Question

1/2(x+2y)+5/3(3x-2y)=-3/2; 5/4(x+2y)-3/5(3x-2y)=61/60. Find x and y?​

Answer 1

Answer:

  (x, y) = (-1/6, 1/4)

Step-by-step explanation:

This is a system of equations in expressions of two variables. You could solve it by expanding the expressions to the base variables, or you could solve it by first finding the values of the expressions.

Let's see what happens when we expand the equations all the way.

First equation:

  $\dfrac{1}{2}(x +2y)+\dfrac{5}{3}(3x-2y)=-\dfrac{3}{2}\\\\3(x+2y)+10(3x-2y)=-9\qquad\text{multiply by 6}\\\\3x+6y+30x-20y=-9\qquad\text{eliminate parentheses}\\\\\boxed{33x-14y+9=0}\qquad\text{collect terms, add 9 to get general form}$

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Second equation:

  $\dfrac{5}{4}(x+2y)-\dfrac{3}{5}(3x-2y)=\dfrac{61}{60}\\\\75(x+2y)-36(3x-2y)=61\qquad\text{multiply by 60}\\\\75x+150y-108x+72y=61\qquad\text{eliminate parentheses}\\\\-33x +222y = 61\qquad\text{collect terms}\\\\\boxed{33x-222y+61=0}\qquad\text{write in general form}$

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The solution process looks like it might be messy for either substitution or elimination, so we choose to solve these equations using the cross-multiplication method.

  Δ1 = (33)(-222) -(33)(-14) = -6864

  Δ2 = (-14)(61) -(-222)(9) = 1144

  Δ3 = (9)(33) -(61)(33) = -1716

  x = Δ2/Δ1 = 1144/-6864 = -1/6

  y = Δ3/Δ1 = -1716/-6864 = 1/4

The solution is (x, y) = (-1/6, 1/4).

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Additional comment

The weird numbers notwithstanding, the solution by elimination might be easily achieved by subtracting the second equation from the first. This would give ...

   208y -52 = 0   ⇒   y = 52/208 = 1/4

Direct substitution of this into either equation will require arithmetic with fractions to obtain the solution for x.

This system can be directly solved by a graphing calculator using the equations as given.

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You can read more about the cross multiplication method of solving a system of linear equations here:

https://brainly.com/question/26397343

Answer 2

Step-by-step explanation:

Given equation are: [1/{2(x+2y)}] + [5/{3(3x-2y)}] = -3/2 →Eqn(1)

[5/{4(x+2y)}] - [3/{5(3x-2y)}] = 61/60 →Eqn(2)

Let {1/(x+2y)} = a ; {1/(3x-2y)} = b then

{1/(2a)} + {5/(3b)} = -3/2 →Eqn(3) * 5/2

{5/(4a)} - {3/(5b)} = 61/60 →Eqn(4) * 1

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{5/(4a)} + {25/(6b)} = -15/4

{5/(4a)} - {3/(5b)} = 61/60

(-) (+) (-)

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{23/(6b)} + {3/(5b)} = -(15/4) - (61/60)

⇛{(125b + 18b)/(30b²)} = {(-225 - 61)/60}

⇛{(143b)/(30b²)} = -286/60

⇛{143/(30b)} = -286/60

⇛1/b = -1

Therefore, b = -1

Substitute the value of b in equation (3) then we have

{1/(2a)} + (5/3) = -3/2

⇛{1/(2a)} = -(3/2) - (5/3)

= (-9-10)/6

⇛{1/(2a)} = -19/6

⇛1/a = -19/3

a = -3/19

Now, {1/(x+2y)} = a ⇛-3/19

⇛x+2y = -19/3. ⇛3x + 6y = -19 →Eqn(5)

{1/(3x-2y)} = 1⇛ 3x - 2y = 1 →Eqn(6)

(-) (+) (-)

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8y = -20

y = -20/8

Therefore, y = -5/2

From equation (6)

3x - 2 * (-5/2) = 1

⇛3x - 5 = 1

⇛3x = 1+5

⇛3x = 6

⇛x = 6/3

Therefore, x = 2/1 = 2

Answer: Hence, the required value of x and y with be -5/2 and 2 respectively.

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