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3 years ago

15

How do you substitute 5x+4y=0 , x-y=9

1 answer:

-BARSIC- [3]3 years ago

5 0

Steps:
1.Change x-y=9 into slope-intercept form.
Subtract x on both sides
-y=-x + 9
Divide everything by -1
Y= x-9
2. Now substitute the y in 5x + 4y=0 for x-9
5x + 4(x-9)= 0
5x + 4x-36=0
3. Now combine like terms
9x-36=0
4. Add 36 on both sides
9x= 36
5. Divide each side by 9
6. X= 4
7. Now substitute the x in x-y=9 for 4
4-y=9
8. Subtract 4 on both sides
-y= 5
9. Divide everything by -1
10. Y= -5
11. Your answer is x= 4 and Y= -5 or (4,-5)

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<u>ANSWER</u>

$(9 {c}^{ - 9} )^{ - 3}=\frac{{c}^{ 27} }{729}$

<u>EXPLANATION</u>

The given expression is

$(9 {c}^{ - 9} )^{ - 3}$

Recall that:

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We use the law of exponents to get

$(9 {c}^{ - 9} )^{ - 3} = 9^{ - 3} {c}^{ - 9 \times - 3}$

Let us multiply in the exponents to get:

$(9 {c}^{ - 9} )^{ - 3} = 9^{ - 3} {c}^{ 27}$

Recall again that:

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This implies that:

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We expand the power to get:

$(9 {c}^{ - 9} )^{ - 3} = \frac{1}{9 \times 9 \times 9} \times {c}^{ 27}$

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We multiply out to get:

$(9 {c}^{ - 9} )^{ - 3}=\frac{{c}^{ 27} }{729}$

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