how can i write these equations in standard form with integer coefficients?
so #1 is: y=1/2x-3/5
and also #2: y=-8/3x+4/9
I know the formula is Ax+By=C..
but I just don't get these two..
plz explain & thanks!
These require two steps.
1. clear the fractions by multiplying by the LCD
2. rearrange, so the x and y terms are on one side and the constant on the other, leading with a positive coefficient for x
1. multiply by 10
10y = 5x - 6
2. 6 = 5x - 10y or 5x - 10y = 6
Now you try the second in the same way.
so would number 2 be:
24x+9y=4 OR 4=24x+9y
??
thnx!
Watch your signs
After you multiply by 9 you get
9y = 24x + 4
then 24x - 9y = -4 or -4 = 24x - 9y
the last two equations are the same thing.
(if A=B then B=A)
in the the general form, the x and y terms are "usually" on the left side of the equation, so 24x - 9y = -4 would be preferred.
To write the equations in standard form with integer coefficients, you need to eliminate any fractions or decimals, and rearrange the equation to have integer coefficients. Here's how you can do it for each equation:
#1: y = (1/2)x - 3/5
1. Multiply both sides of the equation by 10 to eliminate the fractions:
10y = 10 * (1/2)x - 10 * (3/5)
10y = 5x - 6
2. You now have the equation in the form 10y = 5x - 6. However, we want the coefficients to be integers, so we can multiply the entire equation by 2:
2 * 10y = 2 * (5x - 6)
20y = 10x - 12
So, the standard form of the first equation with integer coefficients is 10x - 20y = 12.
#2: y = (-8/3)x + 4/9
1. Multiply both sides of the equation by 9 to eliminate the fraction:
9y = 9 * (-8/3)x + 9 * (4/9)
9y = -24x + 4
2. Dividing the entire equation by 3 will give us integer coefficients:
3 * 9y = 3 * (-24x + 4)
27y = -72x + 12
Hence, the standard form of the second equation with integer coefficients is 72x + 27y = 12.
By following these steps, you can rewrite both equations in standard form with integer coefficients.