help mje sollve this question. alegebra1?

Question

x/3+y/2=1

Answer 1

Convert the fraction equation to standard form

x/3 + y/2 = 1

6(x/3) + 6(y/2) = 6(1)

2x + 3y = 6. . . .This is standard form

- - - - - - - -

solve for the y intercept

2x + 3y = 6

2x + 3y - 2x = - 2x + 6

3y = - 2x + 6

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

y = - 2/3x + 2

- - - - - - - - - s-

Answer 2

And if your solving for y...
x/3 + y/2 =1 [get common denominator]
2x/6 + 3y/6 =1 [put into single fraction]
(2x+3y)/6 = 1 [simplify]
2x + 3y = 6 [isolate 3y]
3y = 6-2x [isolate y]
y = (6-2x)/3 [the end!!]

Answer 3

the least common denominator of x/3 + y/2 is 6

2x/6 + 3y/6 = 1

(2x+3y)/6 = 1

2x + 3y = 6

if you're solving for x
2x = 6 - 3y
x= 3 - 3y/2

if you're solving for y
3y = 6 - 2x
y = 2 - 2x/3

Answer 4

Please be specific - which variable do you need to solve? I'll solve both variables.

1) Solve "x" in the equation.... x/3+y/2=1

First: eliminate fractions - multiply the denominator with each term.

(2)(3)(x/3) + (2)(3)(y/2) = (2)(3)(1)

*Cancel "like" terms & combine the rest.

(2)(x) + (3)(y) = (2)(3)(1)
2x + 3y = 6

Sec: subtract 3y from both sides (when you move a term to the opposite side, always use the opposite sign).

2x + 3y - 3y = - 3y + 6
2x = - 3y + 6

Third: isolate "x" by dividing each term by 2.

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

2) Solve "y" in the equation.... x/3 + y/2 = 1

First: eliminate fractions - multiply the denominator with each term.

(2)(3)(x/3) + (2)(3)(y/2) = (2)(3)(1)

*Cancel "like" terms & combine the rest.

(2)(x) + (3)(y) = (2)(3)(1)
2x + 3y = 6

Sec: subtract 2x from both sides (when you move a term to the opposite side, always use the opposite sign).

2x - 2x + 3y = - 2x + 6
3y = - 2x + 6

Third: isolate "x" by dividing each term by 3.

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

Answer 5

if you're solving for x...

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