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MA_775_DIABLO [31]
2 years ago
8

Can you guys help me solve this equation with step-by-step help, please...

Mathematics
1 answer:
Alex2 years ago
4 0

Step-by-step explanation:

-2x - 5y = 16___(1)

2x - 3y = -16___(2)

(1) + (2) ==> -2x -5y = 16

(+) <u>2x -3y = -16</u>

-8y = 0

y = 0/-8

y = 0

y=0 in (1)

(1)---> -2x-5y =16

-2x - 5(0) = 16

-2x - 0 = 16

-2x = 16

x = 16/-2

x = -8

x = -8 , y = 0

-5x + 2y = 11__(1) ; -3x + 4y = -13___(2)

multiply eqn(1) with 2

2 × (1) : -10x + 4y = 22___(3)

(3) - (2) :. -10x + 4y = 22

(-) <u>-3x </u><u>+</u><u> </u><u>4</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>1</u><u>3</u>

-7x = 35

x = 35/-7

x = -5

x=-5 in (1)

(1) : -5x + 2y = 11

-5(-5) + 2y = 11

25 + 2y = 11

2y = 11 - 25

2y = -14

y = -14/2

y = -7

x = -5 , y = -7

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MakcuM [25]

Answer:

a) w^{13} x^{5} y^{6}

b) \frac{x}{3y^{6} }

Step-by-step explanation:

a) (w^{2} xy^{3} )^{2}(w^{3}x )^{3}

1. Distribute the second power (2) outside the first pair of parenthesis:

(w^{2(2)} x^{2} y^{3(2)} )

= w^{4} x^{2} y^{6} (w^{3}x )^{3}

2. Distribute the third power (3) outside the second pair of parenthesis:

(w^{3(3)} x^{3} )

= w^{4} x^{2} y^{6} w^{9} x^{3}

3. Combine like terms:

w^{13} x^{5} y^{6}

--------------------------------------------

b) \frac{2x^{2} y^{5} }{6xy^{11} }

1. Factor the number 6 (= 2 · 3):

\frac{2x^{2} y^{5} }{2(3)xy^{11} }

2. Cancel the common factor (2):

\frac{x^{2} y^{5} }{3xy^{11} }

3. Cancel out xy^{5} in the numerator an denominator:

\frac{x}{3y^{6} }

hope this helps!

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