What is 9p-q=1,1-3q-5=p ?

The solution to 9p-q=1,1-3q-5=p is p=-1/28 ,q=-37/28

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9p-q=1,1-3q-5=p

9 p - q = 1, 1 - 3 q - 5 = p

p = - \frac{1}{28}, q = - \frac{37}{28}

Solution steps

[9 p - q = 1 1 - 3 q - 5 = p]

Isolate

p

for

9 p - q = 1 : p = \frac{1 + q}{9}

Substitute

p = \frac{1 + q}{9}

[1 - 3 q - 5 = \frac{1 + q}{9}]

Simplify

[- 3 q - 4 = \frac{1 + q}{9}]

Isolate

q

for

- 3 q - 4 = \frac{1 + q}{9} : q = - \frac{37}{28}

For

p = \frac{1 + q}{9}

Substitute

q = - \frac{37}{28}

p = \frac{1 - \frac{37}{28}}{9}

\frac{1 - \frac{37}{28}}{9} = - \frac{1}{28}

p = - \frac{1}{28}

The solutions to the system of equations are:

p = - \frac{1}{28}, q = - \frac{37}{28}

x - 2 y = 1, 2 x - 4 y = 2

[n] y f (y) (y)^{2} = 50 [n], f (y) (y)^{1} = 20

\frac{8}{x} + \frac{5}{y} = - 3, \frac{6}{x} + \frac{25}{y} = 2

y = (\frac{2}{3}) x + 4, - 2 x + 3 y = 6

2 x + y - 2 z = 1, x - 2 y + 3 z = 6, 3 x - 3 y - z = 5