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REY [17]
3 years ago
13

Solve the system of equations by multiplying and adding. Enter the solution as an ordered pair.

Mathematics
1 answer:
GREYUIT [131]3 years ago
3 0

Answer:

(x, y) = (- 3, 4)

Step-by-step explanation:

Given the 2 equations

5x + 2y = - 7 → (1)

4x - y = - 16 → (2)

multiply all terms in (2) by 2

8x - 2y = - 32 → (3)

Add (1) and (3) term by term

(5x + 8x) + (2y - 2y) = (- 7 - 32)

13x = - 39 ( divide both sides by 13 )

x = - 3

Substitute x = - 3 into either (1) or (2) and solve for y

substituting in (1) gives

(5 × - 3) + 2y = - 7

- 15 + 2y = - 7 ( add 15 to both sides )

2y = 8 ( divide both sides by 2 )

y = 4

solution is (- 3, 4 )


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9a^4 b^10(2 +3a^6 b^5)  (2 -3a^6 b^5)

see work below

Step-by-step explanation:

36a^4b^10 - 81a^16b^20

A)  find the GCF

36a^4b^10 = 4*9 a^4b^10  = 2*2*3*3 a*a*a*a*b*b*b*b*b*b*b*b*b*b

81a^16b^20= 9*9a^16b^20= 3*3*3*3* a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a                *b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b

The terms that appear in both terms is the GCF.  The terms that remain are inside the parentheses.

The GCF is 3*3 a*a*a*a*b*b*b*b*b*b*b*b*b*b

36a^4b^10 - 81a^16b^20 = 3*3 a*a*a*a*b*b*b*b*b*b*b*b*b*b *

( 2*2 - 3*3a*a*a*a*a*a*a*a*a*a*a*a b*b*b*b*b*b*b*b*b*b)

Combining like terms

36a^4b^10 - 81a^16b^20 = 9a^4b^10(4-9a^12b^10)

The expression inside the parenthesis can be factored using the difference of squares

let x^2 =4   x =2  

y^2 = 9a^12 b^10   y = 3a^6b^5

(x^2 -y^2) = (x+y)(x-y)

9a^4b^10(4-9a^12b^10) = 9a^4b^10 ( 2+3a^6b^5) ( 2-3a^6b^5)

b) difference of squares  a^2 – b^2 = (a + b)(a – b)

let a^2 = 36a^4b^10

so a = 6a^2b^5

b^2 = 81a^16b^20

b = 9a^8 b^10

a^2 – b^2 = (a + b)(a – b)

36a^4b^10 - 81a^16b^20 = (6a^2b^5 +9a^8 b^10) (6a^2b^5 -9a^8 b^10)

We can factor a 3 a^2 b^5 out of the first term

3 a^2 b^5 (2 +3a^6 b^5) (6a^2b^5 -9a^8 b^10)

3 a^2 b^5 (2 +3a^6 b^5) 3 a^2 b^5 (2 -3a^6 b^5)

Multiply the terms outside the parentheses together

9a^4 b^10(2 +3a^6 b^5)  (2 -3a^6 b^5)

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