What is 23p+27q=10,27p+23q=100 ?

The solution to 23p+27q=10,27p+23q=100 is p= 247/20 ,q=-203/20

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23p+27q=10,27p+23q=100

23 p + 27 q = 10, 27 p + 23 q = 100

p = \frac{247}{20}, q = - \frac{203}{20}

Solution steps

[23 p + 27 q = 10 27 p + 23 q = 100]

Isolate

p

for

23 p + 27 q = 10 : p = \frac{10 - 27 q}{23}

Substitute

p = \frac{10 - 27 q}{23}

[27 \cdot \frac{10 - 27 q}{23} + 23 q = 100]

Simplify

[\frac{270 - 200 q}{23} = 100]

Isolate

q

for

\frac{270 - 200 q}{23} = 100 : q = - \frac{203}{20}

For

p = \frac{10 - 27 q}{23}

Substitute

q = - \frac{203}{20}

p = \frac{10 - 27 ( - \frac{203}{20} )}{23}

\frac{10 - 27 ( - \frac{203}{20} )}{23} = \frac{247}{20}

p = \frac{247}{20}

The solutions to the system of equations are:

p = \frac{247}{20}, q = - \frac{203}{20}

2 x + y = - 1, 0 x - 3 y = 2

3 x - 4 y = 20, x - 12 y = - 20

2 x - y = 90, x - 3 y = - 40

2 x - 3 y = 30, 3 x - 4 y - 5 = 0

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

What is 23p+27q=10,27p+23q=100 ?
The solution to 23p+27q=10,27p+23q=100 is p= 247/20 ,q=-203/20