To solve this system of equations, we can start by solving the second equation for y in terms of x:
5x - y = 9y = 5x - 9
Now we can substitute this expression for y into the first equation:
1/x + 1/(5x - 9) = 1/6
Multiplying through by 6x(5x - 9) to clear the fractions, we get:
6(5x - 9) + 6x = x(5x - 9)30x - 54 + 6x = 5x^2 - 9x36x - 54 = 5x^2 - 9x36x - 54 = 5x^2 - 9x
Rearrange the equation to get a quadratic equation:
5x^2 - 45x + 54 = 0
Now we can solve this quadratic equation for x using the quadratic formula:
x = [-(-45) ± √((-45)^2 - 4(5)(54))] / (2*5)x = [45 ± √(2025 - 1080)] / 10x = [45 ± √945] / 10x = [45 ± 3√105] / 10x = 3.9 or x = 3.0
Now that we have found the possible values for x, we can plug them back into the equation y = 5x - 9 to find the corresponding values for y:
For x = 3.9:y = 5(3.9) - 9y = 19.5 - 9y = 10.5
For x = 3.0:y = 5(3) - 9y = 15 - 9y = 6
Therefore, the solutions to the system of equations are x = 3.9, y = 10.5 and x = 3.0, y = 6.
To solve this system of equations, we can start by solving the second equation for y in terms of x:
5x - y = 9
y = 5x - 9
Now we can substitute this expression for y into the first equation:
1/x + 1/(5x - 9) = 1/6
Multiplying through by 6x(5x - 9) to clear the fractions, we get:
6(5x - 9) + 6x = x(5x - 9)
30x - 54 + 6x = 5x^2 - 9x
36x - 54 = 5x^2 - 9x
36x - 54 = 5x^2 - 9x
Rearrange the equation to get a quadratic equation:
5x^2 - 45x + 54 = 0
Now we can solve this quadratic equation for x using the quadratic formula:
x = [-(-45) ± √((-45)^2 - 4(5)(54))] / (2*5)
x = [45 ± √(2025 - 1080)] / 10
x = [45 ± √945] / 10
x = [45 ± 3√105] / 10
x = 3.9 or x = 3.0
Now that we have found the possible values for x, we can plug them back into the equation y = 5x - 9 to find the corresponding values for y:
For x = 3.9:
y = 5(3.9) - 9
y = 19.5 - 9
y = 10.5
For x = 3.0:
y = 5(3) - 9
y = 15 - 9
y = 6
Therefore, the solutions to the system of equations are x = 3.9, y = 10.5 and x = 3.0, y = 6.