What is 5p-4q+3r=9,2p+q-2r=1,4p+3q+4r=1 ?

The solution to 5p-4q+3r=9,2p+q-2r=1,4p+3q+4r=1 is p=1,q=-1,r=0

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5p-4q+3r=9,2p+q-2r=1,4p+3q+4r=1

5 p - 4 q + 3 r = 9, 2 p + q - 2 r = 1, 4 p + 3 q + 4 r = 1

p = 1, q = - 1, r = 0

Solution steps

5 p - 4 q + 3 r = 9 2 p + q - 2 r = 1 4 p + 3 q + 4 r = 1

Isolate

p

for

5 p - 4 q + 3 r = 9 : p = \frac{9 + 4 q - 3 r}{5}

Substitute

p = \frac{9 + 4 q - 3 r}{5}

[2 \cdot \frac{9 + 4 q - 3 r}{5} + q - 2 r = 1 4 \cdot \frac{9 + 4 q - 3 r}{5} + 3 q + 4 r = 1]

Simplify

[\frac{13 q - 16 r + 18}{5} = 1 \frac{31 q + 8 r + 36}{5} = 1]

Isolate

q

for

\frac{13 q - 16 r + 18}{5} = 1 : q = \frac{16 r - 13}{13}

Substitute

q = \frac{16 r - 13}{13}

[\frac{31 \cdot \frac{16 r - 13}{13} + 8 r + 36}{5} = 1]

Simplify

[\frac{120 r + 13}{13} = 1]

Isolate

r

for

\frac{120 r + 13}{13} = 1 : r = 0

For

q = \frac{16 r - 13}{13}

Substitute

r = 0

q = \frac{16 \cdot 0 - 13}{13}

\frac{16 \cdot 0 - 13}{13} = - 1

q = - 1

For

p = \frac{9 + 4 q - 3 r}{5}

Substitute

q = - 1, r = 0

p = \frac{9 + 4 ( - 1 ) - 3 \cdot 0}{5}

\frac{9 + 4 ( - 1 ) - 3 \cdot 0}{5} = 1

p = 1

The solutions to the system of equations are:

p = 1, q = - 1, r = 0

x + 2 y + 3 z = 2, - x + 4 y + z = 1, 4 x + y + z = 5

5 x - 2 y = 4, x + 8 y = 26

7 x + 6 y = 38, 3 x + 5 y = 1750

x + y = 4, 8 x - 3 y = 4

2 x + 3 y = 0, 3 x + 5 y = 2

What is 5p-4q+3r=9,2p+q-2r=1,4p+3q+4r=1 ?
The solution to 5p-4q+3r=9,2p+q-2r=1,4p+3q+4r=1 is p=1,q=-1,r=0