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Alona [7]
3 years ago
6

Use Cramer's Rule to solve each system. 5x – 3y = 8 -2x - y = 10

Mathematics
2 answers:
mezya [45]3 years ago
8 0

Answer: x = - 2

y = - 6

Step-by-step explanation:

A 2 × 2 matrix would be formed. The steps are shown in the attached photo.

zlopas [31]3 years ago
6 0

Answer:

-2, -6

Step-by-step explanation:

using cramer rule

5    -3    

-2   -1    

calculatiing the determinant = (5 x-1) - (-3x-2) = -5 -(6) = -5-6=-11

using cramer rule

to calculate x we change the coefficient of x with the answer (8,10)

8   -3

10   -1

we calculate determinant = (8x-1)-(-3x10) = -8-(-30) = -8+30 =22

to calculate x

22/-11= -2

to calculate y we change the coefficient of y with the answer (8,10)

5   8

-2   10

we calculate determinant = (5x10)-(8x-2) = 50 -(-16) = 50 +16=66

to calculate y

66/-11= -6

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(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2
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Answer:

\dfrac{m^3}{16p^2q^2}

Step-by-step explanation:

Given:

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}

1.

m^{-1}=\dfrac{1}{m}

2.

q^0=1

3.

2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}

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(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}

5.

m^{-1}=\dfrac{1}{m}

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2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}

7.

\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}

8.

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}

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