how do you hange a slope-intercept form equation into a standard form equation?

Question

this is an algebra1 question

Answer 1

A slope-intercept form equation is in the form y = mx + b.
A standard form equation is in the form of Ax + By = C, where A, B, and C are integers (non-fractions), and A is positive.
So...to change from slope-intercept to standard form...let's say you have the equation y = (3/2)x + (7/4).
Put the x and the y on the same side.
-(3/2)x + y = (7/4)
Then, get rid of the fractions by multiplying each side by 4.
4[-(3/2)x + y] = 4(7/4)
-6x + 4y = 7
Then, make the x positive by multiplying each side by -1.
6x - 4y = -7

This is your standard form equation.

Answer 2

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how do you hange a slope-intercept form equation into a standard form equation?
this is an algebra1 question

Answer 3

Slope-intercept is y = mx + b so you just move the mx to the left and make all coefficients integers to ax + by = c (standard form).

For example, y = -3/2x + 5 is slope intercept for a line with -3/2 and y-intercept at 5. You can move the -3/2 first or multiply all by 2 to get rid of the fraction. (I'd multiply first so you don't make a mistake adding fractions).

2(y) = 2(-3/2x) + 2(5)
2y = -3x + 10
now bring the x term to the left by adding 3x to each side:
3x + 2y = 10 that's the standard form.

Answer 4

A line in slope intercept form would be

y = mx + b

A line in standard form would be

Ax + By + C = 0, for a positive number A.

Example 1:

y = (3/2)x + 4

To convert this to standard form, first multiply both sides by 2.

2y = 3x + 4

Now, move everything to the right hand side.

0 = 3x - 2y + 4

Example 2:
y = (5/2)x - (2/3)

Multiply everything by 6.

6y = 15x - 4

Move everything to the right hand side.

0 = 15x - 6y - 4

And now this is in standard form.

Answer 5

slope intercept form is the stand form

Answer 6

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