please help me solve and check this problem using substitution method?

Question

x = 1/5y
3x + 2y = 26

thank you VERY much for your help.

Answer 1

Substitution Method

x = 1/5y- - - - - - - - -- Equation 1
3x + 2y = 26- - - - - - Equation 2
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Substitute equation 1 into equation 2

3x + 2y = 26

3(1/5y) + 2y = 26

3/5y + 2y = 26

Clear the fraction. Multiply both sides of the equation by 5

5(3/5y) + 5(2y) = 5(26)

3y + 10y = 130

Collect like terms

13y = 130

Divide both sides of the equation by 13

13y / 13 = 130 / 13

y = 130 / 13

y = 10

Insert the y value into equation 1

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x = 1/5y

x = 1/5(10)

x = 10/5

x = 2

Insert the x value into equation 1

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Check for equation 1

x = 1/5y

2 = 1/5(10)

2 = 10/5

2 = 2

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Check for equation 2

3x + 2y = 26

3(2) + 2(10) = 26

6 + 20 = 26

26 = 26

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Both equations balance

The solution set is { 2, 10 }

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Answer 2

If x = 1/5y, then 5x =y.
Substituting 5x for y in the second equation, I get 3x + 2(5x) = 26 which simplifies to x = 2.
Substituting 2 for x in the first or second equation finds that 6 + 2y = 26, or y = 10.
Therefore, the solution to the system of equations is (2, 10)

Answer 3

x = 1/5y

3x + 2y = 26

3/5y + 2y = 26

3 + 10y^2 = 26 x 5y = 130 y

Or 10y^2 - 130y + 3 = 0

This is a quadratic equation and can be solved for y.

If you meant x = y/5 (1/5 . y)

3y/5 + 2y = 26

3y + 10y = 130

13y = 130

y = 10 and x = 10/5 = 2

Answer 4

put x=1/5y in second equation,
3/5y+2y=26
->3+10y^2=130y[quadratic involvd]
->10y^2-130y+3=0
now y=(130+root(130^2-4.10.3)/20 and y=(130-root(130^2-4.10.3)/20
oh god!!!!!i think this is the solution.

Answer 5

SJ is right i solved to make sure he was.