.

y = (7x - 5)/5

Multiply by 5 on both sides.

5y = 7x - 5

Subtract 5y on both sides

0 = 7x - 5y - 5

7x - 5y - 5 = 0

So, the required standard form is 7x - 5y - 5 = 0.

Problem 4 :

y = -x + (11/8)

Solution :

The given equation is in slope intercept form :

y = -x + (11/8)

y = -(x/1) + (11/8)

LCM (8, 1) is 8.

Multiplying the numerator and denominator of the first fraction by 8, we get

y = -(8x/8) + (11/8)

y = (-8x + 11)/8

Multiplying by 8 on both sides, we get

8y = -8x + 11

Add 8x and subtract 11 on both sides.

8x + 8y - 11 = 0

So, the required standard form of given equation is 8x + 8y - 11 = 0.

Problem 5 :

y = -x/3 + (2/9)

Solution :

The given equation is in slope intercept form :

y = -x/3 + (2/9)

LCM (3, 9) is 9.

Multiplying the numerator and denominator of the first fraction by 3, we get

y = (-x/3)(3/3) + (2/9)

y = (-3x/9) + (2/9)

y = (-3x + 2)/9

Multiplying by 9 on both sides.

9y = -3x + 2

Add 3x and subtract 2 on both sides.

3x + y - 2 = 0

So, the required standard form is 3x + y - 2 = 0.