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# what is y= 1/2 + 3 written in standard form using integers

Reviewing for a algebra one exam!

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### 3 Answers

I'm sure you meant    y = 1/2 x + 3

If that's the case then you would follow these steps.

First you will need to eliminate the fraction by multipling everything by 2.

2(y) = (1/2)(2)(x) + 3(2) which in turn equals

2y = x + 6

Now we need to put both the x and y on the same side by subtracting x from both sides

2y - x = x - x + 6 which equals

2y - x = 6

Standard Form is Ax + By = C so rearrange the problem to

-x + 2y = 6

### Comments

The only correction I have to your method is that A must be a positive integer. The last step should be to multiply both sides of the equation by -1, giving:

x - 2y = -6
The standard form for this y-intercept form equation is Ax + By=C

y=1/2x + 3  Multiply both sides of the equations by 2 to get rid of the fraction 1/2

After this is done you get the equation 2y= x + 6

Combine the x's & y's and move the constant to the other side

The result is x- 2y=-6

The Standart for is: Ax+By = C

y= 1/2 + 3, I thing that you missing some variable that could be called "x" or "z"

you expression could be writen y=x/2+3 or y= 1/2 +3x or y=x/2 + 3z, I do not know.

To solve it

step 1: re write the equation eliminating the fraction form, to do this mult it by 2

for y=x/2 + 3 ==> 2y=x+6 ==> 2y-x = 6 ==> x-2y=-6

for y= 1/2 +3x ==> 2y=1+6x ==> 6x-2y=-1 or 2(3x-x)=-1

for y=x/2 + 3z ===> 2y=x+6z ==> x-2y+6z=0, where the standar form is Ax+By+Cz=D