how do i move an equation from standard form to general for? or versa algebra 1

Question

the equation is 5x+3y=9 in standard to general and y=2/3x-4 which is general to standard

Answer 1

You have to be able to manipulate the equations.

5x+3y=9 Subtract 5x from both sides

5x+3y-5x=9-5x simplify

3y = -5x+9 Divide both sides of the equation by 3

3y/3 = -5x/3+9/3 Simplify

y = -5/3x+3

y=2/3x-4 Multiply both sides of the equation by 3 to get rid of the fractions

3(y)=3(2/3x-4) Distribute to simplify

3y=2x-12 Subtract 2x from both sides

3y-2x=2x-12-2x Simplify

-2x+3y = -12 Divide everything by -1, the x term is positive in standard form

-2x/-1+3y/-1 = -12/-1 Simplify

2x-3y = 12

Answer 2

Standard form looks like Ax + By = C

General form looks like Ax + By + C = 0

Slope-intercept form is y = mx + b

To go from standard to general form, move the constant to the left side with the variables so the equation equals 0

5x + 3y = 9

Subtracting 9 from each side, we get5x + 3y - 9 = 0.

y = 2/3x - 4 is in slope-intercept form, not general form.

To go from slope-intercept form to standard form, we first need to get rid of the fraction, second we move the x term to the left side, and last, the x term cannot be negative

First we multiply both sides by 3 to get the 3 out of the denominator

3y = 3(2/3x - 4)

3y = 2x - 12

Subtract 2x from each side

-2x + 3y = -12

Multiply everything by -1 to make the x term positive.

2x - 3y = 12

Answer 3

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