algebra help plz?

Question

solve by the elimination method
what is the solution of the system?


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

(1/2)x - (1/6)y =4

Answer 1

2x/3 + y/2 = 3
x/2 - y/6 = 4
Multiply the second equation by 3

3x/2 - y/2 = 12 (add this to the first equatiion to get:)

2x/3 + 3x/2 = 15 (multiply both sides by 6 to clear fractions)

12x/3 + 18x/2 = 90

4x + 9x = 90

13x + 90

x = 90/13

(2/3) (90/13) + y/2 = 3

180/39 + y/2 = 3

y = 2 (3 - 180/39) (You can simplify it from here.)

Answer 2

Elimination by addition method

(2/3)x + (1/2)y = 3- - - - -Equation 1
(1/2)x - 1/6)y = 4- - - - - - Equation 2
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Clear the fractions in equation 1 and 2

The LCM for equation 1 and 2 is 6

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

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

4x + 3y = 18. . .New equation 1
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(1/2)y - (1/6)y = 4

6(1/2)x - 6(1/6)y = 6(4)

3x - y = 24. ,. .New equation 2

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combine new equation 1 withe new equation 2

4x + 3y = 18
3x - y = 24
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Multiply new equation 2 by 3

3x - y = 24

3(3x) - 3(y) = 3(24)

9x - 3y = 72
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Elimination of y

4x + 3y = 18
9x - 3y = 72
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13x = 90

Divide both sides of the equation by 13

13x / 13 = 90 / 13

x = 90/13

Insert the x value into new equation 1
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4x + 3y = 18

4(90/13) + 3y = 18

360/13 + 3y = 18

Transpose 360/13

360/13 + 3y - 360/13 = 18 - 360/13

3y = 234/13 - 360/13

3y = - 126/13

Divide both sides of the equation by 3

3y / 3 = - 126/13 ÷ 3

y = - 126/12 ÷ 3/1

y = - 126/13 x 1/3

y = - 126/39

Insert the y value into equation 1
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Check for equation 1

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

(2/3)(90/90/13) + (1/2)(- 126/39) = 3

(180/39) + ( - 126/78) = 3

Remove parenthesis

180/39 - 126/78 = 3

The common denominator is 78

360/78 - 126/78 = 3

Subtract the numerator the denominator remains the same.

234 / 78 = 3

3 = 3

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Check for equation 2

(1/2)x - (1/6)y = 4

(1/2)(90/13) - (1/6( - 126/39) = 4

(90/26) - (- 126/234) = 4

(90/26) + (126/234) = 4

The common denominator is 234

810/234 + 126/234 = 4

936/234 = 4

4 = 4

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Both equations balance

The solution set is { 90/13, - 126/39 }

- - - - - - - - -s-

Answer 3

First, let's get rid of the fractions by multiplying the equations through by the common denominators.

6[(2/3)x + (1/2)y = 3] becomes 4x +3y = 18

6[(1/2)x - (1/6)y = 4] becomes 3x - y = 24

Now we can see to multiply the second equation by 3 then add them to eliminate the y terms:

3(3x - y = 24)
9x - 3y = 72
4x + 3y = 18

13x = 90
x = 90/13

Now, substitute this back into one of the equations to find y.

Answer 4

Make one of the values equal. Let's do y:
(2/9)x+(1/6)y=1
(1/2)x-(1/6)y=4
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(13/18)x=5
x=90/13
Then plug this value for x back into one of your equations, and you get y.